Complex Basis
The standard basis for $\mathbb{R}^2$ is $(1,0),(0,1)$. What is the standard basis for $\mathbb{C}^2$? They’re the same basis. Let $x=(x_1,x_2)\in\mathbb{C}^2$ so that $x_1,x_2\in\mathbb{C}$. Then
$$\begin{align*} x=(x_1,x_2)=(x_1,0)+(0,x_2)=x_1(1,0)+x_2(0,1) \end{align*}$$
Hence $(1,0),(0,1)$ spans $\mathbb{C}^2$. And of course $(1,0),(0,1)$ is linearly independent. Similarly, we can show that the standard basis for $\mathbb{R}^n$ is also the standard basis for $\mathbb{C}^n$.
Notice that we’re assuming that $\mathbb{C}^n$ is a vector space over $\mathbb{C}$. Axler defines it this way in 1.10, p.6. And I think this is generally the case. But I’ve seen rare instances where $\mathbb{C}^n$ is a vector space over $\mathbb{R}$. In that case, the standard basis and dimension of $\mathbb{C}^n$ changes.
p.52, Example 3.4, Linear Maps & Lists for $\mathbb{F}^n$ to $\mathbb{F}^m$
Let $(A_{j,k})_{j=1,k=1}^{m,n}$ denote a list of scalars in $\mathbb{F}$. That is, for $j=1,…,m$ and $k=1,…,n$, we have scalars $A_{j,k}\in\mathbb{F}$. Hence $(A_{j,k})_{j=1,k=1}^{m,n}$ is a list of $m\times n$ scalars. Define $T:\mathbb{F}^n\mapsto\mathbb{F}^m$ by
$$\begin{align*} T(x_1,...,x_n) &\equiv (A_{1,1}x_1+\dots+A_{1,n}x_n,...,A_{m,1}x_1+\dots+A_{m,n}x_n) \\ &= \Big(\sum_{k=1}^{n}A_{1,k}x_k,...,\sum_{k=1}^{n}A_{m,k}x_k\Big) \end{align*}$$
Verify additivity:
$$\begin{align*} T(x+y) &= T\big((x_1,...,x_n)+(y_1,...,y_n)\big) \\ &= T\big((x_1+y_1,...,x_n+y_n)\big) \\ &= \Big(\sum_{k=1}^{n}A_{1,k}(x_k+y_k),...,\sum_{k=1}^{n}A_{m,k}(x_k+y_k)\Big) \\ &= \Big(\sum_{k=1}^{n}A_{1,k}x_k+\sum_{k=1}^{n}A_{1,k}y_k,...,\sum_{k=1}^{n}A_{m,k}x_k+\sum_{k=1}^{n}A_{m,k}y_k\Big) \\ &= \Big(\sum_{k=1}^{n}A_{1,k}x_k,...,\sum_{k=1}^{n}A_{m,k}x_k\Big)+ \Big(\sum_{k=1}^{n}A_{1,k}y_k,...,\sum_{k=1}^{n}A_{m,k}y_k\Big) \\ &= Tx+Ty \end{align*}$$
We can similarly verify homogeneity: $T(\lambda x)=\lambda Tx$. Hence $T\in\lnmpsb(\mathbb{F}^n,\mathbb{F}^m)$.
Let $e_1^{(n)},…,e_n^{(n)}$ be the standard basis for $\mathbb{F}^n$. For any map $T:\mathbb{F}^n\mapsto\mathbb{F}^m$, the element $Te_k^{(n)}\in\mathbb{F}^m$ is a unique (Criterion for a basis, 2.29, p.39-40) linear combination of $e_1^{(m)},…,e_m^{(m)}$. That is, there exists a unique list $(A_{j,k})_{j=1,k=1}^{m,n}$ such that
$$ \begin{matrix} Te_1^{(n)}=A_{1,1}e_1^{(m)}+\cdots+A_{m,1}e_m^{(m)}=\sum_{j=1}^mA_{j,1}e_j^{(m)}\\ Te_2^{(n)}=A_{1,2}e_1^{(m)}+\cdots+A_{m,2}e_m^{(m)}=\sum_{j=1}^mA_{j,2}e_j^{(m)}\\ \vdots\\ Te_n^{(n)}=A_{1,n}e_1^{(m)}+\cdots+A_{m,n}e_m^{(m)}=\sum_{j=1}^mA_{j,n}e_j^{(m)}\\ \end{matrix} $$
Proposition W.3.1 A map $T:\mathbb{F}^n\mapsto\mathbb{F}^m$ is linear if and only if there exists a unique list of scalars $(A_{j,k})_{j=1,k=1}^{m,n}$ such that
$$\begin{align*} T(x_1,...,x_n) &= \Big(\sum_{k=1}^{n}A_{1,k}x_k,...,\sum_{k=1}^{n}A_{m,k}x_k\Big) \end{align*}$$
for all $x\in\mathbb{F}^n$. The list is unique to $T$ for the standard bases in $\mathbb{F}^n$ and $\mathbb{F}^m$ but doesn’t depend on $x$.
Proof If $T$ can be written in this form, we showed above that it’s linear. Uniqueness follows because if we change any combination of the $A_{j,k}$’s, then we get a different linear map. For example, suppose $T\in\lnmpsb(\mathbb{F}^2,\mathbb{F})$ and for all $x_1,x_2\in\mathbb{F}$ we have
$$ B_{1}x_1+B_{2}x_2=Tx=A_{1}x_1+A_{2}x_2 $$
This is equivalent to
$$ (B_1-A_1)x_1=B_{1}x_1-A_{1}x_1=A_{2}x_2-B_{2}x_2=(A_2-B_2)x_2 $$
So it must be that $B_{1}=A_{1}$ and $B_{2}=A_{2}$. If not, then this only holds when $x_1$ is a scalar multiple of $x_2$.
In the other direction, suppose $T$ is linear. Then
$$\begin{align*} Tx &= T\Big(\sum_{k=1}^nx_ke_k^{(n)}\Big)\tag{Criterion for a basis} \\ &= \sum_{k=1}^nx_kTe_k^{(n)} \\ &= \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}e_j^{(m)}\Big) \\ &= \sum_{k=1}^nx_k\big(A_{1,k}(1,0,...,0)+A_{2,k}(0,1,...,0)+\dots+A_{m,k}(0,0,...,1)\big) \\ &= \sum_{k=1}^nx_k\big((A_{1,k},0,...,0)+(0,A_{2,k},...,0)+\dots+(0,0,...,A_{m,k})\big) \\\\ &= x_1\big((A_{1,1},0,...,0)+(0,A_{2,1},...,0)+\dots+(0,0,...,A_{m,1})\big) \\ &+ x_2\big((A_{1,2},0,...,0)+(0,A_{2,2},...,0)+\dots+(0,0,...,A_{m,2})\big) \\ &\vdots\\ &+ x_n\big((A_{1,n},0,...,0)+(0,A_{2,n},...,0)+\dots+(0,0,...,A_{m,n})\big) \\\\ &= (A_{1,1}x_1,0,...,0)+(0,A_{2,1}x_1,...,0)+\dots+(0,0,...,A_{m,1}x_1) \\ &+ (A_{1,2}x_2,0,...,0)+(0,A_{2,2}x_2,...,0)+\dots+(0,0,...,A_{m,2}x_2) \\ &\vdots\\ &+ (A_{1,n}x_n,0,...,0)+(0,A_{2,n}x_n,...,0)+\dots+(0,0,...,A_{m,n}x_n) \\\\ &= \Big(\sum_{k=1}^nA_{1,k}x_k,0,...,0\Big)+\Big(0,\sum_{k=1}^nA_{2,k}x_k,...,0\Big)+\dots+\Big(0,0,...,\sum_{k=1}^nA_{m,k}x_k\Big) \\ &= \Big(\sum_{k=1}^nA_{1,k}x_k,\sum_{k=1}^nA_{2,k}x_k,...,\sum_{k=1}^{n}A_{m,k}x_k\Big) \end{align*}$$
Uniqueness for $T$ and these bases follows because we used the unique linear combination $Te_k^{(n)}=\sum_{j=1}^mA_{j,k}e_j^{(m)}$ in the third equality.
$\blacksquare$
Proposition W.3.2 For a given list $(A_{j,k})_{j=1,k=1}^{m,n}$, there exists a unique linear map $T\in\lnmpsb(\mathbb{F}^n,\mathbb{F}^m)$ such that
$$ Te_k^{(n)}=\sum_{j=1}^mA_{j,k}e_j^{(m)}\quad\text{for }k=1,...,n $$
Proof Define
$$\begin{align*} T(x_1,...,x_n) \equiv \Big(\sum_{i=1}^{n}A_{1,i}x_i,...,\sum_{i=1}^{n}A_{m,i}x_i\Big) \end{align*}$$
We know this is linear. Let $x=e_k^{(n)}$ so that $x_k=1$ and $x_i=0$ for $i\neq k$. Then
$$\begin{align*} Te_k^{(n)}=Tx &= \Big(\sum_{i=1}^nA_{1,i}x_i,\sum_{i=1}^nA_{2,i}x_i,...,\sum_{i=1}^{n}A_{m,i}x_i\Big)\\ &= \big(A_{1,k}x_k,A_{2,k}x_k,...,A_{m,k}x_k\big)\tag{$x_i=0$ for $i\neq k$}\\ &= \big(A_{1,k},A_{2,k},...,A_{m,k}\big)\tag{$x_k=1$}\\ &= \big(A_{1,k},0,...,0\big)+\big(0,A_{2,k},...,0\big)+\dots+\big(0,0,...,A_{m,k}\big)\\ &= A_{1,k}\big(1,0,...,0\big)+A_{2,k}\big(0,1,...,0\big)+\dots+A_{m,k}\big(0,0,...,1\big)\\ &= A_{1,k}e_1^{(m)}+A_{2,k}e_2^{(m)}+\dots+A_{m,k}e_m^{(m)}\\ &= \sum_{j=1}^mA_{j,k}e_j^{(m)} \end{align*}$$
To prove uniqueness, let $S\in\lnmpsb(\mathbb{F}^n,\mathbb{F}^m)$ satisfy $Se_k^{(n)}=\sum_{j=1}^mA_{j,k}e_j^{(m)}=Te_k^{(n)}$. Then
$$\begin{align*} Tx-Sx &= T\Big(\sum_{k=1}^nx_ke_k^{(n)}\Big)-S\Big(\sum_{k=1}^nx_ke_k^{(n)}\Big) \\ &= \sum_{k=1}^nx_kTe_k^{(n)}-\sum_{k=1}^nx_kSe_k^{(n)} \\ &= \sum_{k=1}^nx_k\big(Te_k^{(n)}-Se_k^{(n)}\big) \\ &=0 \end{align*}$$
$\blacksquare$
To motivate the next section, note that
$$\begin{align*} T(x_1,...,x_n) &\equiv \Big(\sum_{k=1}^{n}A_{1,k}x_k,...,\sum_{k=1}^{n}A_{m,k}x_k\Big) \\ &= \sum_{k=1}^{n}\big(A_{1,k}x_k,...,A_{m,k}x_k\big) \\ &= \sum_{k=1}^{n}x_k\big(A_{1,k},...,A_{m,k}\big) \\ &= \sum_{k=1}^{n}x_k\Big[\big(A_{1,k},0,...,0\big)+\big(0,A_{2,k},...,0\big)+\dots+\big(0,0,...,A_{m,k}\big)\Big] \\ &= \sum_{k=1}^{n}x_k\Big[A_{1,k}\big(1,0,...,0\big)+A_{2,k}\big(0,1,...,0\big)+\dots+A_{m,k}\big(0,0,...,1\big)\Big] \\ &= \sum_{k=1}^{n}x_k\big(A_{1,k}e_1^{(m)}+A_{2,k}e_2^{(m)}+\dots+A_{m,k}e_m^{(m)}\big) \\ &= \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}e_j^{(m)}\Big) \end{align*}$$
We can alternatively arrive at the same formula this way:
$$\begin{align*} Tx &= T\Big(\sum_{k=1}^nx_ke_k^{(n)}\Big) \\ &= \sum_{k=1}^nx_kTe_k^{(n)} \\ &= \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}e_j^{(m)}\Big) \\ \end{align*}$$
But the two approaches are very different. In the first case, the $A_{j,k}$’s come from the definition of $T$ for any $x\in\mathbb{F}^n$. In the second approach, the $A_{j,k}$’s come from the image of the standard basis of $\mathbb{F}^n$ under $T$. We showed equivalence in the case $\mathbb{F}^n$. But we can generalize this.
Linear Maps & Lists for $V$ and $W$
Let $V$ and $W$ be vector spaces over $\mathbb{F}$. Let $v_1,…v_n$ be a basis for $V$ and let $w_1,…,w_m$ be a basis $W$. For any map $T:V\mapsto W$, the element $Tv_k\in W$ is a unique (Criterion for a basis, 2.29, p.39-40) linear combination of $w_1,…,w_m$. That is, there exists a unique list $(A_{j,k})_{j=1,k=1}^{m,n}$ such that
$$ \begin{matrix} Tv_1=A_{1,1}w_1+\cdots+A_{m,1}w_m=\sum_{j=1}^mA_{j,1}w_j\\ Tv_2=A_{1,2}w_1+\cdots+A_{m,2}w_m=\sum_{j=1}^mA_{j,2}w_j\\ \vdots\\ Tv_n=A_{1,n}w_1+\cdots+A_{m,n}w_m=\sum_{j=1}^mA_{j,n}w_j\\ \end{matrix} $$
Fix an arbitrary $x\in V$. Then $x=x_1v_1+\dots+x_nv_n$ for some scalars $x_k\in\mathbb{F}$. Also fix an arbitrary list $(A_{j,k})_{j=1,k=1}^{m,n}$. Define $T:V\mapsto W$ as
$$\begin{align*} Tx = T(x_1v_1+\dots+x_nv_n) \equiv \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) =\sum_{k=1}^nx_kTv_k \end{align*}$$
Verify additivity:
$$\begin{align*} T(x+y) &= T\Big(\sum_{k=1}^nx_kv_k+\sum_{k=1}^ny_kv_k\Big) \\ &= T\Big(\sum_{k=1}^n(x_k+y_k)v_k\Big) \\ &= \sum_{k=1}^n(x_k+y_k)\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \\ &= \sum_{k=1}^n\Bigg[x_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)+y_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)\Bigg] \\ &= \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)+\sum_{k=1}^ny_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \\ &= Tx+Ty \end{align*}$$
Similarly we can verify homogeneity: $T(\lambda x)=\lambda Tx$. Hence $T\in\lnmpsb(V,W)$.
Proposition W.3.3 A map $T:V\mapsto W$ is linear if and only if, for every pair of bases $v_1,…v_n$ and $w_1,…,w_m$, there exists a unique list of scalars $(A_{j,k})_{j=1,k=1}^{m,n}$ such that
$$\begin{align*} Tx = \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \end{align*}$$
for all $x=\sum_{k=1}^nx_kv_k\in V$. The list is unique to $T$ for given bases (but doesn’t depend on $x$). This means two things:
- For given bases, a different list $(B_{j,k})_{j=1,k=1}^{m,n}\neq(A_{j,k})_{j=1,k=1}^{m,n}$ produces a different linear map.
- The list $(A_{j,k})_{j=1,k=1}^{m,n}$ depends on the bases. If we change bases, we will generally (but not always) get a different list $(B_{j,k})_{j=1,k=1}^{m,n}\neq(A_{j,k})_{j=1,k=1}^{m,n}$ for $T$.
Proof Fix some bases $v_1,…,v_n$ of $V$ and $w_1,…,w_m$ of $W$. First we will prove if and then we’ll prove only if.
If $T$ can be written in this form for some list $(A_{j,k})_{j=1,k=1}^{m,n}$, we showed above that it’s linear. To show that the list $(A_{j,k})_{j=1,k=1}^{m,n}$ is unique to $T$ for the given bases, suppose $T$ can also be written as
$$\begin{align*} Tx = \sum_{k=1}^nx_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big) \end{align*}$$
Hence
$$\begin{align*} \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)=Tx=\sum_{k=1}^nx_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big) \end{align*}$$
and
$$\begin{align*} 0&=\sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)-\sum_{k=1}^nx_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big)\\ &=\sum_{k=1}^n\Bigg[x_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)-x_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big)\Bigg]\\ &=\sum_{k=1}^nx_k\Big[\sum_{j=1}^mA_{j,k}w_j-\sum_{j=1}^mB_{j,k}w_j\Big]\\ &=\sum_{k=1}^nx_k\sum_{j=1}^m\big(A_{j,k}w_j-B_{j,k}w_j\big)\\ &=\sum_{k=1}^nx_k\sum_{j=1}^m\big(A_{j,k}-B_{j,k}\big)w_j\\ &=\sum_{k=1}^n\sum_{j=1}^m\big(A_{j,k}-B_{j,k}\big)x_kw_j\\ &=\sum_{j=1}^m\sum_{k=1}^n\big(A_{j,k}-B_{j,k}\big)x_kw_j\\ &=\sum_{j=1}^mw_j\Big(\sum_{k=1}^n\big(A_{j,k}-B_{j,k}\big)x_k\Big)\\ \end{align*}$$
The linear independence of $w_1,…,w_m$ implies that $0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k$ for $j=1,…m$. Since this is true for any $x=\sum_{k=1}^nx_kv_k\in V$, then we are free to choose the $x_k$’s however we like:
$$\begin{align*} \begin{matrix} x_k=\begin{cases}0&k\neq1\\1&k=1\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,1}-B_{j,1}\implies A_{j,1}=B_{j,1} \\ x_k=\begin{cases}0&k\neq2\\1&k=2\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,2}-B_{j,2}\implies A_{j,2}=B_{j,2} \\ \vdots\\ x_k=\begin{cases}0&k\neq n\\1&k=n\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,n}-B_{j,n}\implies A_{j,n}=B_{j,n} \end{matrix} \end{align*}$$
Hence $(B_{j,k})_{j=1,k=1}^{m,n}=(A_{j,k})_{j=1,k=1}^{m,n}$.
In the other direction, suppose $T\in\lnmpsb(V,W)$. We know that there exists a unique list $(A_{j,k})_{j=1,k=1}^{m,n}$ such that $Tv_k=\sum_{j=1}^mA_{j,k}w_j$ for $k=1,…,n$. Hence
$$\begin{align*} Tx &= T\Big(\sum_{k=1}^nx_kv_k\Big) \\ &= \sum_{k=1}^nx_kTv_k \\ &= \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \end{align*}$$
To verify that the list $(A_{j,k})_{j=1,k=1}^{m,n}$ is unique to $T$ for the given bases, we can run through the same proof as in the converse.
$\blacksquare$
Proposition W.3.4 For a given list $(A_{j,k})_{j=1,k=1}^{m,n}$ and given bases $v_1,…,v_n$ of $V$ and $w_1,…,w_m$ of $W$, there exists a unique linear map $T\in\lnmpsb(V,W)$ such that
$$ Tv_k=\sum_{j=1}^mA_{j,k}w_j\quad\text{for }k=1,...,n $$
Proof For any $x=x_1v_1+\dots+x_nv_n\in V$, define $T:V\mapsto W$ as
$$\begin{align*} Tx = T(x_1v_1+\dots+x_nv_n) \equiv \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \end{align*}$$
We know $T$ is linear. To prove existence, it suffices to show that $Tv_k=\sum_{j=1}^mA_{j,k}w_j$ for $k=1,…,n$. Recall that $Tv_k\in W$ is a unique linear combination of the $w_1,…,w_m$. That is, $Tv_k$ can be written as
$$ Tv_k=\sum_{j=1}^mB_{j,k}w_j\quad\text{for }k=1,...,n $$
for some unique list $(B_{j,k})_{j=1,k=1}^{m,n}$. Hence, to prove existence, we want to show that $(B_{j,k})_{j=1,k=1}^{m,n}=(A_{j,k})_{j=1,k=1}^{m,n}$. Note that
$$\begin{align*} \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) = Tx =T\Big(\sum_{k=1}^nx_kv_k\Big) =\sum_{k=1}^nx_kTv_k =\sum_{k=1}^nx_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big) \end{align*}$$
Equivalently, we have
$$\begin{align*} 0&=\sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)-\sum_{k=1}^nx_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big)\\ &=\sum_{k=1}^n\Bigg[x_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big)-x_k\Big(\sum_{j=1}^mB_{j,k}w_j\Big)\Bigg]\\ &=\sum_{k=1}^nx_k\Big[\sum_{j=1}^mA_{j,k}w_j-\sum_{j=1}^mB_{j,k}w_j\Big]\\ &=\sum_{k=1}^nx_k\sum_{j=1}^m\big(A_{j,k}w_j-B_{j,k}w_j\big)\\ &=\sum_{k=1}^nx_k\sum_{j=1}^m\big(A_{j,k}-B_{j,k}\big)w_j\\ &=\sum_{k=1}^n\sum_{j=1}^m\big(A_{j,k}-B_{j,k}\big)x_kw_j\\ &=\sum_{j=1}^m\sum_{k=1}^n\big(A_{j,k}-B_{j,k}\big)x_kw_j\\ &=\sum_{j=1}^mw_j\Big(\sum_{k=1}^n\big(A_{j,k}-B_{j,k}\big)x_k\Big)\\ \end{align*}$$
The linear independence of $w_1,…,w_m$ implies that $0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k$ for $j=1,…,m$. Since this is true for any $x=\sum_{k=1}^nx_kv_k\in V$, then we are free to choose the $x_k$’s however we like:
$$\begin{align*} \begin{matrix} x_k=\begin{cases}0&k\neq1\\1&k=1\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,1}-B_{j,1}\implies A_{j,1}=B_{j,1} \\ x_k=\begin{cases}0&k\neq2\\1&k=2\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,2}-B_{j,2}\implies A_{j,2}=B_{j,2} \\ \vdots\\ x_k=\begin{cases}0&k\neq n\\1&k=n\end{cases}\implies0=\sum_{k=1}^n(A_{j,k}-B_{j,k})x_k=A_{j,n}-B_{j,n}\implies A_{j,n}=B_{j,n} \end{matrix} \end{align*}$$
Hence $(B_{j,k})_{j=1,k=1}^{m,n}=(A_{j,k})_{j=1,k=1}^{m,n}$ and we have shown existence.
To prove that $T$ is the unique linear map such that $Tv_k=\sum_{j=1}^mA_{j,k}w_j$, let $S\in\lnmpsb(V,W)$ satisfy $Sv_k=\sum_{j=1}^mA_{j,k}w_j$ as well. Then $Tv_k=Sv_k$ for $k=1,…,n$ and
$$\begin{align*} Tx-Sx &= T\Big(\sum_{k=1}^nx_kv_k\Big)-S\Big(\sum_{k=1}^nx_kv_k\Big) \\ &= \sum_{k=1}^nx_kTv_k-\sum_{k=1}^nx_kSv_k \\ &= \sum_{k=1}^nx_k\big(Tv_k-Sv_k\big) \\ &=0 \end{align*}$$
$\blacksquare$
Linear Maps & Matrices
Fix a basis $v_1,…,v_n$ for $V$ and a basis $w_1,…,w_m$ for $W$. In propositions W.3.3 and W.3.4, we established a 1-to-1 correspondence between $\lnmpsb(V,W)$ and the set of all lists $(A_{j,k})_{j=1,k=1}^{m,n}$. For each $T\in\lnmpsb(V,W)$, there is a unique list $(A_{j,k})_{j=1,k=1}^{m,n}$ such that
$$\begin{align*} Tx = \sum_{k=1}^nx_k\Big(\sum_{j=1}^mA_{j,k}w_j\Big) \end{align*}$$
And for each list $(A_{j,k})_{j=1,k=1}^{m,n}$, there is a unique $T\in\lnmpsb(V,W)$ such that
$$ Tv_k=\sum_{j=1}^mA_{j,k}w_j\quad\text{for }k=1,...,n $$
We can modify this for matrices. Let $[A_{j,k}]_{j=1,k=1}^{m,n}$ denote the matrix of $m$ rows and $n$ columns such that the element in the $j^{th}$ row and $k^{th}$ column is $A_{j,k}\in\mathbb{F}$. We now have a 1-to-1 correspondence between $\lnmpsb(V,W)$ and the set of all matrices $[A_{j,k}]_{j=1,k=1}^{m,n}$. For $m$ and $n$ positive integers, let $\mathbb{F}^{m,n}$ denote set of all $m$-by-$n$ matrices with elements in $\mathbb{F}$. That is, for positive integers $m$ and $n$, we define
$$ \mathbb{F}^{m,n}\equiv\big\{[A_{j,k}]_{j=1,k=1}^{m,n}:A_{j,k}\in\mathbb{F}\big\} $$
Axler goes on to prove in 3.40, p.74 that $\mathbb{F}^{m,n}$ is a vector space with dimension $mn$. And he proves in 3.60, p.83 that $\lnmpsb(V,W)$ and $\mathbb{F}^{m,n}$ are isomorphic.
On p.70-71, you can read the definition for the matrix of linear map. We denote the matrix $\mtrxof{T}$ for $T\in\lnmpsb(V,W)$. The notes after the definition are helpful. Particularly note that
$$\begin{align*} \begin{matrix} \begin{matrix} \begin{matrix}\text{ }\end{matrix}\\ \mtrxof{T}\equiv \begin{matrix}w_1\\w_2\\\vdots\\w_m\end{matrix} \end{matrix} \begin{matrix} \begin{matrix}v_1&\dots&v_k&\dots&v_n\end{matrix}\\ \begin{bmatrix}&&&&A_{1,k}&&&&\\&&&&A_{2,k}&&&&\\&&&&\vdots&&&&\\&&&&A_{m,k}&&&&\end{bmatrix} \end{matrix} \end{matrix}\tag{W.3.5} \end{align*}$$
We have written across the top of the matrix the basis $v_1,…,v_n$ for the domain $V$ of $T$ and we have written down the left side the basis $w_1,…,w_m$ for $W$, the vector space into which $T$ maps. We have only shown the $k^{th}$ column in the matrix. This illustrates a key point: The $k^{th}$ column of $\mtrxof{T}$ consists of the scalars needed to write $Tv_k$ as a linear combination of the basis $w_1,…,w_m$:
$$ Tv_k=\sum_{j=1}^mA_{j,k}w_j\tag{W.3.6} $$
On p.84, 3.62, we can read the definition of a matrix of a vector. It’s just an $n\times1$ matrix of the scalars of the vector’s unique linear combination of the basis. That is, if $x=x_1v_1+\dots+x_nv_n\in V$, then
$$ \mtrxof{x}\equiv\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} $$
For emphasis, recall that the matrix of a linear map depends on the bases of the domain and space into which it maps. Similarly, we note that the matrix of a vector depends on the basis. Obviously if $u_1,…,u_n$ is a different basis for $V$, then the same $x$ will probably have different scalars for this basis: $x=c_1u_1+\dots+c_nu_n$. Hence
$$ \mtrxof{x}\equiv\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix} $$
Glancing back at W.3.5 and W.3.6, we see that $\mtrxof{T}_{:,k}=\mtrxof{Tv_k}$. Misprints aside, Axler notes this in 3.64. p.85.
On p.74-75, Axler gives excellent motivation for why matrix multiplication is done the way that it is. After that he discusses alternatives to the standard matrix multiplication. In particular, on p.77, we have an example 3.51 and an unproved statement 3.52 that matrix-vector multiplication is just a linear combination of the columns of the matrix.
In 3.65, p.85, Axler uses all of this to prove that matrix multiplication acts like linear maps and vice versa. The subsequent paragraph is helpful.
$$ \mtrxof{Tx}=\mtrxof{T}\mtrxof{x} $$
That is, for
$$ \mtrxof{T}\equiv\begin{bmatrix}A_{1,1}&\dots&A_{1,n}\\\vdots&\ddots&\vdots\\A_{m,1}&\dots&A_{m,n}\end{bmatrix} \quad\quad\text{and}\quad\quad \mtrxof{x}\equiv\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} $$
we have
$$\begin{align*} \mtrxof{Tx}=\mtrxof{T}\mtrxof{x} &=\begin{bmatrix}A_{1,1}&\dots&A_{1,n}\\\vdots&\ddots&\vdots\\A_{m,1}&\dots&A_{m,n}\end{bmatrix}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} \\\\ &=\begin{bmatrix}A_{1,1}x_1+\dots+A_{1,n}x_n\\\vdots\\A_{m,1}x_1+\dots+A_{m,n}x_n\end{bmatrix} \\\\ &= \begin{bmatrix}\sum_{k=1}^{n}A_{1,k}x_k\\\vdots\\\sum_{k=1}^{n}A_{m,k}x_k\end{bmatrix} \\\\ \end{align*}$$
And we can verify the additivity of $\mtrxofsb$:
$$\begin{align*} \mtrxof{T(x+y)}=\mtrxof{T}\mtrxof{x} &= \begin{bmatrix}A_{1,1}&\dots&A_{1,n}\\\vdots&\ddots&\vdots\\A_{m,1}&\dots&A_{m,n}\end{bmatrix}\Bigg(\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}+\begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}\Bigg) \\\\ &= \begin{bmatrix}A_{1,1}&\dots&A_{1,n}\\\vdots&\ddots&\vdots\\A_{m,1}&\dots&A_{m,n}\end{bmatrix}\begin{bmatrix}x_1+y_1\\\vdots\\x_n+y_n\end{bmatrix} \\\\ &= \begin{bmatrix}A_{1,1}(x_1+y_1)+\dots+A_{1,n}(x_n+y_n)\\\vdots\\A_{m,1}(x_1+y_1)+\dots+A_{m,n}(x_n+y_n)\end{bmatrix} \\\\ &= \begin{bmatrix}A_{1,1}x_1+A_{1,1}y_1+\dots+A_{1,n}x_n+A_{1,n}y_n\\\vdots\\A_{m,1}x_1+A_{m,1}y_1+\dots+A_{m,n}x_n+A_{m,n}y_n\end{bmatrix} \\\\ &= \begin{bmatrix}A_{1,1}x_1+\dots+A_{1,n}x_n\\\vdots\\A_{m,1}x_1+\dots+A_{m,n}x_n\end{bmatrix} + \begin{bmatrix}A_{1,1}y_1+\dots+A_{1,n}y_n\\\vdots\\A_{m,1}y_1+\dots+A_{m,n}y_n\end{bmatrix} \\\\ &= \mtrxof{Tx}+\mtrxof{Ty} \end{align*}$$
p.66, 3.26 Homogeneous system of linear equations
Fix positive integers $m$ and $n$, and let $A_{j,k}\in\mathbb{F}$ for $j=1,…,m$ and $k=1,…,n$. Given the homogeneous system of linear equations
$$ \begin{matrix}\sum_{k=1}^nA_{1,k}x_k=0\\\vdots\\\sum_{k=1}^nA_{m,k}x_k=0\end{matrix}\tag{W.3.7} $$
Obviously $0=x_1=\dots=x_n$ is a solution of this system. The question is whether any other solutions exist. Define $T:\mathbb{F}^n\mapsto\mathbb{F}^m$ by
$$ T\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\equiv\begin{bmatrix}\sum_{k=1}^nA_{1,k}x_k\\\vdots\\\sum_{k=1}^nA_{m,k}x_k\end{bmatrix} $$
Then W.3.7 can be expressed as
$$ Tx=T\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}\sum_{k=1}^nA_{1,k}x_k\\\vdots\\\sum_{k=1}^nA_{m,k}x_k\end{bmatrix}=\begin{bmatrix}0\\\vdots\\0\end{bmatrix}=0 $$
Hence the above question “are there other solutions besides $0=x_1=\dots=x_n$” is equivalent to the question “is $\mathscr{N}(T)$ strictly bigger than $\{0\}$”. Recall 3.16 (p.61): Injectivity is equivalent to $\mathscr{N}(T)=\{0\}$. And recall 3.23 (p.64): A map to a smaller dimensional space is not injective. Hence a homogeneous system of linear equations with more variables ($n$) than equations ($m$) has nonzero solutions.
p.71, Example 3.33 and Proposition 3.65, p.85
Suppose $T\in\lnmpsb(\mathbb{F}^2,\mathbb{F}^3)$ is defined by
$$ T(x,y)=(x+3y,2x+5y,7x+9y) $$
Find the matrix of $T$ with respect to the standard bases $\mathbb{F}^2$ and $\mathbb{F}^3$.
Solution The entries $A_{j,k}$ of $\mtrxof{T}$ for the standard bases are defined by
$$ Tv_k=A_{1,k}w_1+A_{2,k}w_2+A_{3,k}w_3\quad\text{for }k=1,2 $$
where $v_1,v_2$ is the standard basis of $\mathbb{F}^2$ and $w_1,w_2,w_3$ is the standard basis of $\mathbb{F}^3$:
$$ \mathbb{F}^2\text{ basis}=\Bigg(v_1=\begin{bmatrix}1\\0\end{bmatrix},v_2=\begin{bmatrix}0\\1\end{bmatrix}\Bigg) \quad\quad\quad \mathbb{F}^3\text{ basis}=\Bigg(w_1=\begin{bmatrix}1\\0\\0\end{bmatrix},w_2=\begin{bmatrix}0\\1\\0\end{bmatrix},w_3=\begin{bmatrix}0\\0\\1\end{bmatrix}\Bigg) $$
Hence
$$\begin{align*} Tv_1&=(1\cdot1+3\cdot0,2\cdot1+5\cdot0,7\cdot1+9\cdot0) \\ &=(1,2,7) \\ &=(1,0,0)+(0,2,0)+(0,0,7) \\ &=1(1,0,0)+2(0,1,0)+7(0,0,1) \\ &=1w_1+2w_2+7w_3 \end{align*}$$
Hence $A_{1,1}=1,A_{2,1}=2,A_{3,1}=7$. Similarly
$$\begin{align*} Tv_2&=(1\cdot0+3\cdot1,2\cdot0+5\cdot1,7\cdot0+9\cdot1) \\ &=(3,5,9) \\ &=(3,0,0)+(0,5,0)+(0,0,9) \\ &=3(1,0,0)+5(0,1,0)+9(0,0,1) \\ &=3w_1+5w_2+9w_3 \end{align*}$$
Hence $A_{1,2}=3,A_{2,2}=5,A_{3,2}=9$. Hence
$$ \mtrxof{T}\equiv\begin{bmatrix}A_{1,1}&A_{1,2}\\A_{2,1}&A_{2,2}\\A_{3,1}&A_{3,2}\end{bmatrix}=\begin{bmatrix}1&3\\2&5\\7&9\end{bmatrix} $$
Define $u\equiv(11,-7)\in\mathbb{R}^2$. Then
$$\begin{align*} u&= 11v_1-7v_2 \\ &= 11(1,0)-7(0,1) \end{align*}$$
Above we computed $Tv_1=1w_1+2w_2+7w_3$ and $Tv_2=3w_1+5w_2+9w_3$. Since we have computed the values of this map for all of the basis vectors in the domain, then proposition $3.5, p.54$ tells us that we have a unique linear map defined on the entire domain. In particular, we can compute $Tu$:
$$\begin{align*} Tu &= T\big(11v_1-7v_2\big) \\ &= 11Tv_1-7Tv_2 \\ &= 11(w_1+2w_2+7w_3)-7(3w_1+5w_2+9w_3) \\ &= 11w_1+22w_2+77w_3-21w_1-35w_2-63w_3) \\ &= (11-21)w_1+(22-35)w_2+(77-63)w_3 \\ &= -10w_1-13w_2+14w_3 \\ \end{align*}$$
And the corresponding matrix computation is
$$ \mtrxof{T}\mtrxof{u} =\begin{bmatrix}1&3\\2&5\\7&9\end{bmatrix}\begin{bmatrix}11\\-7\end{bmatrix} =\begin{bmatrix}11-21\\22-35\\77-63\end{bmatrix} =\begin{bmatrix}-10\\-13\\14\end{bmatrix} =\mtrxof{Tu} $$
In proposition 3.65, p.85, Axler proves that linear maps act like matrix multiplication. This example demonstrates exactly what he means. Executing the map $T$ on the vector $u$ gives us the vector $Tu$ and this is precisely the action of $\mtrxof{T}\mtrxof{u}=\mtrxof{Tu}$. Notice how $\mtrxof{Tu}$ is simply the matrix of scalars of $Tu\in\mathbb{R}^3$ in terms of its basis $w_1,w_2,w_3$. Similarly $\mtrxof{u}$ is the matrix of scalars of $u\in\mathbb{R}^2$ in terms of its basis $v_1,v_2$.
Also notice that the scalars from the computed $Tv_k$’s are the columns of $\mtrxof{T}$:
$$\begin{matrix} Tv_1=1w_1+2w_2+7w_3&1^{st}\text{ column}\\ Tv_2=3w_1+5w_2+9w_3&2^{nd}\text{ column}\\ \end{matrix}$$
This example illustrates this statement in the text: “If you think of elements of $\mathbb{F}^m$ as columns of $m$ numbers, then you can think of the $k^{th}$ column of $\mtrxof{T}$ as $T$ applied to the $k^{th}$ standard basis vector”.
p.72, Example 3.34 and Proposition 3.65, p.85
Suppose $D\in\lnmpsb(\mathscr{P}_{3}(\mathbb{R}),\mathscr{P}_{2}(\mathbb{R}))$ is the differentiation map defined by $Dp=p’$. Find the matrix of $D$ with respect to the standard bases of $\mathscr{P}_{3}(\mathbb{R})$ and $\mathscr{P}_{2}(\mathbb{R})$.
Solution The entries $A_{j,k}$ of $\mtrxof{D}$ for the standard bases are defined by
$$ Dv_k=A_{1,k}w_1+A_{2,k}w_2+A_{3,k}w_3\quad\text{for }k=1,2,3,4 $$
where $v_1,v_2,v_3,v_4$ is the standard basis of $\mathscr{P}_{3}(\mathbb{R})$ and $w_1,w_2,w_3$ is the standard basis of $\mathscr{P}_{2}(\mathbb{R})$:
$$ \mathscr{P}_{3}(\mathbb{R})\text{ basis}=\Big(v_1=1,v_2=x,v_3=x^2,v_4=x^3\Big) \quad\quad\quad \mathscr{P}_{2}(\mathbb{R})\text{ basis}=\Big(w_1=1,w_2=x,w_3=x^2\Big) $$
Hence
$$ Dv_1=0=0w_1+0w_2+0w_3 $$
Hence $A_{1,1}=A_{2,1}=A_{3,1}=0$. Similarly
$$ Dv_2=1=1w_1+0w_2+0w_3 $$
Hence $A_{1,2}=1$ and $A_{2,2}=A_{3,2}=0$. Similarly
$$ Dv_3=2x=0w_1+2w_2+0w_3 $$
Hence $A_{2,3}=2$ and $A_{1,3}=A_{3,3}=0$. Similarly
$$ Dv_4=3x^2=0w_1+0w_2+3w_3 $$
Hence $A_{3,4}=3$ and $A_{1,4}=A_{2,4}=0$. Hence
$$ \mtrxof{D}\equiv\begin{bmatrix}A_{1,1}&A_{1,2}&A_{1,3}&A_{1,4}\\A_{2,1}&A_{2,2}&A_{2,3}&A_{2,4}\\A_{3,1}&A_{3,2}&A_{3,3}&A_{3,4}\end{bmatrix}=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\\end{bmatrix} $$
Define $p\in\mathscr{P}_{3}(\mathbb{R})$ as $p(x)\equiv 41+7x-19x^2+8x^3$. Then $Dp(x)=7-38x+24x^2$. Above we slightly abused notation by letting $x^i$ denote a function rather than a value. That is, our standard basis $1,x,x^2,x^3$ for $\mathscr{P}_{3}(\mathbb{R})$ is a list of functions, not a list of values. When we do the same thing here, then we get
$$\begin{align*} p &= 41v_1+7v_2-19v_3+8v_4 \\ &= 41\cdot1+7x-19x^2+8x^3 \end{align*}$$
Above we computed $Dv_1=0$, $Dv_2=w_1$, $Dv_3=2w_2$, and $Dv_4=3w_3$. Since we have computed the values of this map for all of the basis vectors in the domain, then proposition $3.5, p.54$ tells us that we have a unique linear map defined on the entire domain. In particular, we can compute $Dp$:
$$\begin{align*} Dp &= D\big(41v_1+7v_2-19v_3+8v_4\big) \\ &= 41Dv_1+7Dv_2-19Dv_3+8Dv_4 \\ &= 41\cdot0+7w_1-19\cdot2w_2+8\cdot3w_3 \\ &= 7w_1-38w_2+24w_3 \\ &= 7\cdot1-38x+24x^2 \end{align*}$$
And the corresponding matrix computation is
$$ \mtrxof{D}\mtrxof{p} =\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\\end{bmatrix}\begin{bmatrix}41\\7\\-19\\8\end{bmatrix} =\begin{bmatrix}7\\-38\\24\end{bmatrix} =\mtrxof{Dp} $$
In proposition 3.65, p.85, Axler proves that linear maps act like matrix multiplication. This example demonstrates exactly what he means. Executing the map $D$ on the polynomial $p$ gives us the vector $Dp$ and this is precisely the action of $\mtrxof{D}\mtrxof{p}=\mtrxof{Dp}$. Notice how $\mtrxof{Dp}$ is simply the matrix of scalars of $Dp\in\mathscr{P}_{2}(\mathbb{R})$ in terms of its basis $1,x,x^2$. Similarly $\mtrxof{p}$ is the matrix of scalars of $p\in\mathscr{P}_{3}(\mathbb{R})$ in terms of its basis $1,x,x^2,x^3$.
Also notice that the scalars from the computed $Dv_k$’s are the columns of $\mtrxof{D}$:
$$\begin{matrix} Dv_1=0w_1+0w_2+0w_3&1^{st}\text{ column}\\ Dv_2=1w_1+0w_2+0w_3&2^{nd}\text{ column}\\ Dv_3=0w_1+2w_2+0w_3&3^{rd}\text{ column}\\ Dv_4=0w_1+0w_2+3w_3&4^{th}\text{ column}\\ \end{matrix}$$
p.80-81, Proposition 3.56
In the proof, it reads “For each $w\in W$, define $Sw$ to be the unique element of $V$ such that $T(Sw)=w$ (the existence and uniqueness of such an element follow from the surjectivity and injectivity of $T$). Clearly $T\circ S$ equals the identity map on $W$. To prove that $S\circ T$ equals the identity map on $V$, let $v\in V$. Then
$$ T\big((S\circ T)v\big)=(T\circ S)(Tv)=I(Tv)=Tv $$
This equation implies that $(S\circ T)v=v$ because $T$ is injective. Thus $S\circ T$ equals the identity map on $V$.”
This is confusing to me. What is $S$? Where did it come from? IF such an $S$ exists, then I’ll buy that $T\circ S$ is the identity map on $W$ since $T(Sw)=w$ by supposition. But that doesn’t help explain what $S$ is. Let’s look at this differently.
Let $T:V\mapsto W$ be surjective and injective. For each $w\in W$, the surjectivity of $T$ implies that there exists an element $v\in V$ such that $Tv=w$. The injectivity of $T$ implies that $v$ is the unique element in $V$ such that $Tv=w$. Hence, for each $w\in W$, there is exactly one element $v\in V$ such that $Tv=w$. In the other direction, for each $v\in V$, there exists a unique element $w=Tv\in W$ (since $T$ is a function).
Hence we have shown that $(w,v)=(Tv,v)$ is always a unique pair for any $w\in W$ or for any $v\in V$. Hence (see exercise 3.E.1), we can define a function, graph, or map
$$\begin{align*} S &\equiv\{(Tv,v)\in W\times V:v\in V\} \\ &=\{(w,v):w\in W, v\in V\text{ such that }Tv=w\} \end{align*}$$
We can define the composite $S\circ T=S(T)$ as the identity map on $V$:
$$ (S\circ T)v\equiv S(Tv)\equiv v $$
Note that the identity map must satisfy $Iv_1\neq Iv_2$ if $v_1\neq v_2$. The above definition is justified because $Tv$ is defined and unique for each $v\in V$. That is, if $v_1\neq v_2$, then $Tv_1\neq Tv_2$ and $(S\circ T)v_1=v_1\neq v_2=(S\circ T)v_2$.
Note that $S$ is surjective and injective: let $v\in V$. Since $T$ is defined on $V$, then $Tv=w$ for some $w=Tv\in W$. Hence there exists some $w\in W$ such that $Sw=S(Tv)=(S\circ T)v=v$. Hence $\text{range}(S)=V$ and $S$ is surjective. Since $T$ is a function, $w\in W$ is unique. Hence $S$ is injective. Hence $(w,v)=(Tv,v)$ is well defined for any $v\in V$.
Hence $Sw\equiv v$ is the unique element in $V$ such $Tv=w$. And $T\circ S$ is the identity map on $W$:
$$ (T\circ S)w\equiv T(Sw)\equiv w $$
p.83, 3.60, $\lnmpsb(V,W)$ and $\mathbb{F}^{m,n}$ are isomorphic
The proof reads “… suppose $A\in\mathbb{F}^{m,n}$. Let $T$ be the linear from $V$ to $W$ such that
$$ Tv_k=\sum_{j=1}^mA_{j,k}w_j $$
for $k=1,…,n$ (see 3.5, p.54).”
I was initially very confused by this because I couldn’t see how proposition 3.5 gave this result. Hence I proved proposition W.3.4 in these notes. Proposition W.3.4 proves this directly without the use of proposition 3.5 from the text.
But now I can see how proposition 3.5 gives the result: define
$$ u_k\equiv\sum_{j=1}^mA_{j,k}w_j\quad\text{for }k=1,...,n $$
Then, by 3.60’s supposition, $v_1,…,v_n$ is a basis of $V$. And, by the above definition, we have $u_1,…,u_n\in W$. Hence we have satisified the hypotheses of proposition 3.5. Hence there exists a unique linear map $T:V\mapsto W$ such that
$$ Tv_k=u_k=\sum_{j=1}^mA_{j,k}w_j $$
for $k=1,…,n$. This use of proposition 3.5 illustrates its power. It also illustrates this statement from p.53:
The existence part of the next result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of the next result means that a linear map is completely determined by its values on a basis.
Injectivity and Linear Independence
Proposition W.3.8 Let $v_1,…,v_n$ be linearly independent in $V$ and let $T\in\lnmpsb(V,W)$ be injective. Then $Tv_1,…,Tv_n$ is linearly independent in $W$.
Proof Consider scalars $a_1,…,a_n$ such that
$$ 0=\sum_{k=1}^na_kTv_k=T\Big(\sum_{k=1}^na_kv_k\Big) $$
Hence $\sum_{k=1}^na_kv_k\in\mathscr{N}(T)$. Since $T$ is injective, proposition 3.16, p.61 gives that $\mathscr{N}(T)=\{0\}$. Hence
$$ 0=\sum_{k=1}^na_kv_k $$
And the linear independence of $v_1,…,v_n$ implies that $0=a_1=\dots=a_n$. $\blacksquare$
Proposition W.3.9 Let $v_1,…,v_n$ be a basis for $V$. Then $T\in\lnmpsb(V,W)$ is injective if and only if $Tv_1,…,Tv_n$ is linearly independent in $W$.
Proof Suppose $T$ is injective. Proposition W.3.8 gives that $Tv_1,…,Tv_n$ is linearly independent in $W$. Conversely, suppose $Tv_1,…,Tv_n$ is linearly independent in $W$. We will show that $\mathscr{N}(T)=\{0\}$. Let $v\in\mathscr{N}(T)\subset V$ so that for some scalars $a_1,…,a_n$, we have $v=\sum_{k=1}^na_kv_k$ and
$$ 0=Tv=T\Big(\sum_{k=1}^na_kv_k\Big)=\sum_{k=1}^na_kTv_k $$
The linear independence of $Tv_1,…,Tv_n$ gives $0=a_1=\dots=a_n$ and $v=0$. $\blacksquare$
Note the difference in hypotheses between W.3.8 and W.3.9. Prop W.3.8 assumes that $v_1,…,v_n$ is linearly independent but makes no assumption that it spans $V$. Prop W.3.9 assumes $v_1,…,v_n$ spans $V$ in addition to linear independence. We need the stronger assumption in W.3.9 because we write $v=\sum_ka_kv_k$ as a linear combination of the basis. And we can produce a counterexample if we don’t assume that $v_1,…,v_n$ spans $V$:
Let $V=\mathbb{R}^3$ and let $v_1,v_2,v_3$ be the standard basis there. Let $W=\mathbb{R}^4$ and let $w_1,w_2,w_3,w_4$ be the standard basis there. Then proposition 3.5, p.54 gives a unique linear map $T:V\mapsto W$ such that
$$ Tv_1=w_1\quad\quad Tv_2=w_2\quad\quad Tv_3=0 $$
It’s clear that $v_1,v_2$ is linearly independent but that $\text{span}(v_1,v_2)\neq V$. And it’s clear that $Tv_1,Tv_2$ is linearly independent in $W$. But $T$ is not injective since $\{0,v_3\}\subset\mathscr{N}(T)$.
Surjectivity and Span
Proposition W.3.10 Let $v_1,…,v_n$ span $V$ and let $T\in\lnmpsb(V,W)$. Then $Tv_1,…,Tv_n$ spans $\mathscr{R}(T)$.
Proof Let $w\in\mathscr{R}(T)$. Then there exists a vector $v\in V$ such that $Tv=w$. Because $v_1,…,v_n$ spans $V$, there exist scalars $a_1,…,a_n$ such that $v=\sum_{k=1}^na_kv_k$. Then
$$ w=Tv=T\Big(\sum_{k=1}^na_kv_k\Big)=\sum_{k=1}^na_kTv_k\in\text{span}(Tv_1,...,Tv_n) $$
Conversely, let $w\in\text{span}(Tv_1,…,Tv_n)$. Then there exist scalars $a_1,…,a_n$ such that
$$ w=\sum_{k=1}^na_kTv_k=T\Big(\sum_{k=1}^na_kv_k\Big)\in\mathscr{R}(T)\quad\blacksquare $$
Corollary W.3.10.b Let $v_1,\dots,v_n$ span $V$ and let $T\in\lnmpsb(V,W)$. Let $L$ denote the length of any of the largest linearly independent sublists of $Tv_1,\dots,Tv_n$. Then $\dim{\mathscr{R}(T)}=L$.
Proof Proposition W.3.10 gives the first equality and proposition W.2.20 gives the second equality:
$$ \dim{\mathscr{R}(T)}=\dim{\text{span}(Tv_1,\dots,Tv_n)}=L\quad\blacksquare $$
Proposition W.3.11 Let $v_1,…,v_n$ span $V$ and let $T\in\lnmpsb(V,W)$. Then $T$ is surjective if and only if $Tv_1,…,Tv_n$ spans $W$.
In words: If $T$ is linear, then $T$ is surjective if and only if it maps any spanning list of the domain to a spanning list of the codomain.
Proof Suppose $T$ is surjective. Proposition W.3.10 gives that $Tv_1,…,Tv_n$ spans $\mathscr{R}(T)=W$. Conversely, suppose $Tv_1,…,Tv_n$ spans $W$. Let $w\in W=\text{span}(Tv_1,…,Tv_n)$ so that there exist scalars $a_1,…,a_n$ such that
$$ w=\sum_{k=1}^na_kTv_k=T\Big(\sum_{k=1}^na_kv_k\Big)\in\mathscr{R}(T)\quad\blacksquare $$
General
Proposition W.3.12 Let $v_1,…,v_n$ be a basis for $V$ and let $T\in\lnmpsb(V,W)$. Then $T=0$ if and only if $Tv_k=0$ for $k=1,…,n$.
Proof If $T=0$, then $Tv_k=0$ for $k=1,…,n$. If $Tv_k=0$ for $k=1,…,n$ and $v=\sum_k^na_kv_k\in V$, then $Tv=\sum_k^na_kTv_k=0$. $\blacksquare$
Proposition W.3.13 Let $v_1,…,v_n$ be a list of vectors in some vector space. And let $L$ denote the number of linearly independent vectors in $v_1,…,v_n$. Then $L=\dim{\text{span}(v_1,…,v_n)}$.
Proof The process in the proof of proposition 2.31 starts with $n$ vectors, discards all of the linearly dependent ones, and keeps all of the linearly independent ones. What remains is a list of length $L$ that forms a basis for $\text{span}(v_1,…,v_n)$. $\blacksquare$
Proposition W.3.14 Let $v_1,…,v_n$ and $w_1,…,w_m$ be bases for $V$ and $W$. Let $T\in\lnmpsb(V,W)$. Then column $k$ of $\mtrxof{T}$ is zero if and only if $Tv_k=0$.
Proof Recall the definition (3.32, p.70) of the $m\times n$ matrix $\mtrxof{T}$ with entries $A_{j,k}$. For $k=1,…,n$, the equalities $Tv_k=\sum_{j=1}^mA_{j,k}w_j$ completely determine the entries $A_{j,k}$. The criterion for basis (proposition 2.29, p.40) implies that the $A_{j,k}$ are unique since $Tv_k\in W$ and $w_1,…,w_m$ is a basis of $W$.
Suppose column $k$ in $\mtrxof{T}$ is zero. Then $A_{:,k}$ is zero hence $A_{j,k}=0$ for $j=1,…,m$. Hence
$$ Tv_k=\sum_{j=1}^mA_{j,k}w_j=\sum_{j=1}^m0w_j=0 $$
Suppose $0=Tv_k=\sum_{j=1}^mA_{j,k}w_j$. The linear independence of $w_1,…,w_m$ implies that $A_{j,k}=0$ for $j=1,…,m$. Hence $A_{:,k}=0$. $\blacksquare$.
Proposition W.3.15 Let $V$ be finite-dimensional and let $T\in\lnmpsb(V,W)$. Then any basis of $V$ contains at most $\dim{\mathscr{N}(T)}$ vectors that are in $\mathscr{N}(T)$.
Proof Let $v_1,…,v_n$ be a basis for $V$. Let $N$ be the number of $v_k$ such that $Tv_k\neq0$ so that $n-N$ is the number of basis vectors in $\mathscr{N}(T)$. The Fundamental Theorem of Linear Maps gives
$$ \dim{\mathscr{N}(T)}=V-\dim{\mathscr{R}(T)} $$
Let $L$ denote the length of any of the largest linearly independent sublists of $Tv_1,…,Tv_n$. Proposition W.3.10.b gives that $\dim{\mathscr{R}(T)}=L$. Since $L\leq N$ and
$$ \dim{\mathscr{N}(T)}=V-\dim{\mathscr{R}(T)}=n-L\geq n-N\quad\blacksquare $$
Example W.3.16 In the previous proposition, we showed that any basis of $V$ contains at most $\dim{\mathscr{N}(T)}$ vectors that are in $\mathscr{N}(T)$. This example demonstrates that the number of basis vectors of $V$ that are in $\mathscr{N}(T)$ can be strictly less than $\dim{\mathscr{N}(T)}$.
Let $v_1=(1,0)$ and $v_2=(0,1)$. Proposition 3.5, p.54 gives the existence of $T\in\lnmpsb(\mathbb{R}^2,\mathbb{R})$ such that
$$ Tv_1=1\quad\quad\quad Tv_2=-1 $$
We see that $\mathscr{R}(T)=\mathbb{R}$: Let $r\in\mathbb{R}$. Then
$$ T(r,0)=T(rv_1)=rTv_1=r\cdot1=r $$
Hence $r\in\mathscr{R}(T)$. And there’s another vector in $\mathbb{R}^2$ that $T$ maps to $r\in\mathbb{R}$:
$$ T(0,-r)=T(-rv_2)=-rTv_2=-r\cdot-1=r $$
Hence $T$ is not injective. Let’s find $\mathscr{N}(T)$:
$$ 0=T(u_1,u_2)=T(u_1v_1+u_2v_2)=u_1Tv_1+u_2Tv_2=u_1-u_2 $$
Hence
$$ \mathscr{N}(T)=\big\{(u_1,u_2)\in\mathbb{R}^2:u_1=u_2\big\}=\big\{(x,x)\in\mathbb{R}^2:x\in\mathbb{R}\big\} $$
Let’s find a basis for $\mathscr{N}(T)$. Since $(x,x)=x(1,1)$, then $(1,1)$ spans $\mathscr{N}(T)$. And a single, nonzero vector is linearly independent. Hence $(1,1)$ is a basis for $\mathscr{N}(T)$ and $\dim{\mathscr{N}(T)}=1$. But there are zero basis vectors in $\mathscr{N}(T)$.
Above we saw that $\mathscr{R}(T)=\mathbb{R}$. Hence $\dim{\mathscr{R}(T)}=1$. Proposition W.3.10 gives that $\text{span}(Tv_1,Tv_2)=\mathscr{R}(T)$. Since $Tv_1,Tv_2=1,-1$, then $Tv_2=-1\cdot Tv_1$ is a linear combination of $Tv_1$ and this all agrees:
$$ 1=\dim{\mathbb{R}}=\dim{\mathscr{R}(T)}=\dim{\text{span}(Tv_1,Tv_2)}=\dim{\text{span}(1,-1)} $$
Proposition W.3.17 A linear map that maps a basis to a basis is an isomorphism. That is, let $v_1,…,v_n$ be a basis for $V$. Then $T\in\lnmpsb(V,W)$ is an isomorphism if and only if $Tv_1,…,Tv_n$ is a basis of $W$.
Proof Suppose $Tv_1,…,Tv_n$ is a basis of $W$. Then $Tv_1,…,Tv_n$ spans $W$ and proposition W.3.11 gives that $T$ is surjective. Also, $Tv_1,…,Tv_n$ is linearly independent and proposition W.3.9 gives that $T$ is injective. Since $T$ is surjective and injective, then proposition 3.56, p.80-81 gives that $T$ is invertible.
Suppose $T$ is an isomorphism. Then proposition 3.56 gives that $T$ is injective and surjective. Then proposition W.3.9 gives that $Tv_1,…,Tv_n$ is linearly independent and proposition W.3.11 gives that $W=\text{span}(Tv_1,…,Tv_n)$. Hence $Tv_1,…,Tv_n$ is a basis of $W$. $\blacksquare$
p.112, 3.117 Dimension of $\mathscr{R}(T)$ equals column rank of $\mtrxof{T}$
The proof reads “The function that takes $w\in\text{span}(Tv_1,\dots,Tv_n)$ to $\mtrxof{w}$ is easily seen to be an isomorphism from $\text{span}(Tv_1,\dots,Tv_n)$ onto $\text{span}(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n})$.”
The function he refers to is
$$ \mtrxofsb:\text{span}(Tv_1,\dots,Tv_n)\mapsto\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big) $$
Because he wants surjectivity and injectivity, he is careful to specify this domain and codomain. And we’ll see why in a moment. For now, let’s recognize the parent vector spaces.
The domain $\text{span}(Tv_1,\dots,Tv_n)$ is a subspace of $W$ since $Tv_k\in W$. Hence, since the dimension of $W$ is $m$, then $\mtrxof{Tv_k}\in\mathbb{F}^{m,1}$ if we’re using the given basis $w_1,…,w_m$. We could alternatively use a basis of $\text{span}(Tv_1,\dots,Tv_n)$. But the proposition intentionally makes no assumption on the dimension of $\text{span}(Tv_1,\dots,Tv_n)$.
Since $\mtrxof{Tv_k}\in\mathbb{F}^{m,1}$ in the basis $w_1,\dots,w_m$, then the codomain $\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)$ is a subspace of $\mathbb{F}^{m,1}$. We point out that if a basis of $\text{span}(Tv_1,\dots,Tv_n)$ can be determined, let’s say $\phi_1,\dots,\phi_p$, then $\mtrxof{Tv_k}\in\mathbb{F}^{p,1}$ if we use the basis $\phi_1,\dots,\phi_p$. In that case, the codomain $\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)$ is a subspace of $\mathbb{F}^{p,1}$. We present this point for clarity but we’ll stick with the basis $w_1,\dots,w_m$ throughout the discussion of this proposition.
Let’s prove injectivity. Suppose $\mtrxof{w}=0\in\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)\subset\mathbb{F}^{m,1}$ for some $w\in\text{span}(Tv_1,\dots,Tv_n)\subset W$. Recall the definition of the matrix of a vector (3.62, p.84): If $w=\sum_j^mc_jw_j$, then
$$ \mtrxof{w}\equiv\begin{bmatrix}c_1\\\vdots\\c_m\end{bmatrix} $$
So $0=\mtrxof{w}$ means
$$ \begin{bmatrix}0\\\vdots\\0\end{bmatrix}=\mtrxof{w}=\begin{bmatrix}c_1\\\vdots\\c_m\end{bmatrix} $$
Hence $0=c_1=\dotsb=c_m$ and $w=\sum_j^mc_jw_j=0$. And $\mtrxofsb$ is injective on the given domain and codomain.
Let’s prove surjectivity. Let $c\in\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)$. We want to show that $c\in\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big)$:
$$\begin{align*} c&=\sum_{k=1}^na_k\mtrxof{Tv_k} \\ &=\sum_{k=1}^na_k\mtrxof{T}_{:,k}\tag{by Prop 3.64, p.85} \\ &=\mtrxof{T}\mtrxof{a}\tag{by Prop 3.52, p.77} \\ &=\mtrxof{Ta}\tag{by Prop 3.65, p.85} \\ &\in\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big) \end{align*}$$
where we define $a\in V$ by $a\equiv\sum_{k=1}^na_kv_k$. Hence $Ta=\sum_{k=1}^na_kTv_k\in\text{span}(Tv_1,\dots,Tv_n)$. Hence $\mtrxof{Ta}\in\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big)$ and
$$ \text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)\subset\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big) $$
Notice that if $c\in\mathbb{F}^{m,1}\setminus\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)$, then we cannot find coefficients $a_1,\dots,a_n$ such that $c=\sum_k^na_k\mtrxof{Tv_k}$. And then there’s no guarantee that $c\in\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big)$. This is the reason that the author restricted the domain and codomain.
In the other direction, let $c\in\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big)$. Then for some $b\in\text{span}(Tv_1,\dots,Tv_n)$, we have
$$\begin{align*} c&=\mtrxof{b} \\ &=\mtrxofsb\Big(\sum_{k=1}^nb_kTv_k\Big) \\ &=\sum_{k=1}^nb_k\mtrxof{Tv_k}\tag{by 3.36 and 3.38, p.73} \\ &\in\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big) \end{align*}$$
Hence
$$ \mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big)\subset\text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big) $$
and
$$ \text{span}\big(\mtrxof{Tv_1},\dots,\mtrxof{Tv_n}\big)=\mathscr{R}\big(\mtrxofsb\bar_{\text{span}(Tv_1,\dots,Tv_n)}\big) $$
and $\mtrxofsb$ is surjective on this domain and codomain.
Row rank and column rank (Incomplete)
Proposition W.3.18 Let $T\in\lnmpsb(V)$ and let $\phi\equiv u_1,\dots,u_n$ and $\psi\equiv v_1,\dots,v_n$ both be bases of $V$. Then the following are equivalent:
(A) $T$ is invertible.
(B) The columns of $\mtrxof{T}$ are linearly independent in $\mathbb{F}^{n,1}$.
(C) The columns of $\mtrxof{T}$ span $\mathbb{F}^{n,1}$.
(D) The rows of $\mtrxof{T}$ are linearly independent in $\mathbb{F}^{1,n}$.
(E) The rows of $\mtrxof{T}$ span $\mathbb{F}^{1,n}$.
Proof Incomplete We take $\mtrxof{T}=\mtrxofsb(T,\phi,\psi)$ throughout this proof.
(A)$\implies$(B): Suppose $T$ is invertible. Then $T$ is an isomorphism. Proposition W.3.17 gives that $T$ maps a basis to a basis. Hence $Tv_1,\dots,Tv_n$ is a basis of $V$. Define $\beta\equiv Tv_1,\dots,Tv_n$. Suppose
$$ \begin{bmatrix}0\\\vdots\\0\end{bmatrix}=0=\sum_{k=1}^na_k\mtrxofsb(T,\phi,\beta)_{:,k}=\sum_{k=1}^na_k\mtrxofsb(Tv_k,\beta)=\mtrxofsb\Big(\sum_{k=1}^na_kTv_k,\beta\Big)=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix} $$
Hence $0=a_1=\dotsb=a_n$ and the columns $\mtrxof{T}_{:,k}=\mtrxof{Tv_k}$ of $\mtrxof{T}$ are linearly independent in $\mathbb{F}^{n,1}$. The last equality follows from the definition of the matrix of a vector (3.62, p.84) since $Tv_1,\dots,Tv_n$ is a basis of $V$.
(B)$\implies$(C): Suppose the columns $\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}$ of $\mtrxof{T}$ are linearly independent in $\mathbb{F}^{n,1}$. Since
$$ \dim{\mathbb{F}^{n,1}}=n\cdot1=n=\text{len}\big(\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}\big) $$
then proposition 2.39, p.45 gives that $\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}$ is a basis of and hence spans $\mathbb{F}^{n,1}$.
(C)$\implies$(D): Suppose the columns $\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}$ of $\mtrxof{T}$ span $\mathbb{F}^{n,1}$. Since
$$ \dim{\mathbb{F}^{n,1}}=n\cdot1=n=\text{len}\big(\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}\big) $$
then proposition 2.42, p.46 gives that $\mtrxof{T}_{:,1},\dotsb,\mtrxof{T}_{:,n}$ is a basis of and hence is linearly independent in $\mathbb{F}^{n,1}$.
General
Proposition W.3.19 Suppose $S,T\in\linmap{V}{W}$ and $\lambda\in\wF$. Then $S+T\in\linmap{V}{W}$ and $\lambda T\in\linmap{V}{W}$, as defined in 3.6.
Proof Let $u,v\in V$ and $\beta\in\wF$. To show that $S+T\in\linmap{V}{W}$, we must show that $(S+T)(u+v)=(S+T)u+(S+T)v$ and $(S+T)(\beta v)=\beta\prn{(S+T)v}$.
$$\align{ (S+T)(u+v) &= S(u+v)+T(u+v)\tag{by 3.6} \\ &= Su+Sv+Tu+Tv\tag{by 3.2.additivity} \\ &= Su+Tu+Sv+Tv\tag{by 1.19.commutativity} \\ &= (S+T)u+(S+T)v\tag{by 3.6} }$$
and
$$\align{ (S+T)(\beta v) &= S(\beta v)+T(\beta v)\tag{by 3.6} \\ &= \beta(Sv)+\beta(Tv)\tag{by 3.2.homogeneity} \\ &= \beta(Sv+Tv)\tag{by 1.19.distributivity} \\ &= \beta\prn{(S+T)v}\tag{by 3.6} }$$
Hence $S+T\in\linmap{V}{W}$.
To show that $\lambda T\in\linmap{V}{W}$, we must show that $(\lambda T)(u+v)=(\lambda T)u+(\lambda T)v$ and $(\lambda T)(\beta v)=\beta\prn{(\lambda T)v}$.
$$\align{ (\lambda T)(u+v) &= \lambda\prn{T(u+v)}\tag{by 3.6} \\ &= \lambda(Tu+Tv)\tag{by 3.2.additivity} \\ &= \lambda(Tu)+\lambda(Tv)\tag{by 1.19 distributivity} \\ &= (\lambda T)u+(\lambda T)v\tag{by 3.6} }$$
and
$$\align{ (\lambda T)(\beta v) &= \lambda\prn{T(\beta v)}\tag{by 3.6} \\ &= \lambda\prn{\beta(Tv)}\tag{by 3.2.homogeneity} \\ &= (\lambda\beta)Tv\tag{by 1.19 associativity} \\ &= (\beta\lambda)Tv\tag{by 1.3 commutativity} \\ &= \beta\prn{\lambda(Tv)}\tag{by 1.19 associativity} \\ &= \beta\prn{(\lambda T)v\tag{by 3.6}} }$$
Hence $\lambda T\in\linmap{V}{W}$. $\blacksquare$
Proposition W.3.20 Distribution of the Product of Linear Maps If $S\in\linmap{V}{U}$ and $Q,R\in\linmap{U}{W}$, then
$$ (Q+R)S=QS+RS \tag{W.3.20.1} $$
If $R,S\in\linmap{V}{U}$ and $Q\in\linmap{U}{W}$, then
$$ Q(R+S)=QR+QS \tag{W.3.20.2} $$
If $S,T\in\linmap{V}{U}$ and $Q,R\in\linmap{U}{W}$, then
$$ (Q+R)(S+T)=QS+QT+RS+RT \tag{W.3.20.3} $$
Proof Let $v\in V$. Then W.3.20.1 is proved by
$$\align{ \prn{(Q+R)S}v &= (Q+R)(Sv)\tag{by 3.8} \\ &= Q(Sv)+R(Sv)\tag{by 3.6} \\ &= (QS)v+(RS)v\tag{by 3.8} \\ &= (QS+RS)v\tag{by 3.6} }$$
And W.3.20.2 is proved by
$$\align{ \prn{Q(R+S)}v &= Q\prn{(R+S)v}\tag{by 3.8} \\ &= Q(Rv+Sv)\tag{by 3.6} \\ &= Q(Rv)+Q(Sv)\tag{by 3.2 additivity} \\ &= (QR)v+(QS)v\tag{by 3.8} \\ &= (QR+QS)v\tag{by 3.6} }$$
Define $X\equiv Q+R$. Then W.3.19 gives $X\in\linmap{U}{W}$. Hence
$$\align{ \prn{(Q+R)(S+T)}v &= \prn{X(S+T)}v \\ &= (XS+XT)v\tag{by W.3.20.2} \\ &= \prn{(Q+R)S+(Q+R)T}v \\ &= (QS+RS+QT+RT)v\quad\blacksquare\tag{by W.3.20.1} \\ }$$
Proposition W.3.21 Composition of Linear is Linear If $T\in\linmap{U}{V}$ and $S\in\linmap{V}{W}$, then $ST\in\linmap{U}{W}$.
Proof Let $a,b\in U$. Then
$$\align{ (ST)(a+b) &= S\prn{T(a+b)}\tag{by 3.8} \\ &= S(Ta+Tb)\tag{by 3.2.additivity} \\ &= S(Ta)+S(Tb)\tag{by 3.2.additivity} \\ &= (ST)a+(ST)b\tag{by 3.8} }$$
Let $\lambda\in\wF$. Then
$$\align{ (ST)(\lambda a) &= S\prn{T(\lambda a)}\tag{by 3.8} \\ &= S\prn{\lambda(Ta)}\tag{by 3.2.homogeneity} \\ &= \lambda S(Ta)\tag{by 3.2.homogeneity} \\ &= \lambda (ST)a\tag{by 3.8}\quad\blacksquare }$$
Proposition W.3.22 Let $T\in\linmap{V}{U}$, $S\in\linmap{U}{W}$, and $\lambda\in\wF$. Then
$$ S\lambda T=\lambda ST $$
Proof Let $v\in V$. Then
$$\align{ (S\lambda T)v &= \prn{S(\lambda T)}v \\ &= S\prn{(\lambda T)v}\tag{by 3.8} \\ &= S\prn{\lambda(Tv)}\tag{by 3.6.scalar multiplication} \\ &= \lambda S(Tv)\tag{by 3.2.homogeneity} \\ &= \lambda(ST)v\tag{by 3.8} }$$
$\blacksquare$
Proposition W.3.23 Let $T\in\linmap{V}{W}$ and let $a,b\in\wF$. Then
$$ aT+bT=(a+b)T $$
Proof For any $v\in V$, we have
$$\align{ (aT+bT)v &= (aT)v+(bT)v\tag{by 3.6.sum} \\ &= a(Tv)+b(Tv)\tag{by 3.6.scalar multiplication} \\ &= (a+b)(Tv)\tag{by 1.19.distributive properties} }$$
$\blacksquare$
Proposition W.3.24 Let $T\in\oper{V}$ and let $a,b\in\wF$. Then
$$ (T+aI)(T+bI)=T^2+(a+b)T+abI=(T+bI)(T+aI) $$
Proof We have
$$\align{ (T+aI)(T+bI) &= TT+T(bI)+(aI)T+(aI)(bI)\tag{by W.3.20.3} \\ &= T^2+bTI+aIT+b(aI)I\tag{by W.3.22} \\ &= T^2+bT+aT+b(aI)\tag{by 3.9.identity} \\ &= T^2+(a+b)T+abI\tag{by W.3.23} \\ }$$
$\blacksquare$
Proposition W.3.25 Let $v_1,…,v_n$ be a basis for $V$ and let $S,T\in\linmap{V}{W}$. Then $S=T$ if and only if $Sv_k=Tv_k$ for $k=1,…,n$.
Proof Note that
$$ S=T \iff S-T=0 \iff (S-T)v_k=0\text{ for }k=1,\dots,n \iff Sv_k=Tv_k\text{ for }k=1,\dots,n $$
where the second equivalence follows from W.3.12. $\blacksquare$
Proposition W.3.26 Any two diagonal matrices commute and their product is diagonal.
Proof Let $A$ and $C$ be $n\times n$ diagonal matrices. Then $0=A_{j,r}$ for $j\neq r$ and $0=C_{r,k}$ for $r\neq k$. Hence $0=A_{j,r}C_{r,k}$ unless $r=j=k$. Hence, for $j\neq k$, we have
$$ (AC)_{j,k}=\sum_{r=1}^nA_{j,r}C_{r,k}=\sum_{r=1}^n0=0=\sum_{r=1}^n0=\sum_{r=1}^nC_{j,r}A_{r,k}=(CA)_{j,k} $$
Also note that
$$ (AC)_{j,j}=\sum_{r=1}^nA_{j,r}C_{r,j}=A_{j,j}C_{j,j}=\sum_{r=1}^nC_{j,r}A_{r,j}=(CA)_{j,j} $$
$\wes$
Proposition W.3.27 <–> 3.43 <–> 10.4 Suppose that $u\equiv u_1,\dots,u_p$ is a basis of $U$, that $v\equiv v_1,\dots,v_n$ is a basis of $V$, and that $w\equiv w_1,\dots,w_m$ is a basis of $W$. Let $T\in\linmap{U}{V}$ and $S\in\linmap{V}{W}$. Then
$$\align{ \mtrxof{ST,u,w} &= \mtrxof{S,v,w}\mtrxof{T,u,v}\tag{W.3.27.1} \\\\ &= \sum_{j=1}^n\mtrxof{S,v,w}_{:,j}\mtrxof{T,u,v}_{j,:}\tag{W.3.27.2} }$$
Proof Since $w=w_1,\dots,w_m$ is a basis of $W$, then there exist scalars $s_{i,j}$ for $i=1,\dots,m$ and $j=1,\dots,n$ such that
$$ Sv_j=\sum_{i=1}^ms_{i,j}w_i $$
Similarly there exist scalars $t_{j,k}$ for $j=1,\dots,n$ and $k=1,\dots,p$ such that
$$ Tu_k=\sum_{j=1}^nt_{j,k}v_j $$
Hence
$$ (ST)u_k=S(Tu_k)=S\Prn{\sum_{j=1}^nt_{j,k}v_j}=\sum_{j=1}^nt_{j,k}Sv_j=\sum_{j=1}^nt_{j,k}\sum_{i=1}^ms_{i,j}w_i=\sum_{j=1}^n\sum_{i=1}^ms_{i,j}t_{j,k}w_i=\sum_{i=1}^m\sum_{j=1}^ns_{i,j}t_{j,k}w_i $$
In particular
$$ (ST)u_1=\sum_{i=1}^m\sum_{j=1}^ns_{i,j}t_{j,1}w_i=\sum_{j=1}^ns_{1,j}t_{j,1}w_1+\dotsb+\sum_{j=1}^ns_{m,j}t_{j,1}w_m $$
and
$$ (ST)u_p=\sum_{i=1}^m\sum_{j=1}^ns_{i,j}t_{j,p}w_i=\sum_{j=1}^ns_{1,j}t_{j,p}w_1+\dotsb+\sum_{j=1}^ns_{m,j}t_{j,p}w_m $$
Hence
$$\align{ \mtrxof{S,v,w}\mtrxof{T,u,v} &= \pmtrx{\mtrxof{S,v,w}_{:,1}&\dotsb&\mtrxof{S,v,w}_{:,n}}\pmtrx{\mtrxof{T,u,v}_{:,1}&\dotsb&\mtrxof{T,u,v}_{:,p}} \\ &= \pmtrx{\mtrxof{Sv_1,w}&\dotsb&\mtrxof{Sv_n,w}}\pmtrx{\mtrxof{Tu_1,v}&\dotsb&\mtrxof{Tu_p,v}} \\ &= \pmtrx{s_{1,1}&\dotsb&s_{1,n}\\\vdots&\ddots&\vdots\\s_{m,1}&\dotsb&s_{m,n}}\pmtrx{t_{1,1}&\dotsb&t_{1,p}\\\vdots&\ddots&\vdots\\t_{n,1}&\dotsb&t_{n,p}} \\ &= \pmtrx{\sum_{j=1}^ns_{1,j}t_{j,1}&\dotsb&\sum_{j=1}^ns_{1,j}t_{j,p}\\\vdots&\ddots&\vdots\\\sum_{j=1}^ns_{m,j}t_{j,1}&\dotsb&\sum_{j=1}^ns_{m,j}t_{j,p}} \\ &= \pmtrx{\mtrxof{STu_1,w}&\dotsb&\mtrxof{STu_p,w}} \\ &= \pmtrx{\mtrxof{ST,u,w}_{:,1}&\dotsb&\mtrxof{ST,u,w}_{:,p}} \\ &= \mtrxof{ST,u,w} }$$
and
$$\align{ \mtrxof{S,v,w}\mtrxof{T,u,v} &= \pmtrx{\sum_{j=1}^ns_{1,j}t_{j,1}&\dotsb&\sum_{j=1}^ns_{1,j}t_{j,p}\\\vdots&\ddots&\vdots\\\sum_{j=1}^ns_{m,j}t_{j,1}&\dotsb&\sum_{j=1}^ns_{m,j}t_{j,p}} \\ &= \sum_{j=1}^n\pmtrx{s_{1,j}t_{j,1}&\dotsb&s_{1,j}t_{j,p}\\\vdots&\ddots&\vdots\\s_{m,j}t_{j,1}&\dotsb&s_{m,j}t_{j,p}} \\ &= \sum_{j=1}^n\pmtrx{s_{1,j}\\\vdots\\s_{m,j}}\pmtrx{t_{j,1}&\dotsb&t_{j,p}} \\ &= \sum_{j=1}^n\mtrxof{S,v,w}_{:,j}\mtrxof{T,u,v}_{j,:} }$$
$\wes$
Proposition W.3.28 <–> 10.5 Suppose that $u\equiv u_1,\dots,u_n$ is a basis of $U$ and that $v\equiv v_1,\dots,v_n$ is a basis of $V$. Then the matrices $\mtrxof{I,u,v}$ and $\mtrxof{I,v,u}$ are invertible (in the sense of 10.3, not 3.53) and
$$ \inv{\mtrxof{I,u,v}}=\mtrxof{I,v,u} \dq\dq \inv{\mtrxof{I,v,u}}=\mtrxof{I,u,v} $$
Proof W.3.27 gives
$$ \mtrxof{I,v,u}\mtrxof{I,u,v}=\mtrxof{II,u,u}=I_n \\ \mtrxof{I,u,v}\mtrxof{I,v,u}=\mtrxof{II,v,v}=I_n $$
$\wes$
Proposition W.3.29 <–> 3.65 Suppose $v\equiv v_1,\dots,v_n$ is a basis of $V$ and $w\equiv w_1,\dots,w_m$ is a basis of $W$. Let $T\in\linmap{V}{W}$. Then, for any $u\in V$, we have
$$ \mtrxof{Tu,w}=\mtrxof{T,v,w}\mtrxof{u,v} $$
Proof Let $u\in V$. Then $u=\sum_{j=1}^nu_iv_i$ for some scalars $u_1,\dots,u_n\in\wF$. Then 3.62 gives
$$ \mtrxof{u,v}=\pmtrx{u_1\\\vdots\\u_n} $$
We also have
$$ Tu=T\Prn{\sum_{j=1}^nu_jv_j}=\sum_{j=1}^nu_jTv_j $$
Hence
$$\align{ \mtrxof{Tu,w} &= \mtrxofB{\sum_{j=1}^nu_jTv_j,w} \\ &= \sum_{j=1}^nu_j\mtrxof{Tv_j,w}\tag{by 3.36 and 3.38} \\ &= \sum_{j=1}^nu_j\mtrxof{T,v,w}_{:,j} \\ &= \mtrxof{T,v,w}\mtrxof{u,v}\tag{by 3.52} }$$
$\wes$
Proposition W.3.30 Let $m$ and $n$ be positive integers and let $u\in\wF^n$ and $v\in\wF^m$ be nonzero vectors. Then $uv^t\in\wFnm$ has rank one.
Proof Note that
$$ uv^t = \pmtrx{u_{1}\\\vdots\\u_{n}}\pmtrx{v_{1}&\dotsb&v_{m}} = \pmtrx{u_{1}v_{1}&\dotsb&u_{1}v_{m}\\\vdots&\ddots&\vdots\\u_{n}v_{1}&\dotsb&u_{n}v_{m}} $$
Since $v\neq0$, then there exists an integer $\ell\in\set{1,\dots,m}$ such that $v_{\ell}\neq0$. Put $c_k\equiv\frac{v_{k}}{v_{\ell}}$ for $k=1,\dots,m$. Then
$$ c_k(uv^t)_{:,\ell} = \frac{v_{k}}{v_{\ell}}\pmtrx{u_{1}v_{\ell}\\\vdots\\u_{n}v_{\ell}} = \pmtrx{u_{1}v_{\ell}\frac{v_{k}}{v_{\ell}}\\\vdots\\u_{n}v_{\ell}\frac{v_{k}}{v_{\ell}}} = \pmtrx{u_{1}v_{k}\\\vdots\\u_{n}v_{k}} = (uv^t)_{:,k} $$
Hence every column of $uv^t$ is a scalar multiple of the $\ell^{th}$ column. Hence $uv^t$ has rank one. $\wes$
Proposition W.3.32<–>3.69 Suppose $U$ and $V$ are finite-dimensional with $\dim{U}=\dim{V}$. Let $T\in\linmap{U}{V}$. Then the following are equivalent:
(a) $T$ is invertible
(b) $T$ is injective
(c) $T$ is surjective
Proof Suppose $T$ is invertible. Let $u_1,u_2\in U$ such that $Tu_1=Tu_2$. Then
$$ u_1=(\inv{T}T)u_1=\inv{T}(Tu_1)=\inv{T}(Tu_2)=(\inv{T}T)u_2=u_2 $$
Hence $T$ is injective and (a)$\implies$(b).
Suppose $T$ is injective. Then $\nullsp{T}=\set{0}$ (by 3.16) and hence $0=\dim{\nullsp{T}}$. From the Fundamental Theorem of Linear Maps (3.22), we have
$$ \dim{\rangsp{T}}=\dim{U}-\dim{\nullsp{T}}=\dim{V} $$
Hence $\rangsp{T}=V$ (by W.2.12) and (b)$\implies$(c).
Suppose $T$ is surjective so that $\rangsp{T}=V$. From the Fundamental Theorem of Linear Maps, we get
$$ \dim{\nullsp{T}}=\dim{U}-\dim{\rangsp{T}}=\dim{V}-\dim{V}=0 $$
Hence $\nullsp{T}=\set{0}$ and $T$ is injective. Since $T$ is surjective and injective, then 3.56 implies that $T$ is invertible. Hence (c)$\implies$(a). $\wes$
Proposition W.3.33<–>Exercise 3.D.9 Suppose $U$ and $V$ are finite-dimensional with $\dim{U}=\dim{V}$. And let $S\in\linmap{U}{V}$ and $T\in\linmap{V}{U}$. Then $ST\wiov$ is invertible if and only if both $S$ and $T$ are invertible.
Proof Suppose $ST$ is invertible. Then there exists $R\in\lnmpsb(V)$ such that $R(ST)=(ST)R=I_V$.
Let $v_1\in\mathscr{N}(T)\subset V$. Then
$$ v_1=I_Vv_1=R(ST)v_1=(RS)Tv_1=(RS)0=0 $$
Hence $\mathscr{N}(T)=\{0\}$ and $T$ is injective. Then W.3.32 gives that $T$ is invertible.
Let $v_2\in V$. Then
$$ v_2=I_Vv_2=(ST)Rv_2=S(TR)v_2\in\mathscr{R}(S) $$
Hence $V\subset\rangsp{S}$, hence $V=\rangsp{S}$, and hence $S$ is surjective. Then W.3.32 gives that $S$ is invertible.
Conversely, suppose $S$ and $T$ are both invertible. Then
$$ (ST)(T^{-1}S^{-1})=S(TT^{-1})S^{-1}=SI_US^{-1}=SS^{-1}=I_V $$
and
$$ (T^{-1}S^{-1})(ST)=T^{-1}(S^{-1}S)T=T^{-1}I_UT=T^{-1}T=I_V $$
Hence $T^{-1}S^{-1}$ satisfies the properties required for an inverse of $ST$. Hence $ST$ is invertible and $(ST)^{-1}=T^{-1}S^{-1}$. $\wes$
Proposition W.3.34<–>Exercise 3.D.10 Suppose $U$ and $V$ are finite-dimensional with $\dim{U}=\dim{V}$. And let $S\in\linmap{U}{V}$ and $T\in\linmap{V}{U}$. Then $ST=I_V$ if and only if $TS=I_U$.
Proof Suppose $ST=I_V$. Note that $I_V$ is invertible ($I_V^{-1}=I_V$ because $I_VI_V=I_V$). Hence $ST$ is invertible. Then W.3.33 implies that both $S$ and $T$ are invertible. Multiplying both sides of $ST=I_V$ by $T^{-1}$ on the right, we get
$$ S=STT^{-1}=I_VT^{-1}=T^{-1} $$
Now multiply both sides of this equation by $T$ on the left:
$$ TS=TT^{-1}=I_U $$
Conversely, we can simply reverse the roles of $S$ and $T$ and walk through the same steps. $\wes$
Example & Counterexample W.3.35 Define $U\equiv\setb{(x,y,0)\in\wR^3:x,y\in\wR}$ so that $\dim{U}=2$. Let $e\equiv e_1,e_2$ be the standard basis of $\wR^2$ and let $f\equiv f_1,f_2,f_3$ be the standard basis of $\wR^3$. Define $f_U\equiv f_1,f_2$ so that $f_U$ is a basis of $U$.
Define $S\in\linmap{U}{\wR^2}$ by
$$ \left. \begin{array}{l} Sf_1 \equiv-2e_1+\frac32e_2 \\ Sf_2 \equiv e_1-\frac12e_2 \end{array} \right. \dq\iff\dq \mtrxof{S,f_U,e}=\pmtrx{-2&1\\\frac32&-\frac12} $$
Define $T\in\linmap{\wR^2}{U}$ by
$$ \left. \begin{array}{l} Te_1 \equiv f_1+3f_2 \\ Te_2 \equiv2f_1+4f_2 \end{array} \right. \dq\iff\dq \mtrxof{T,e,f_U}=\pmtrx{1&2\\3&4} $$
Then
$$\align{ \mtrxof{ST,e} &= \mtrxof{S,f_U,e}\mtrxof{T,e,f_U} = \pmtrx{-2&1\\\frac32&-\frac12}\pmtrx{1&2\\3&4} \\ &= \pmtrx{-2+3&-4+4\\\frac32-\frac32&3-2} = \pmtrx{1&0\\0&1} = \mtrxof{I_{\wR^2},e} }$$
and
$$\align{ \mtrxof{TS,f_U} &= \mtrxof{T,e,f_U}\mtrxof{S,f_U,e} = \pmtrx{1&2\\3&4}\pmtrx{-2&1\\\frac32&-\frac12} \\ &= \pmtrx{-2+3&1-1\\-6+6&3-2} = \pmtrx{1&0\\0&1} = \mtrxof{I_U,f_U} }$$
Note that $\mtrxof{\cdot,a,b}$ is an isomorphism (3.60) for any bases $a$ and $b$. Hence $ST=I_{\wR^2}$ and $TS=I_U$. We can also see this by computing as follows:
$$\align{ STe_1 &= S(f_1+3f_2)=Sf_1+3Sf_2=-2e_1+\frac32e_2+3\Prn{e_1-\frac12e_2}=-2e_1+\frac32e_2+3e_1-\frac32e_2=e_1 \\\\ STe_2 &= S(2f_1+4f_2)=2Sf_1+4Sf_2=2\Prn{-2e_1+\frac32e_2}+4\Prn{e_1-\frac12e_2}=-4e_1+3e_2+4e_1-2e_2=e_2 }$$
and
$$\align{ TSf_1 &= T\Prn{-2e_1+\frac32e_2}=-2Te_1+\frac32Te_2=-2(f_1+3f_2)+\frac32(2f_1+4f_2)=-2f_1-6f_2+3f_1+6f_2=f_1 \\\\ TSf_2 &= T\Prn{e_1-\frac12e_2}=Te_1-\frac12Te_2=(f_1+3f_2)-\frac12(2f_1+4f_2)=f_1+3f_2-f_1-2f_2=f_2 }$$
The result $ST=I_{\wR^2}\iff TS=I_U$ holds from W.3.34 because $\dim{\wR^2}=\dim{U}$. Whereas, if we define $S’\in\linmap{\wR^3}{\wR^2}$ by
$$ \left. \begin{array}{l} S'f_1 \equiv-2e_1+\frac32e_2 \\ S'f_2 \equiv e_1-\frac12e_2 \\ S'f_3 \equiv 0 \end{array} \right. \dq\iff\dq \mtrxof{S',f,e}=\pmtrx{-2&1&0\\\frac32&-\frac12&0} $$
and define $T’\in\linmap{\wR^2}{\wR^3}$ by
$$ \left. \begin{array}{l} T'e_1 \equiv f_1+3f_2+0f_3 \\ T'e_2 \equiv2f_1+4f_2+0f_3 \end{array} \right. \dq\iff\dq \mtrxof{T',e,f}=\pmtrx{1&2\\3&4\\0&0} $$
then
$$\align{ \mtrxof{S'T',e} &= \mtrxof{S',f,e}\mtrxof{T',e,f} = \pmtrx{-2&1&0\\\frac32&-\frac12&0}\pmtrx{1&2\\3&4\\0&0} \\ &= \pmtrx{-2+3&-4+4\\\frac32-\frac32&3-2} = \pmtrx{1&0\\0&1} = \mtrxof{I_{\wR^2},e} }$$
but
$$\align{ \mtrxof{T'S',f} &= \mtrxof{T',e,f}\mtrxof{S',f,e} = \pmtrx{1&2\\3&4\\0&0}\pmtrx{-2&1&0\\\frac32&-\frac12&0} \\ &= \pmtrx{-2+3&1-1&0\\-6+6&3-2&0\\0&0&0} = \pmtrx{1&0&0\\0&1&0\\0&0&0} \neq \mtrxof{I_{\wR^3},f} }$$
Hence $S’T’=I_{\wR^2}$ but $T’S’\neq I_{\wR^3}$. We can also see this from these computations:
$$\align{ S'T'e_1 &= S'(f_1+3f_2)=S'f_1+3S'f_2=-2e_1+\frac32e_2+3\Prn{e_1-\frac12e_2}=-2e_1+\frac32e_2+3e_1-\frac32e_2=e_1 \\\\ S'T'e_2 &= S'(2f_1+4f_2)=2S'f_1+4S'f_2=2\Prn{-2e_1+\frac32e_2}+4\Prn{e_1-\frac12e_2}=-4e_1+3e_2+4e_1-2e_2=e_2 }$$
but
$$\align{ T'S'f_1 &= T'\Prn{-2e_1+\frac32e_2}=-2T'e_1+\frac32T'e_2=-2(f_1+3f_2)+\frac32(2f_1+4f_2)=-2f_1-6f_2+3f_1+6f_2=f_1 \\\\ T'S'f_2 &= T'\Prn{e_1-\frac12e_2}=T'e_1-\frac12T'e_2=(f_1+3f_2)-\frac12(2f_1+4f_2)=f_1+3f_2-f_1-2f_2=f_2 \\\\ T'S'f_3 &= T'(0)=0 }$$
Note that W.3.34 doesn’t aaply here because $\dim{\wR^2}\neq\dim{\wR^3}$. $\wes$
Example & Counterexample W.3.36 Let $f_3\equiv(0,0,1)$ be the standard third basis vector in $\wR^3$. Arbitrarily pick two more linearly independent vectors in $\wR^3$. Say $f_1\equiv(1,1,1)\in\wR^3$ and $f_2\equiv(1,-1,1)\in\wR^3$. Define $U\equiv\span{f_1,f_2}$ so that $\dim{U}=2$. Define $S$ and $T$ as in the previous example. Then we will get the same matrices, matrix products, and results. And we’ll get the same results for $S’$ and $T’$. $\wes$