(1) Prove that if $v_1,…,v_n$ spans $V$ then so does the list $v_1-v_2,v_2-v_3,…,v_{n-1}-v_n,v_n$ obtained by subtracting from each vector (except the last one) the following vector.
Proof We first show that $\text{span}(v_1,…,v_n)\subset\text{span}(v_1-v_2,…,v_{n-1}-v_n,v_n)$. For any $v\in\text{span}(v_1,…,v_n)$, there exist $a_1,…,a_n\in\mathbb{F}$ such that
$$\begin{align*} v &= a_1v_1+a_2v_2+a_3v_3+a_4v_4+\dots+a_nv_n\\\\ &= a_1v_1-a_1v_2+a_1v_2+a_2v_2+a_3v_3+a_4v_4+\dots+a_nv_n\\\\ &= a_1(v_1-v_2)+(a_1+a_2)v_2+a_3v_3+a_4v_4+\dots+a_nv_n\\\\ &= a_1(v_1-v_2)+(a_1+a_2)v_2-(a_1+a_2)v_3+(a_1+a_2)v_3+a_3v_3+a_4v_4+\dots+a_nv_n\\\\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)v_3+a_4v_4+\dots+a_nv_n\\\\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)v_3-(a_1+a_2+a_3)v_4+(a_1+a_2+a_3)v_4+a_4v_4+\dots+a_nv_n\\\\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)(v_3-v_4)+(a_1+a_2+a_3+a_4)v_4+\dots+a_nv_n\\\\ &= \sum_{i=1}^{n-1}\Bigg[\Bigg(\sum_{j=1}^{i}a_j\Bigg)(v_i-v_{i+1})\Bigg]+a_nv_n\\\\ &\in\text{span}(v_1-v_2,...,v_{n-1}-v_n,v_n) \end{align*}$$
Conversely for any $v\in\text{span}(v_1-v_2,…,v_{n-1}-v_n,v_n)$, there exist $a_1,…,a_n\in\mathbb{F}$ such that
$$\begin{align*} v &= a_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+\dots+a_nv_n\\\\ &= a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+\dots+(a_n-a_{n-1})v_n\\\\ &\in\text{span}(v_1,...,v_n) \end{align*}$$
Hence we have $\text{span}(v_1-v_2,…,v_{n-1}-v_n,v_n)\subset\text{span}(v_1,…,v_n)$ as well. $\blacksquare$
(1b) This problem isn’t given but it’s similar to (1) and useful. If $v_1,v_2,…,v_n$ spans $V$, then so does $v_1+v_n,v_2+v_n,…,v_{n-1}+v_n,v_n$ (add the last vector to all the previous vectors).
Proof We first show that $\text{span}(v_1,v_2,…,v_n)\subset\text{span}(v_1+v_n,v_2+v_n,…,v_{n-1}+v_n,v_n)$. For any $v\in\text{span}(v_1,…,v_n)$, there exist $a_1,…,a_n\in\mathbb{F}$ such that
$$\begin{align*} v &= a_1v_1+a_2v_2+\dots+a_{n-1}v_{n-1}+a_nv_n\\\\ &= a_1v_1+a_1v_n-a_1v_n+a_2v_2+\dots+a_{n-1}v_{n-1}+a_nv_n\\\\ &= a_1(v_1+v_n)+a_2v_2+\dots+a_{n-1}v_{n-1}+(a_n-a_1)v_n\\\\ &= a_1(v_1+v_n)+a_2v_2+a_2v_n-a_2v_n+\dots+a_{n-1}v_{n-1}+(a_n-a_1)v_n\\\\ &= a_1(v_1+v_n)+a_2(v_2+v_n)+\dots+a_{n-1}v_{n-1}+(a_n-a_1-a_2)v_n\\\\ &\vdots\\\\ &= a_1(v_1+v_n)+a_2(v_2+v_n)+\dots+a_{n-1}v_{n-1}+\Big(a_n-\sum_{i=1}^{n-2}a_i\Big)v_n\\\\ &= a_1(v_1+v_n)+a_2(v_2+v_n)+\dots+a_{n-1}v_{n-1}+a_{n-1}v_n-a_{n-1}v_n+\Big(a_n-\sum_{i=1}^{n-2}a_i\Big)v_n\\\\ &= a_1(v_1+v_n)+a_2(v_2+v_n)+\dots+a_{n-1}(v_{n-1}+v_n)+\Big(a_n-\sum_{i=1}^{n-1}a_i\Big)v_n\\\\ &\in\text{span}(v_1+v_n,v_2+v_n,...,v_{n-1}+v_n,v_n) \end{align*}$$
Conversely for any $v\in\text{span}(v_1+v_n,v_2+v_n,…,v_{n-1}+v_n,v_n)$, there exist $a_1,a_2,…,a_n\in\mathbb{F}$ such that
$$\begin{align*} v &= a_1(v_1+v_n)+a_2(v_2+v_n)+\dots+a_{n-1}(v_{n-1}+v_n)+a_nv_n\\\\ &= a_1v_1+a_1v_n+a_2v_2+a_2v_n+\dots+a_{n-1}v_{n-1}+a_{n-1}v_n+a_nv_n\\\\ &= a_1v_1+a_2v_2+\dots+a_{n-1}v_{n-1}+\Big(\sum_{i=1}^{n}a_i\Big)v_n\\\\ &\in\text{span}(v_1,v_2,...,v_n) \end{align*}$$
Hence we have $\text{span}(v_1+v_n,v_2+v_n,…,v_{n-1}+v_n,v_n)\subset\text{span}(v_1,v_2,…,v_n)$ as well. $\blacksquare$
(2) Prove that if $v_1,…,v_n$ is linearly independent in $V$, then so is the list $v_1-v_2,v_2-v_3,…,v_{n-1}-v_n,v_n$ obtained by subtracting from each vector (except the last one) the following vector.
Proof
$$ 0=\sum_{i=1}^{n-1}a_i(v_i-v_{i+1})+a_nv_n=a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+\dots+(a_n-a_{n-1})v_n $$
Then $0=a_1=a_2-a_1=a_3-a_2=\dots=a_n-a_{n-1}$ by the linear independence of $v_1,…,v_n$. But this implies that $a_2=0$ which implies that $a_3=0$ and so on. That is $0=a_1=a_2=\dots=a_n$. $\blacksquare$
(2b) This problem isn’t given but it’s similar to (2) and useful. If $v_1,v_2,…,v_n$ is linearly independent in $V$, then so is $v_1+v_n,v_2+v_n,…,v_{n-1}+v_n,v_n$ (add the last vector to all the previous vectors).
Proof
$$ 0=\sum_{i=1}^{n-1}a_i(v_i+v_n)+a_nv_n=\sum_{i=1}^{n-1}a_iv_i+\sum_{i=1}^{n-1}a_iv_n+a_nv_n=\sum_{i=1}^{n-1}a_iv_i+\sum_{i=1}^{n}a_iv_n=\sum_{i=1}^{n-1}a_iv_i+\Big(\sum_{i=1}^{n}a_i\Big)v_n $$
Then $0=a_1=a_2=\dots=a_{n-1}=\sum_{i=1}^{n}a_i$ by the linear independence of $v_1,…,v_n$. But this implies that $a_n=0$ as well. $\blacksquare$
(3) Suppose $v_1,…,v_n$ is linearly independent in $V$ and $w\in V$. Prove that if $v_1+w,…,v_n+w$ is linearly dependent, then $w\in\text{span}(v_1,…,v_n)$.
Proof If $v_1+w,…,v_n+w$ is linearly dependent, then there exists $a_1,…,a_n\in\mathbb{F}$ not all zero such that
$$ 0=\sum_{i=1}^{n}a_i(v_i+w)=\sum_{i=1}^{n}a_iv_i+w\sum_{i=1}^{n}a_i $$
or
$$ w=\sum_{i=1}^{n}\frac{-a_i}{\sum_{j=1}^{n}a_j}v_i\in\text{span}(v_1,...,v_n) $$
$\blacksquare$
(4) Suppose $m$ is a positive integer. Is the set $U$ consisting of $0$ and all polynomials with coefficients in $\mathbb{F}$ and with degree equal to $m$ a subspace of $\mathscr{P}(\mathbb{F})$?
Solution No. Define $p(z)\equiv az^{m-1}+az^m$ and $q(z)\equiv az^{m-1}-az^m$ for any $a\neq0\in\mathbb{F}$. Then $p,q\in U$ but $(p+q)(z)=2az^{m-1}$ hence $p+q\notin U$. Hence $U$ isn’t closed under addition and it’s not a subspace.
(5) Prove that $\mathbb{F}^{\infty}$ is infinite-dimensional.
We present two different proofs.
Proof 1 Suppose $\mathbb{F}^{\infty}$ is finite-dimensional. Recall that every linearly independent list of vectors in a finite dimensional vector space can be extended to a basis of the vector space. Define $e_i\equiv(0,0,…,0,1,0,…)$ to be the vector with $1$ in the $i^{th}$ component and $0$ in all other components. The list $e_1,e_2,…,e_n$ is linearly independent for every positive integer $n$. Let’s extend it to a basis $e_1,e_2,…,e_b$ for $V$. But $e_{b+1}\notin\text{span}(e_1,e_2,…,e_b)$, a contradiction. $\blacksquare$
Proof 2 In our ch.2 notes, we proved that a vector space $V$ is infinite-dimensional if and only if there exists a sequence of vectors $v_1,v_2,…$ in $V$ such that the list $v_1,v_2,…,v_n$ is linearly independent for every positive integer $n$. Define $e_i\equiv(0,0,…,0,1,0,…)$ to be the vector with $1$ in the $i^{th}$ component and $0$ in all other components. For the sequence $e_1,e_2,…$ in $\mathbb{F}^{\infty}$, the list $e_1,e_2,…,e_n$ is linearly independent for every positive integer $n$. $\blacksquare$
(6) Prove that the real vector space $\mathscr{C}^{[0,1]}$ consisting of all continuous real-valued functions on the interval $[0,1]$ is infinite dimensional.
We present two different proofs.
Proof 1 Consider the set $\mathscr{P}(\mathbb{F})^{[0,1]}\equiv\{p(z)\in\mathscr{P}(\mathbb{F}):z\in[0,1]\}$. This set is a subspace of $\mathscr{C}^{[0,1]}$ - we can show this in the same way that we showed in our ch.2 notes that $\mathscr{P}(\mathbb{F})$ is a subspace of $\mathbb{F}^{\mathbb{F}}$. And in the same way as the text (ex.2.16, p.32) showed that $\mathscr{P}(\mathbb{F})$ is infinite-dimensional, we can show that $\mathscr{P}(\mathbb{F})^{[0,1]}$ is infinite-dimensional. Hence $\mathscr{C}^{[0,1]}\supset\mathscr{P}(\mathbb{F})^{[0,1]}$ is infinite-dimensional. $\blacksquare$
Proof 2 For $n=1,2,…$, define
$$ f_n(x)\equiv\begin{cases}x-\frac{1}{n}&x\in[\frac{1}{n},1]\\0&x\in[0,\frac{1}{n}]\end{cases} $$
Clearly $f_n$ is piecewise-continuous on $[0,1]$. But at the one potential point of discontinuity $x=\frac{1}{n}$, we see that the left and right limits are both $0$. Hence $f_n$ is continuous on $[0,1]$. Hence $f_n\in\mathscr{C}^{[0,1]}$ for $n=1,2,…$.
In our ch.2 notes, we proved that a vector space $V$ is infinite-dimensional if and only if there exists a sequence of vectors $v_1,v_2,…$ in $V$ such that the list $v_1,v_2,…,v_n$ is linearly independent for every positive integer $n$. Hence it suffices to show that $f_1,f_2,…,f_n$ is linearly independent for every positive integer $n$.
By the contrapositive of the Linear Dependence Lemma, it suffices to show that $f_{n+1}\notin\text{span}(f_1,…,f_n)$. Suppose that is not the case. Then for some $a_1,…,a_n$ not all zero and all $x\in[0,1]$, we have
$$ f_{n+1}(x)=a_1f_1(x)+\dots+a_nf_n(x) $$
For $i=1,…,n$, we have $n\geq i\implies\frac{1}{n}\in[0,\frac{1}{i}]$ and hence $f_i(\frac{1}{n})=0$. Hence the right-hand side of the above equality is $0$.
But $n<n+1\implies\frac{1}{n}\in[\frac{1}{n+1},1]$ and hence $f_{n+1}(\frac{1}{n})=\frac{1}{n}-\frac{1}{n+1}>0$. Hence the left-hand side of the above equality is positive. Contradiction. $\blacksquare$
(7) Prove that a vector space $V$ is infinite-dimensional if and only if there exists a sequence of vectors $v_1,v_2,…$ in $V$ such that the list $v_1,v_2,…,v_n$ is linearly independent for every positive integer $n$.
Proof We proved this in the ch.2 notes.
(8) Let $U$ be the subspace of $\mathbb{R}^5$ defined by
$$ U\equiv\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5:x_1=3x_2\text{ and }x_3=7x_4\} $$
Find a basis of $U$.
Solution Let’s write any vector in $U$ as a linear combination of some linearly independent list. By definition, such a list will be a basis for $U$.
$$\begin{align*} (x_1,x_2,x_3,x_4,x_5) &= (3x_2,x_2,7x_4,x_4,x_5)\\ &= (3x_2,x_2,0,0,0)+(0,0,7x_4,x_4,0)+(0,0,0,0,x_5)\\ &= x_2(3,1,0,0,0)+x_4(0,0,7,1,0)+x_5(0,0,0,0,1)\\ &= x_2u_1+x_4u_2+x_5u_3 \end{align*}$$
where the last equality defines $u_1,u_2,$ and $u_3$. Hence the list $u_1,u_2,u_3$ spans $U$. To show that $u_1,u_2,u_3$ is linearly independent, we set the linear combination to zero and solve for the coefficients:
$$ 0=au_1+bu_2+cu_3 $$
or
$$ \begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}=a\begin{bmatrix}3\\1\\0\\0\\0\end{bmatrix}+b\begin{bmatrix}0\\0\\7\\1\\0\end{bmatrix}+c\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}=\begin{bmatrix}3a\\1a\\7b\\1b\\1c\end{bmatrix} $$
This implies $0=a=b=c$. $\blacksquare$
(9) There exists a basis $p_0,p_1,p_2,p_3$ of $\mathscr{P}_{3}(\mathbb{F})$ such that none of the polynomials $p_0,p_1,p_2,p_3$ has a degree $2$.
Proof Recall that $1,x,x^2,x^3$ spans and is linearly independent in $\mathscr{P}_{3}(\mathbb{F})$ (we proved both facts in our ch.2 notes). Hence, from problems (1b) and (2b) in this solution set, we know that $1+x^3,x+x^3,x^2+x^3,x^3$ spans and is linearly independent in $\mathscr{P}_{3}(\mathbb{F})$. $\blacksquare$
(11) Suppose that $V$ is finite dimensional and $U$ is a subspace of $V$ such that $\dim{U}=\dim{V}$. Then $U=V$.
Contrapositive Suppose that $V$ is finite dimensional and $U$ is a subspace of $V$. If $U\neq V$, then $\dim{U}<\dim{V}$.
Proof Let $u_1,…,u_n$ be a basis for $U$. Then $n=\dim{U}=\dim{V}$ and $u_1,…,u_n\in V$. By 2.39 (p.45), $u_1,…,u_n$ is a basis for $V$. Hence $V=\text{span}(u_1,…,u_n)=U$. $\blacksquare$
(12) Suppose that $p_0,p_1,…,p_m$ are polynomials in $\mathscr{P}_{m}(\mathbb{F})$ such that $p_j(2)=0$ for each $j$. Then $p_0,p_1,…,p_m$ is not linearly independent in $\mathscr{P}_{m}(\mathbb{F})$.
Proof Recall from our ch.2 notes that $1,z,z^2,…,z^m$ spans and is linearly independent in $\mathscr{P}_{m}(\mathbb{F})$. Hence $1,z,z^2,…,z^m$ is a basis for $\mathscr{P}_{m}(\mathbb{F})$. Hence $\dim{\mathscr{P}_{m}(\mathbb{F})}=m+1$.
Hence if $p_0,p_1,…,p_m$ is linearly independent in $\mathscr{P}_{m}(\mathbb{F})$, then it’s a basis for $\mathscr{P}_{m}(\mathbb{F})$, by 2.39 (p.45). Hence for any $p\in\mathscr{P}_{m}(\mathbb{F})$, we have $p(2)=\sum_{i=0}^{m}c_ip_i(2)=0$. But this is a contradiction since there exists $p\in\mathscr{P}_{m}(\mathbb{F})$ such that $p(2)\neq0$. For example, $p(z)=1,z,z^2,…,z^m\neq0$ for $z=2$. $\blacksquare$
Exercises 2.C
(10) Suppose $p_0,p_1,…,p_m\in\mathscr{P}(\mathbb{F})$ are such that each $p_j$ has degree $j$. Then $p_0,p_1,…,p_m$ is a basis for $\mathscr{P}_m(\mathbb{F})$.
Proof Proposition W.2.2 gives that $1,z,…,z^m$ is a basis for $\mathscr{P}_m(\mathbb{F})$. Note that
$$ p_0=c_01\quad\quad\quad p_j=c_jz^j+\sum_{i=0}^{j-1}d_{j,i}z^i\quad\text{for }j=1,...,m $$
where $c_0,c_1,…,c_m$ are some nonzero scalars and the $d_{j,i}$ are some scalars (zero or not) for $j=1,…,m$ and $i=0,1,…,j-1$. Then proposition W.2.18 gives that $p_0,p_1,…,p_m$ is a basis for $\mathscr{P}_m(\mathbb{F})$. $\blacksquare$
(11) Suppose $U$ and $W$ are subspaces of $\mathbb{R}^{8}$ such that $\dim{U}=3,\dim{W}=5$, and $U+W=\mathbb{R}^{8}$. Then $U\cap W=\{0\}$ and $\mathbb{R}^{8}=U\oplus W$.
Proof By 2.43 (p.47), $\dim{(U\cap W)}=\dim{U}+\dim{W}-\dim{(U+W)}=3+5-8=0$. By 1.45 (p.23), $U\oplus W=U+W=\mathbb{R}^{8}$. $\blacksquare$
(12) Suppose $U$ and $W$ are both five-dimensional subspaces of $\mathbb{R}^{9}$. Then $U\cap W\neq\{0\}$.
Proof Suppose $U\cap W=\{0\}$. By 2.43 (p.47), $\dim{(U+W)}=\dim{U}+\dim{W}-\dim{(U\cap W)}=5+5-0=10$. But $U+W$ is a collection of linear combinations of vectors from $\mathbb{R}^{9}$. Hence $U+W\subset\mathbb{R}^{9}$ and $10=\dim{(U+W)}\leq\dim{\mathbb{R}^{9}}=9$ where the inequality follows from 2.38 (p.45). Contradiction. $\blacksquare$
(15) Suppose that $V$ is finite dimensional and $\dim{V}=n$. Then there exist one-dimensional subspaces $U_1,…,U_n$ of $V$ such that $V=U_1\oplus\dots\oplus U_n$.
Proof Let $v_1,…,v_n$ be a basis for $V$ and define $U_i\equiv\text{span}(v_i)$ for $i=1,…,n$. Note that by the definition of dimension, $\dim{U_i}=1$.
We could use proposition W.2.11 from our ch.2 notes to show that $V=U_1\oplus\dots\oplus U_n$. But it’s helpful to write this out:
Let $u\in U_1+\dots+U_n$. Then $u=u_1+\dots+u_n$ for some $u_1\in U_1,…,u_n\in U_n$. And $u_i=a_iv_i$ for some $a_i\in\mathbb{F}$ since $U_i=\text{span}(v_i)$. Hence $u=a_1v_1+\dots+a_nv_n\in V$ and $U_1+\dots+U_n\subset V$.
In the other direction, let $v\in V$. Then we can write $v$ as $v=\sum_{i=1}^{n}a_iv_i$. Note that $a_iv_i\in U_i$ since $U_i=\text{span}(v_i)$. Hence $v=\sum_{i=1}^{n}a_iv_i\in U_1+\dots+U_n$ and $V\subset U_1+\dots+U_n$.
Hence $V=U_1+\dots+U_n$ and it suffices to show that $U_1+\dots+U_n$ is a direct sum. Let $u_1\in U_1,…,u_n\in U_n$ be such that $0=\sum_{i=1}^{n}u_i$. By 1.44 (p.23), it suffices to show that $u_i=0$ for $i=1,…,n$. Note that $u_i=a_iv_i$ for some $a_i\in\mathbb{F}$. Hence $0=\sum_{i=1}^{n}a_iv_i$. But $v_1,…,v_n$ is linearly independent. Hence the $a_i$’s are zero and hence the $u_i$’s are zero. $\blacksquare$
(14) & (16) If $U_1,…,U_m$ are finite-dimensional subspaces of $V$, then $U_1+\dots+U_m$ is finite-dimensional and
$$ \dim{(U_1+\dots+U_m)}\leq\dim{U_1}+\dots+\dim{U_m} $$
Equality holds if and only if $U_1+\dots+U_m=U_1\oplus\dots\oplus U_m$ is a direct sum.
Proof Let $u_1^{(i)},…,u_{n_i}^{(i)}$ be a basis for $U_i$ for $i=1,…,m$. Then $\dim{U_i}=n_i$ and
$$ \dim{U_1}+\dots+\dim{U_m}=n_1+\dots+n_m $$
Define the list $W\equiv u_1^{(1)},…,u_{n_1}^{(1)},u_1^{(2)},…,u_{n_2}^{(2)},…,u_1^{(m)},…,u_{n_m}^{(m)}$. Then, by proposition W.2.11 from our ch.2 notes, $U_1+\dots+U_m=\text{span}(W)$.
Hence, by 2.31 (p.40), we know that a basis for $U_1+\dots+U_m$ is contained in $W$. Hence $\dim{(U_1+\dots+U_m)}\leq\text{len}(W)$ where $\text{len}(W)$ denotes the number of vectors in the list $W$. But $\text{len}(W)=n_1+\dots+n_m$. Putting it all together, we have
$$ \dim{(U_1+\dots+U_m)}\leq\text{len}(W)=n_1+\dots+n_m=\dim{U_1}+\dots+\dim{U_m} $$
Note that equality holds if and only if $\dim{(U_1+\dots+U_m)}=\text{len}(W)$. In proposition W.2.11 of our ch.2 notes, we proved $\dim{(U_1+\dots+U_m)}=\text{len}(W)$ if and only if $U_1+\dots+U_m=U_1\oplus\dots\oplus U_m$ is a direct sum. $\blacksquare$
(17) You might guess, by analogy with the formula for the number of elements in the union of three subsets of a finite set, that if $U_1,U_2,U_3$ are subspaces of a finite-dimensional vector space, then
$$\begin{align*} \dim{(U_1+U_2+U_3)} &= \dim{U_1}+\dim{U_2}+\dim{U_3} \\ &-\dim{(U_1\cap U_2)}-\dim{(U_1\cap U_3)}-\dim{(U_2\cap U_3)} \\ &+\dim{(U_1\cap U_2\cap U_3)} \end{align*}$$
But this formula doesn’t hold. Here’s a counterexample: In our ch.1 notes, we showed $U\oplus W=\mathbb{R}^{2}$ where
$$ U\equiv\{u=(u_1,u_2)\in\mathbb{R}^2: u_1=u_2\} \quad\quad\text{and}\quad\quad W\equiv\{w=(w_1,w_2)\in\mathbb{R}^2: w_1=-w_2\} $$
If we also define $Y\equiv \{y=(y_1,y_2)\in\mathbb{R}^2: y_2=0\}$, then we can similarly show that $U\oplus W\oplus Y=\mathbb{R}^{2}$ and that $\{0\}=U\cap W=U\cap Y=W\cap Y=U\cap W\cap Y$. Then if the proposed formula holds, we have
$$\begin{align*} \dim{(U+W+Y)} &= \dim{U}+\dim{W}+\dim{Y} \\ &-\dim{(U\cap W)}-\dim{(U\cap Y)}-\dim{(W\cap Y)} \\ &+\dim{(U\cap W\cap Y)} \\ &=1+1+1-0-0-0-0 \\ &=3 \end{align*}$$
But this is a contradiction since $\dim{(U+W+Y)}=\dim{(U\oplus W\oplus Y)}=\dim{\mathbb{R}^2}=2$.