If $A$ and $B$ are similar, why does $\text{rank}(A) = \text{rank}(B)$?

Question

Suppose $A$ and $B$ are similar matrices over $\mathbb{C}^n$. Why do we have $\text{rank}(A) = \text{rank}(B)$?

Answer 1

There is an invertible matrix $P$ such that $PA = BP$, so there is a well-defined map $\operatorname{Col}(A) \to \operatorname{Col}(B)$ sending $x\mapsto Px$. It is a linear transformation that is bijective (since $P$ is invertible). Therefore the columnspaces $\operatorname{Col}(A)$ and $\operatorname{Col}(B)$ have the same dimension, i.e., $\operatorname{rank}(A) = \operatorname{rank}(B)$.

Answer 2

I will show that by premultiplication and postmultiplication by invertible matrices, rank is unchanged.

Let $A=PBQ$ where $P,Q$ are invertible.

$Rank(A)=Rank(PBQ)\leq Rank(PB)\leq Rank(B)= Rank(P^{-1}AQ^{-1})\leq Rank(A)$ so all inequalities must be equalities.

So $Rank(A)=Rank(B)$.

In particular, if $A$ and $B$ are similar, then $Q=P^{-1}$ which is still invertible.