Proof of a matrix and it's transpose being similar.

Question

Prove or disprove: If $A$ is an $n\times n$, then $A$ and $A^T$ are similar.

Definition of Two Similar Matrices: Let $A$ and $B$ be two $n\times n$ matrices. Then $A$ and $B$ are similar if there exists a nonsingular matrix $S$, such that $A=S^{-1}BS$.

I feel like there aren't similar. But I am not sure. Can I get some help?

Edit: So I have been googling an answer. And I see that most theorems state that this claim is true if the entries of $A$ are complex. Is it true if they were? i.e. Is $A$ and $A^T$ similar, when $A$ has real components?

Hint: similarity is an equivalence relation. You know every matrix is similar to a matrix in Jordan normal form. — Zhen Lin, Dec 08 '21 at 23:41

Vítězslav Štembera · Answer 1 · 2021-12-08T23:37:40.160

1

They are always similar, the proof is given in this article: ON THE SIMILARITY TRANSFORMATION BETWEEN A MATRIX AND ITS TRANSPOSE, OLGA TAUSSKY AND HANS ZASSENHAUS, Caltech, 1959.

edited Dec 08 '21 at 23:37

answered Dec 08 '21 at 23:28

Vítězslav Štembera

2,213

I don't see a proof for Thm 1 in that article. https://msp.org/pjm/1959/9-3/pjm-v9-n3-p25-s.pdf – Username Unknown Dec 08 '21 at 23:46
You are right, that is strange, I thought that the proof, which follows, proves both Theorem 1 and Theorem 2. However they state that it is so. – Vítězslav Štembera Dec 08 '21 at 23:56
Is the proof avaliable somewhere else? – Username Unknown Dec 09 '21 at 00:11
Look here: https://math.stackexchange.com/questions/62497/matrix-is-conjugate-to-its-own-transpose – Vítězslav Štembera Dec 09 '21 at 00:15

Proof of a matrix and it's transpose being similar.

1 Answers1