# 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 $L_{i}$ 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 $f$ is one-to-one that means the inputs to $f$ 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 ${\text{Mat}}_{2\times 2}(\mathbb {R} )$ 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.