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 $K\circ L=0$?

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 $x=0$.

Proof:

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 $x=0$. 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 $L_{i}$ 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 (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 L_i} are one-to-one, 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 L_1 \circ \cdots \circ L_n \ne 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 (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 L_1,...,L_n} are all one-to-one. 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 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 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 0} from problem 4. 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 L_1 \circ \cdots \circ L_n \ne 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 (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,g} are one-to-one functions. Consider the function 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} . Then to show this new function is one-to-one assume 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 f \circ g(x) = f \circ g (y)} . 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 f(g(x)) = f(g(y))} . But since 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} is one-to-one that means the inputs to 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} must be the same or in other words 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 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 ${\text{Mat}}_{2\times 2}(\mathbb {R} )$ that does not come from $\mathbb {C}$.

Proof:

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