Why is it that if $A$ and $B$ are similar matrices then they must have the same rank?
I know that there exists a matrix $P$: $B=P^{-1}AP$ (the same applies for matrix $A$ with $B$ instead of $A$) where $P$ is a matrix of order $n$ and $rank(P)=n$ since $P$ is invertible.
I also know that $rank(P^{-1})=rank(P)$ does that mean that $rank(B) \leq rank(A)$? Is that enough evidence to conclude that the rank of both matrices is the same?
Thanks in Advance !
Best Answer
Yes, they must have the same rank. Rank can never increase as the result of matrix multiplication (i.e. $\operatorname{rank}(CD) \leq \min(\operatorname{rank}(C), \operatorname{rank}(D))$ for any matrices $C, D$ where the dimensions match up), so $$ B = P^{-1}AP \implies \operatorname{rank}(B)\geq \operatorname{rank}(A) $$ Now do the same resoning with $A = PBP^{-1}$.