Section 1.10

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

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

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

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

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

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

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

Proof:
Call this map ${\displaystyle {\hat {L}}:V\to G_{L}}$. That is ${\displaystyle {\hat {L}}(x)=(x,L(x))}$. First I will show this map is linear:

${\displaystyle {\hat {L}}(x_{1}+x_{2})=(x_{1}+x_{2},L(x_{1}+x_{2}))=(x_{1}+x_{2},L(x_{1})+L(x_{2}))=(x_{1},L(x_{1}))+(x_{2},L(x_{2}))={\hat {L}}(x_{1})+{\hat {L}}(x_{2})}$ and ${\displaystyle {\hat {L}}(\alpha x)=(\alpha x,L(\alpha x))=(\alpha x,\alpha L(x))=\alpha (x,L(x))=\alpha {\hat {L}}(x)}$. Thus ${\displaystyle {\hat {L}}}$ is linear. Now to show ${\displaystyle {\hat {L}}}$ is bijective. If ${\displaystyle (x,L(x))\in G_{L}}$, then ${\displaystyle {\hat {L}}(x)=(x,L(x))}$ so ${\displaystyle {\hat {L}}}$ is trivially onto. In fact, we essentially chose to the codomain of our function ${\displaystyle {\hat {L}}}$ to just be the image/range of the map to ensure it was onto. Now to show ${\displaystyle {\hat {L}}}$ is one-to-one. Suppose ${\displaystyle {\hat {L}}(x_{1})={\hat {L}}(x_{2})}$. Then ${\displaystyle (x_{1},L(x_{1}))=(x_{2},L(x_{2})}$. But two ordered pairs are equal if and only if both components are equal. That is, ${\displaystyle x_{1}=x_{2}}$. Thus ${\displaystyle {\hat {L}}}$ is one-to-one. Therefore ${\displaystyle {\hat {L}}}$ is an isomorphism.