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If $A$ and $B$ are similar, can the same be said of $A^t$ and $B^t$?

$A$ similar to $B$ $\iff \exists P$ such that $A = PBP^{-1}$

Apply the transpose on both sides

$A^t = (PBP^{-1})^t = (P^{-1})^tB^tP^t = (P^t)^{-1}B^tP^t$

Can I conclude that they are similar?

Kevin Zakka
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  • It is as you say, if $A$ and $B$ are similar, so too will $A^t$ and $B^t$ be similar. Your proof is sound. – hardmath Dec 12 '15 at 14:09

2 Answers2

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Your reasoning looks good. In case you're not certain about $(P^t)^{-1} = (P^{-1})^t$:

$$I = P P^{-1} \iff I^t =(PP^{-1})^t = (P^{-1})^t P ^t.$$

Since $I = I^t,$ we have that $(P^{-1})^t$ is a left inverse for $P^t$. Use $I = P^{-1}P$ for the right inverse and you have shown that it's ok to exchange transposing and taking the inverse (for an invertible matrix).

Roland
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A different and more high-brow approach: Every matrix is similar to its transpose. Hence $A^t$ is similar to $A$ is similar to $B$ is similar to $B^t$. By transitivity of similarity, the claim follows.

Community
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MonadBoy
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    I see that my answer was accepted. While that is appreciated, I believe I should give the warning that the key step, "Every matrix is similar to its transpose" is nowhere near as easy to prove as intuition might let one believe. It is only after that (quite huge) step that the rest of my proof falls out. – MonadBoy Dec 12 '15 at 15:11
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    I added a link confirming that "high-brow approach". The premise is especially difficult to establish if the field of matrix entries is not algebraically closed. – hardmath Dec 12 '15 at 16:29