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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\times N\subset V\times W} is also a subspace.

Proof:

Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x_{1},y_{1}),(x_{2},y_{2})\in M\times N} . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1},x_{2}\in M} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y_{1},y_{2}\in N} . But Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M} is a subspace and so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}+x_{2}\in M} . Also $N$ is a subspace so $y_{1}+y_{2}\in N$. This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x_{1}+x_{2},y_{1}+y_{2})\in M\times N} . On the other hand Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})} . Thus, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\times N} is closed under vector addition. Now suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x,y)\in M\times N} and $\alpha \in \mathbb {F}$. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\in M} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y\in N} . But Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M} and $N$ are subspaces so $\alpha x\in M$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha y\in N} . That means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\alpha x,\alpha y)\in M\times N} . This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha (x,y)=(\alpha x,\alpha y)\in M\times N} . Hence Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\times N} is closed under scalar multiplication. Therefore Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\times N} is a subspace of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V\times W} .

12. Let $L:V\to W$ be a linear map and consider the graph Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}=\{(x,L(x)):x\in V\}\subset V\times W} (a) Show that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}} is a subspace.

Proof:

Suppose $(x_{1},L(x_{1}),(x_{2},L(x_{2})\in G_{L}$. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $L$ is linear which means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x_{1})+L(x_{2})=L(x_{1}+x_{2})} . Thus, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}} is closed under vector addition. Now suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x,L(x))\in G_{L}} and $\alpha \in \mathbb {F}$. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha L(x)=L(\alpha x)} . Hence Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}} is closed under scalar multiplication. Therefore Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}} is a subspace of $V\times W$.

(b) Show that the map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V\to G_{L}} that sends $x$ to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x,L(x)} is an isomorphism.

Proof:

Call this map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}:V\to G_{L}} . That is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}(x)=(x,L(x))} . First I will show this map is linear: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}(\alpha x)=(\alpha x,L(\alpha x))=(\alpha x,\alpha L(x))=\alpha (x,L(x))=\alpha {\hat {L}}(x)} . Thus Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} is linear. Now to show Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} is bijective. If $(x,L(x))\in G_{L}$, then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}(x)=(x,L(x))} so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} is trivially onto. In fact, we essentially chose to the codomain of our function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} to just be the image/range of the map to ensure it was onto. Now to show Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} is one-to-one. Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}(x_{1})={\hat {L}}(x_{2})} . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}=x_{2}} . Thus Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}} is one-to-one. Therefore ${\hat {L}}$ is an isomorphism.