# Section 1.10

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 ${\hat {L}}$ to just be the image/range of the map to ensure it was onto. Now to show ${\hat {L}}$ is one-to-one. Suppose ${\hat {L}}(x_{1})={\hat {L}}(x_{2})$ . Then $(x_{1},L(x_{1}))=(x_{2},L(x_{2})$ . But two ordered pairs are equal if and only if both components are equal. That is, $x_{1}=x_{2}$ . Thus ${\hat {L}}$ is one-to-one. Therefore ${\hat {L}}$ is an isomorphism.