Section 1.8

1. Let $L,K:V\to V$ be linear maps between finite-dimensional vector spaces that satisfy $L\circ K=0$. Is it true that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle K\circ L=0} ?

Solution:

No. in general composition of functions is not commutative. By the theorem that any linear map can be expressed as a matrix, finding a counterexample comes down to finding two matrices $A,B$ such that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle AB=0} but $BA\neq 0$. Here is one example of functions: $V=\mathbb {R} ^{2}$. $L(x,y)=(x,0),K(x,y)=(0,x+y)$. Then we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L\circ K(x,y)=L(0,x+y)=(0,0)} but Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle K\circ L(x,y)=K(x,0)=(0,x+0)} so $L\circ K=0$ but Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle K\circ L\neq 0} .

4. Show that a linear map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L:V\to W} is one-to-one if and only if $L(x)=0$ implies $x=0$.

Proof:

First note that for any linear map $L(0)=0$ because Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(0)=L(0\cdot 0)=0\cdot L(0)=0} . Proof: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\Rightarrow )} Suppose that $L$ is one-to-one. Then if $L(x)=0$ we have $L(x)=L(0)$ by the note above so that we must have $x=0$. Therefore $L(x)=0$ implies $x=0$. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (\Leftarrow )} Now suppose that $L(x)=0$ implies $x=0$. If $L(x)=L(y)$ then by linearity of $L$ we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x-y)=L(x)-L(y)=0} . But then by hypothesis that means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x-y=0} which implies Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=y} . Therefore $L$ is one-to-one.

6. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V\neq \{0\}} be finite-dimensional and assume that

$L_{1},L_{2},...,L_{n}:V\to V$

are linear operators. Show that if Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L_{1}\circ L_{2}\circ \cdots \circ L_{n}=0} then at least one of the $L_{i}$ are not one-to-one.

Proof:

I will use proof by contrapositive. The equivalent statement would then be "`If all of the $L_{i}$ are one-to-one, then $L_{1}\circ \cdots \circ L_{n}\neq 0$. Then this becomes very easy if you know the fact from set theory that the composition of one-to-one functions is a one-to-one function. This gives the following. Suppose that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L_{1},...,L_{n}} are all one-to-one. Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L_{1}\circ \cdots \circ L_{n}} is also a one-to-one function and so the only input that will give an output of 0 is the input $0$ from problem 4. Therefore $L_{1}\circ \cdots \circ L_{n}\neq 0$ and we are done. If you don’t know the fact from set theory you can prove it as follows. Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f,g} are one-to-one functions. Consider the function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f\circ g} . Then to show this new function is one-to-one assume that $f\circ g(x)=f\circ g(y)$. Then $f(g(x))=f(g(y))$. But since $f$ is one-to-one that means the inputs to $f$ must be the same or in other words Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=g(y)} . But then $g$ is one-to-one so that means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=y} and therefore 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 f \circ g} is one-to-one.

13. Consider the map 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 \Psi: \mathbb{C} \to \text{Mat}_{2 \times 2} (\mathbb{R})} defined by 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 \Psi(\alpha + i \beta) = \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix}} ( a) Show that this 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 \mathbb{R}} -linear and one-to-one, but not onto. Find an example of a matrix in 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{Mat}_{2 \times 2}(\mathbb{R})} that does not come from 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 \mathbb{C}} .

Proof:

To show this 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 \mathbb{R}-} linear let 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 z_1 = \alpha_1+i \beta_1, z_2 = \alpha_2 + i \beta_2 \in \mathbb{C}} . Then: 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 \Psi(z_1+z_2) = \Psi(\alpha_1+i\beta_1 + \alpha_2+i\beta_2) = \Psi(\alpha_1 + \alpha_2 + i(\beta_1 + \beta_2)) = \begin{bmatrix} \alpha_1+\alpha_2 & -\beta_1-\beta_2 \\ \beta_1+\beta_2 & \alpha_1+\alpha_2 \end{bmatrix} = \begin{bmatrix} \alpha_1 & -\beta_1 \\ \beta_1 & \alpha_1 \end{bmatrix} +\begin{bmatrix} \alpha_2 & -\beta_2 \\ \beta_2 & \alpha_2 \end{bmatrix} = \Psi(z_1) + \Psi(z_2)} Similarly if 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 z = \alpha + i \beta \in \mathbb{C}} 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 a \in \mathbb{R}} then: 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 \Psi(a z) = \Psi (a \alpha + i a\beta) = \begin{bmatrix} a\alpha & -a\beta \\ a\beta & a\alpha \end{bmatrix} = a \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix} = a \Psi(z)} Therefore 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 \Psi} 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 \mathbb{R}} -linear. Now to show 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 \Psi} is not onto we notice that any matrix in the image of 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 \Psi} has top left and bottom right coordinate the same. So the simple matrix 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} 1 & 2 \\ 3 & 4 \end{bmatrix}} cannot possibly be in the image of 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 \Psi} . Therefore 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 \Psi} is not onto.