Section 1.10 Homework

3. Let $L:V\to W$ be a linear map and $N\subset W$ a subspace. Show that: $L^{-1}(N)=\{x\in V:L(x)\in N\}$ is a subspace of $V$.

Proof: Suppose $x_{1},x_{2}\in L^{-1}(N)$. Then $L(x_{1}),L(x_{2})\in N$. But $N$ is a subspace and so $L(x_{1})+L(x_{2})\in N$. But $L$ is linear so that $L(x_{1}+x_{2})=L(x_{1})+L(x_{2})\in N$ so that $x_{1}+x_{2}\in L^{-1}(N)$. Thus, $L^{-1}(N)$ is closed under vector addition. Now suppose $x\in L^{-1}(N)$ and $\alpha \in \mathbb {F}$. Then $L(x)\in N$ and since $N$ is a subspace, $\alpha L(x)\in N$. But again $L$ is linear so $L(\alpha x)=\alpha L(x)\in N$. This means $\alpha x\in L^{-1}(N)$. Hence $L^{-1}(N)$ is closed under scalar multiplication. Therefore $L^{-1}(N)$ is a subspace of $V$.

10. Show that if $M\subset V$ and $N\subset W$ are subspaces, then $M\times N\subset V\times W$ is also a subspace.

Proof: Suppose $(x_{1},y_{1}),(x_{2},y_{2})\in M\times N$. Then $x_{1},x_{2}\in M$ and $y_{1},y_{2}\in N$. But $M$ is a subspace and so $x_{1}+x_{2}\in M$. Also $N$ is a subspace so $y_{1}+y_{2}\in N$. This means $(x_{1}+x_{2},y_{1}+y_{2})\in M\times N$. On the other hand $(x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})$. Thus, $M\times N$ is closed under vector addition. Now suppose $(x,y)\in M\times N$ and $\alpha \in \mathbb {F}$. Then $x\in M$ and $y\in N$. But $M$ and $N$ are subspaces so $\alpha x\in M$ and $\alpha y\in N$. That means $(\alpha x,\alpha y)\in M\times N$. This means $\alpha (x,y)=(\alpha x,\alpha y)\in M\times N$. Hence $M\times N$ is closed under scalar multiplication. Therefore $M\times N$ is a subspace of $V\times W$.

12. Let $L:V\to W$ be a linear map and consider the graph $G_{L}=\{(x,L(x)):x\in V\}\subset V\times W$ (a) Show that $G_{L}$ is a subspace.

Proof: Suppose $(x_{1},L(x_{1}),(x_{2},L(x_{2})\in G_{L}$. Then $(x_{1},L(x_{1})+(x_{2},L(x_{2})=(x_{1}+x_{2},L(x_{1})+L(x_{2}))=(x_{1}+x_{2},L(x_{1}+x_{2})\in G_{L}$. Here I used the fact that $L$ is linear which means $L(x_{1})+L(x_{2})=L(x_{1}+x_{2})$. Thus, $G_{L}$ is closed under vector addition. Now suppose $(x,L(x))\in G_{L}$ and $\alpha \in \mathbb {F}$. Then $\alpha (x,L(x))=(\alpha x,\alpha L(x))=(\alpha x,L(\alpha x)\in G_{L}$. Again I used the linearity property to conclude $\alpha L(x)=L(\alpha x)$. Hence $G_{L}$ is closed under scalar multiplication. Therefore $G_{L}$ is a subspace of $V\times W$.

(b) Show that the map $V\to G_{L}$ that sends $x$ to $(x,L(x)$ is an isomorphism.

Proof: Call this map ${\hat {L}}:V\to G_{L}$. That is ${\hat {L}}(x)=(x,L(x))$. First I will show this map is linear: ${\hat {L}}(x_{1}+x_{2})=(x_{1}+x_{2},L(x_{1}+x_{2}))=(x_{1}+x_{2},L(x_{1})+L(x_{2}))=(x_{1},L(x_{1}))+(x_{2},L(x_{2}))={\hat {L}}(x_{1})+{\hat {L}}(x_{2})$ and ${\hat {L}}(\alpha x)=(\alpha x,L(\alpha x))=(\alpha x,\alpha L(x))=\alpha (x,L(x))=\alpha {\hat {L}}(x)$. Thus ${\hat {L}}$ is linear. Now to show ${\hat {L}}$ is bijective. If $(x,L(x))\in G_{L}$, then ${\hat {L}}(x)=(x,L(x))$ so ${\hat {L}}$ is trivially onto. In fact, we essentially chose to the codomain of our function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} to just be the image/range of the map to ensure it was onto. Now to show Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is one-to-one. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(x_1) = \hat{L}(x_2)} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1,L(x_1)) = (x_2,L(x_2)} . But two ordered pairs are equal if and only if both components are equal. That is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_1 = x_2} . Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is one-to-one. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is an isomorphism.