Section 1.11 Homework

7. Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.

Proof: (by contradiction) Suppose $M,N$ are both 2-dimensional subspaces of a 3-dimension vector space Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V} and assume that $M,N$ have trivial intersection. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M+N} is also a subspace of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V} , and since $M,N$ have a trivial intersection $M+N=M\oplus N$. But then:
$\dim(M+N)=\dim M+\dim N=2+2$. However subspaces must have a smaller dimension than the whole vector space and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 4>3} . This is a contradiction and so $M,N$ must have trivial intersection.

8. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M_{1},M_{2}\subset V} be subspaces of a finite dimensional vector space Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V} . Show that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}\cap M_{2})+\dim(M_{1}\cup M_{2})=\dim M_{1}+\dim M_{2}} .

Proof: Define the linear map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L:M_{1}\times M_{2}\to V} by $L(x_{1},x_{2})=x_{1}-x_{2}$. Then by dimension formula Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}\times M_{2})=\dim \ker(L)+\dim {\text{im}}(K)} First note that in general Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(V\times W)=\dim V+\dim W} . This fact I won’t prove here but is why Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim \mathbb {R} ^{2}=1+1=2} . Now $\ker(L)=\{(x_{1},x_{2}):L(x_{1},x_{2})=0\}$. That is, $(x_{1},x_{2})\in \ker(L)$ iff Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}-x_{2}=0\Rightarrow x_{1}=x_{2}} . But since $x_{1}\in M_{1}$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{2}\in M_{2}} and they are actually the same vector, $x_{1}=x_{2}$, then we must have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}=x_{2}\in M_{1}\cap M_{2}} . That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in $M_{1}\cap M_{2}$. Then we can write Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ker(L)=\{(x,x):x\in M_{1}\cap M_{2}\}} . I claim that this is isomorphic to $M_{1}\cap M_{2}$. To prove this consider the function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi :M_{1}\cap M_{2}\to \ker(L)} as $\phi (x)=(x,x)$. This map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi } is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}\cap M_{2})=\dim(\ker(L))} . Finally let us examine ${\text{im}}(L)=\{x_{1}-x_{2}:x_{1}\in M_{1},x_{2}\in M_{2}\}$. I claim that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{im}}(L)=M_{1}+M_{2}} . Note, this is equal and not just isomorphic. To see this, we note that if Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{2}\in M_{2}} then $-x_{2}\in M_{2}$ by subspace property. So then any Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}+x_{2}\in M_{1}+M_{2}} is also equal to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}-(-x_{2})\in {\text{im}}(L)} . So these sets do indeed contain the exact same elements. That means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}+M_{2})=\dim {\text{im}}(L)} . Putting this all together gives:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim M_{1}+\dim M_{2}=\dim(M_{1}\times M_{2})=\dim \ker(L)+\dim {\text{im}}(L)=\dim(M_{1}\cap M_{2})+\dim(M_{1}+M_{2})} .

8. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M_{1},M_{2}\subset V} be subspaces of a finite dimensional vector space Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V} . Show that $\dim(M_{1}\cap M_{2})+\dim(M_{1}\cup M_{2})=\dim M_{1}+\dim M_{2}$.

Proof: Define the linear map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L:M_{1}\times M_{2}\to V} by $L(x_{1},x_{2})=x_{1}-x_{2}$. Then by dimension formula Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}\times M_{2})=\dim \ker(L)+\dim {\text{im}}(K)} First note that in general Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(V\times W)=\dim V+\dim W} . This fact I won’t prove here but is why Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim \mathbb {R} ^{2}=1+1=2} . Now $\ker(L)=\{(x_{1},x_{2}):L(x_{1},x_{2})=0\}$. That is, $(x_{1},x_{2})\in \ker(L)$ iff Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}-x_{2}=0\Rightarrow x_{1}=x_{2}} . But since $x_{1}\in M_{1}$ and $x_{2}\in M_{2}$ and they are actually the same vector, $x_{1}=x_{2}$, then we must have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}=x_{2}\in M_{1}\cap M_{2}} . That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in $M_{1}\cap M_{2}$. Then we can write Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ker(L)=\{(x,x):x\in M_{1}\cap M_{2}\}} . I claim that this is isomorphic to $M_{1}\cap M_{2}$. To prove this consider the function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi :M_{1}\cap M_{2}\to \ker(L)} as $\phi (x)=(x,x)$. This map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \phi } is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}\cap M_{2})=\dim(\ker(L))} . Finally let us examine ${\text{im}}(L)=\{x_{1}-x_{2}:x_{1}\in M_{1},x_{2}\in M_{2}\}$. I claim that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{im}}(L)=M_{1}+M_{2}} . Note, this is equal and not just isomorphic. To see this, we note that if $x_{2}\in M_{2}$ then $-x_{2}\in M_{2}$ by subspace property. So then any $x_{1}+x_{2}\in M_{1}+M_{2}$ is also equal to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}-(-x_{2})\in {\text{im}}(L)} . So these sets do indeed contain the exact same elements. That means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim(M_{1}+M_{2})=\dim {\text{im}}(L)} . Putting this all together gives:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \dim M_{1}+\dim M_{2}=\dim(M_{1}\times M_{2})=\dim \ker(L)+\dim {\text{im}}(L)=\dim(M_{1}\cap M_{2})+\dim(M_{1}+M_{2})} .

16. Show that the matrix
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{bmatrix}0&1\\0&1\end{bmatrix}}} as a linear map satisfies Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ker(L)={\text{im}}(L)} .

Proof: The matrix is already in eschelon form and has one pivot in the second column. That means that a basis for the column space which is the same as the image would be the second column. In other words, ${\text{im}}(L)={\text{Span}}\left({\begin{bmatrix}1\\0\end{bmatrix}}\right)$. Now for the kernel space. Writing out the equation $Lx=0$ reads Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0x_{1}+1x_{2}=0} or in other words Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{2}=0} . Then an arbitrary element of the kernel Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}=x_{2}{\begin{bmatrix}1\\0\end{bmatrix}}} . So again $\ker(L)={\text{Span}}\left({\begin{bmatrix}1\\0\end{bmatrix}}\right)$. In other words, $\ker(L)={\text{im}}(L)$.

17. Show that
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{bmatrix}0&0\\\alpha &1\end{bmatrix}}} defines a projection for all $\alpha \in \mathbb {F}$. Compute the kernel and image.

First I will deal with the case 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 \alpha = 0} . In this case the matrix 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 \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}} and we see by the procedure in the last problem that: 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 \text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )} and 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 \ker(L) = \text{Span} \left ( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )} .

Now for the case 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 \alpha \ne 0} . Then we still have only one pivot and either column can form a basis for the image. Using the second column makes it look nicer, and is the same as the previous case. 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 \text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )} . The difference is when we write out the equation 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 Lx = 0} to find the kernel, we get 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 \alpha x_1 + x_2 = 0} . With 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_2} as our free variable this means 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 = -\frac{1}{\alpha} x_2 } so that a basis for the kernel 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 \ker(L) = \text{Span} \left ( \begin{bmatrix} -\frac{1}{\alpha} \\ 1 \end{bmatrix} \right )} .