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 !
On
If $P$ is s.t $PAP^{-1}=B$, then $P$ is invertible, which is an iff. with being injective and therefore bijective, as $P$ is linear. So $P$ cannot collapse a dimension as its kernel is of dimension 0. In fact, $P$ is just changing the basis of your vector space. In plain english, $P$ preserves all the data encoded in $A$ or $B$, and just changes the basis of your vector space that $A$ or $B$ act on.
It is then clear that $A$ and $B$ encode the same amount of data, as they are the same mapping with respect to some other basis.
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 $A = PBP^{-1}$.