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Quick question:

I was asked if there exists an invertible matrix $P$ over the complex numbers such that for any matrix $A$:

$PAP^{-1} = A^{T}$

I don't know how to prove it, but I don't think this is true. I know every matrix is similair to its transpose, but it can't be the same matrix $P$ for all matrices...So my gut feeling tells me no, but how do I show it?

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  • $\begingroup$ Don't you mean to ask if for any matrix $A$ there exists $P$ with the given property? $\endgroup$
    – Git Gud
    Commented Dec 8, 2013 at 21:06
  • $\begingroup$ No. I mean exactly the opposite. That there is a matrix $P$ that transposes ALL matrices. $\endgroup$
    – Oria Gruber
    Commented Dec 8, 2013 at 21:07
  • $\begingroup$ In fact there is, and the answer is here: projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – DonAntonio
    Commented Dec 8, 2013 at 21:07
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    $\begingroup$ @DonAntonio That paper is about the statement in my comment. $\endgroup$
    – Git Gud
    Commented Dec 8, 2013 at 21:09
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    $\begingroup$ Ok I think I have a solution, tell me what u guys think of it: Let's assume that there is such a matrix. so $PAP^{-1} = A^{T}$, or in other words $PA=A^{T}P$. Now lets look at $PABP^{-1}=B^{T}A^{T}$, we can write it as: $A^{T}PBP^{-1}=B^{T}A^{T}$, but $PBP^{-1}=B^{T}$ so we get: $A^{T}B^{T}=B^{T}A^{T}$ which is CLEARLY not true for all matrices...What do you guys think? $\endgroup$
    – Oria Gruber
    Commented Dec 8, 2013 at 21:19

4 Answers 4

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(promoted from a comment)

This solution was essentially given by the asker himself in a comment to the question.

First note that for $1\times 1$ matrices any invertible $P$ trivially works.

Then suppose such a magical $P$ also existed in general. Then for any matrices $A$ and $B$,

$$AB=P^{−1}A^TPP^{−1}B^TP=P^{−1}A^TB^TP=P^{−1}(BA)^TP=BA$$

where at first we used that $P$ works for both $A$ and $B$, and in the last equality we used that $P$ works for $BA$ too.

For 1×1 matrices this is no contradiction, but for matrices of greater size multiplication is not commutative.

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As said in the comments, this solution seems to work.

Assume there is such a magical $P$ so that for any matrix $A$: $PAP^{-1}=A^{T}$

we can also write it as: $PA=A^{T}P$ (simply multiply by $P$ in the right side).

now lets look at $PABP^{-1}=(AB)^{T}=B^{T}A^{T}$.

$PABP^{-1}=A^{T}PBP^{-1} = A^{T}B^{T}$

Since matrix multiplication is not commutative, $A^{T}B^{T}=B^{T}A^{T}$ does not hold for all matrices. contradiction.

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First, if you let $A=P,$ then you see that $P$ is symmetric. That means that it has an orthogonal basis of eigenvectors $e_1, \dots, e_n.$ Let $A$ be a matrix which sends $e_1$ to $e_2,$ $e_2$ to $e_3,$ etc ($e_n$ can go to $0.$) What happens to your matrices $P^{-1} A P$ and $A^t$ when applied to that basis?

EDIT As @julien points out, this argument works well if the ground field is the reals OR instead of transpose we use the hermitian adjoint (conjugate transpose). The argument does not work over $\mathbb{C}$ as stated.

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  • $\begingroup$ It is over complex numbers. So a symmetric matrix need not be diagonalizable. $\endgroup$
    – Julien
    Commented Dec 8, 2013 at 21:23
  • $\begingroup$ I wrote a solution I thought of in the comments. Would you be so kind as to check the validity of what I wrote? I am interested to know if the logic is valid. $\endgroup$
    – Oria Gruber
    Commented Dec 8, 2013 at 21:27
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    $\begingroup$ Sorry to insist but for instance $A=\pmatrix{2i&1\\1&0}$ is symmetric yet not diagonalizable. Your argument works nicely if this is over real matrices, or if the OP meant the transconjugate $A^*$ when s/he wrote $A^T$. $\endgroup$
    – Julien
    Commented Dec 8, 2013 at 21:35
  • $\begingroup$ @julien you are correct on both counts. I will add something to my answer. $\endgroup$
    – Igor Rivin
    Commented Dec 8, 2013 at 21:42
  • $\begingroup$ @OriaGruber This answer is marked as accepted but is not quite correct. Add julien points out, his matrix $A$ is symmetric (though not "Hermitean"), but it has eigenvalues $\lambda_1 = \lambda_2 = i$. So it is not diagonlizable since clearly it can't be similar to the matrix $\lambda_1 I = iI$. So this answer has a problem. I think you should mark your own answer as accepted (mine is really just a copy of yours). $\endgroup$
    – Jeppe Stig Nielsen
    Commented Dec 9, 2013 at 15:33
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You may also prove that, if $PSP^{-1}=S$ for all symmetric matrices $S$, then $P$ must be a scalar multiple of the identity matrix $I$ and hence $PAP^{-1}\equiv A$ for any (symmetric or not) matrix $A$. Since $A\ne A^T$ in general, the answer to your question is negative when $n\ge2$.

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