Section 1.8 homework

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 $K\circ L=0$?

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 $AB=0$ 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 $L\circ K(x,y)=L(0,x+y)=(0,0)$ but $K\circ L(x,y)=K(x,0)=(0,x+0)$ so $L\circ K=0$ but $K\circ L\neq 0$.

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

Solution First note that for any linear map $L(0)=0$ because $L(0)=L(0\cdot 0)=0\cdot L(0)=0$.

Proof: $(\Rightarrow )$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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=0} . Therefore $L(x)=0$ implies $x=0$. $(\Leftarrow )$ Now suppose that $L(x)=0$ implies $x=0$. If $L(x)=L(y)$ then by linearity of $L$ we have $L(x-y)=L(x)-L(y)=0$. But then by hypothesis that means $x-y=0$ which implies $x=y$. Therefore $L$ is one-to-one.

6. Let $V\neq \{0\}$ be finite-dimensional and assume that

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

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

Proof: I will use proof by contrapositive. The equivalent statement would then be "`If all of the Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 $L_{1},...,L_{n}$ are all one-to-one. Then $L_{1}\circ \cdots \circ L_{n}$ 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 $f,g$ are one-to-one functions. Consider the function $f\circ g$. 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f} is one-to-one that means the inputs to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f} must be the same or in other words $g(x)=g(y)$. But then $g$ is one-to-one so that means $x=y$ and therefore $f\circ g$ is one-to-one.

13. Consider the map $\Psi :\mathbb {C} \to {\text{Mat}}_{2\times 2}(\mathbb {R} )$ defined by $\Psi (\alpha +i\beta )={\begin{bmatrix}\alpha &-\beta \\\beta &\alpha \end{bmatrix}}$ ( a) Show that this is $\mathbb {R}$-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 $\mathbb {R} -$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.