I have got in trouble with one of the problems I'm solving for my university studies. This is the Problem:
Let $A \in \Bbb R^{n \times n}$ and $U\in \Bbb R^{n \times n}$ such that $\ker U = \{0\}$. Define $B := UAU^{-1}$. Show that $f_A$ is injective if and only if $f_B$ is injective. What about surjectivity?
This is how far I have come:
$ f_U $ is injective and surjective and therefor bijective since $ \ker A = \{0\} \land \operatorname{ran} U = \Bbb R^{n} $. And I have got to show that $ \ker A = \{0\}$ iff $\ker B = \{0\}$. That means I have to show that $\ker A = \{o\} \Rightarrow \ker B = \{o\} \land \ker B = \{o\} \Rightarrow \ker A = \{o\}$.
Some help of you guys would come in handy.
More generally let $f:X\to X$ and $g:Y\to Y$ be functions, and let $h:Y\to X$ be an invertible function with $g=h\circ f\circ h^{-1}$.
Then $h$ and $h^{-1}$ are both bijective.
So if $f$ is injective then $g$ is a composition of injective functions and if $f$ is surjective then $g$ is a composition of surjective functions.
So we can apply the following lemma's:
1) If $u:A\to B$ and $v:B\to C$ are injective then so is $v\circ u:A\to C$
Proof: $v\circ u\left(a\right)=v\circ u\left(a'\right)\stackrel{v\text{ injective}}{\implies}u\left(a\right)=u\left(a'\right)\stackrel{u\text{ injective}}{\implies}a=a'$
2) If $u:A\to B$ and $v:B\to C$ are surjective then so is $v\circ u:A\to C$
Proof: Let $c\in C$. Then $c=v\left(b\right)$ for some $b\in B$ and $b=u\left(a\right)$ for some $a\in A$. So $c=u\circ v\left(a\right)$.
Also we have $f=h^{-1}\circ g\circ h$ so in both cases also the converse is true. That means that we end up with:$$f\text{ is injective }\iff g\text{ is injective }$$and:$$f\text{ is surjective }\iff g\text{ is surjective }$$