Question

prove that equivalent matrices have the same rank​

Answer 1

Answer:

rank(AB)=rank((AB)t)=rank(BtAt)=rank(At)=rank(A), since Bt is nonsingular. Thus, similar matrices have the same rank. ... By definition, A∼B iff there exists an invertible matrix C such that A=CBC−1. If v is in the image of B, say v=Bw, then Cv=ACw is in the image of A.

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Answer 2

Answer:

rank(AB)=rank((AB)t)=rank(BtAt)=rank(At)=rank(A), since Bt is nonsingular. Thus, similar matrices have the same rank. ... By definition, A∼B iff there exists an invertible matrix C such that A=CBC−1. If v is in the image of B, say v=Bw, then Cv=ACw is in the image of A.