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: Failed to parse (syntax error): {\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 ${\hat {L}}(\alpha x)=(\alpha x,L(\alpha x))=(\alpha x,\alpha L(x))=\alpha (x,L(x))=\alpha {\hat {L}}(x)$. 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.