Proves using the definition of matrix similarity

Question

Using the definition of matrix similarity if A and B are similar , prove that:

1) $A^t$ and $B^t$ are similar

2) $A$ has an inverse $<=>$ $B$ has an inverse

3) $A^k$ and $B^k$ área similar for all $k$ in $N$

Answer 1

Suppose that $A,B\in \mathbb{R}^{n\times n}$ are similar, i.e. there exists a square invertible matrix $P$ of the same size such that $B=P^{-1}AP$.

Then $B^t=(P^{-1}AP)^t=P^tA^t(P^{-1})^t=P^tA^t(P^t)^{-1}$. It follows that $A^t$ and $B^t$ are similar as well.

Now suppose that $A$ is invertible, i.e. $\det(A)\neq 0$. Then $\det(B)=\det(P^{-1}AP)=\det(A)$, hence $\det(B)\neq 0$ and $B$ is also invertible.

For the last one, notice that $B^k=(P^{-1}AP)^k=P^{-1}APP^{-1}AP\dots PP^{-1}AP=P^{-1}A^kP$.

You can do the second one without using determinants, try it!

Answer 2

From the comments:

Suppose we have that $A=PBP^{-1}$. Then $A^{T}=(P^{-1})^{T}B^{T}P^{T}=(P^{T})^{-1}B^TP^T$.

The last equality follows from the fact that interchanging rows and columns "commutes" with taking inverses.

More abstractly, the transpose can be thought of as the induced linear map by $f:V \to W$, so it is $f^*:W^* \to V^*$, and so if $f$ has an inverse, then $id^*=(ff^{-1})^*=(f^{-1})^*f^*$, meaning that $(f^*)^{-1}=(f^{-1})^*$.

For part $2$:

$AA^{-1}=PBP^{-1}(PBP^{-1})^{-1}$ and use the fact that $(AB)^{-1}=B^{-1}A^{-1}$.

for part 3:

note that $A^2=(PBP^{-1})(PBP^{-1})=PB^2P^{-1}$ and generalize.