Similar matrices have the same row and column ranks

Let A and B be similar square matrices over a field F. Prove that A and B have the same row and column ranks.

Since A and B are similar, there exists a linear transformation \varphi and two bases E and F (of F^n, where n is the dimension of A and B) such that A = M^E_E(\varphi) and B = M^F_F(\varphi). In the previous exercise we saw that \mathsf{col\ rank}\ A = \mathsf{rank}\ \varphi = \mathsf{col\ rank}\ B.

Recall that the transpose of an n \times m matrix A = [a_{i,j}], denoted A^\mathsf{T}, is the m \times n matrix [a_{j,i}].

Lemma: If R is a commutative ring, A an n \times k matrix over R, and B a k \times m matrix over R, then (AB)^\mathsf{T} = B^\mathsf{T} A^\mathsf{T}. Proof: We have (AB)^\mathsf{T} = [\sum_{t=1}^k a_{i,t}b_{t,j}]^\mathsf{T} = [\sum_{t=1}^k a_{j,t}b_{t,i}] = B^\mathsf{T} A^\mathsf{T}. \square

Clearly \mathsf{col\ span}\ A = \mathsf{row\ span}\ A^\mathsf{T} and \mathsf{row\ span}\ A = \mathsf{col\ span}\ A^\mathsf{T}. Moreover, if AB = I, then I = I^\mathsf{T} = (AB)^\mathsf{T} = B^\mathsf{T} A^\mathsf{T}. In particular, if A is invertible, then so is A^\mathsf{T}, and (A^\mathsf{T})^{-1} = (A^{-1})^\mathsf{T}.

Now PAP^{-1} = B for some invertible matrix P. Thus (P^\mathsf{T})^{-1} A^\mathsf{T} P^\mathsf{T} = B^\mathsf{T}. By the above argument, A^\mathsf{T} and B^\mathsf{T} have the same column rank. Thus A and B have the same row rank.

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