- #1

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^{-1}AP=A

^{T}. Does such a matrix exist? How do I find it?

What if I have two matrices A,B. Does there exist a matrix P, that transforms both of them to their transposes?

Thanks

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- Thread starter Leo321
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What if I have two matrices A,B. Does there exist a matrix P, that transforms both of them to their transposes?

Thanks

- #2

Fredrik

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^{T}.

No, I need X

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HallsofIvy

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But it's exactly the same thing. If X is the matrix with ones on the "anti-diagonal" (running from upper right to lower left) then [itex]X= X^{-1}[/itex].

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Fredrik

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I'm also looking for an X such that X^-1AX=A^T. Shouldn't there be a method for generically finding a similarity transformation that gives you a matrix B, whether it's the transpose of A or something else?

- #7

Fredrik

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I don't know any such method.

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A X - X A^T = 0 is just a linear equation for the coefficients of X.

Experimentation shows that in general, for a n*n system, only n (n-1) components of X are fixed. So there is some freedom in the choice of X.

Some experimentation also shows that, in general, it's not possible to find an X that transposes two matrices, ie such that

A X = X A^T and B X = X B^T.

I'm not sure what conditions A and B would have to satisfy such that a solution exists.

Note that if it does hold true for A and B, ie

A X = X A^T and B X = X B^T

then you must have a strange antihomorphism

A B X = A X X^{-1} B X = X A^T B^T = X (B A)^T

I'm not sure what that means. But it does extend to the entire group generated by A and B...

- #9

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[tex]C=\begin{bmatrix} b & Ab & \dots & A^{n-1}b \end{bmatrix}[/tex]

[tex]O=\begin{bmatrix} c & A'c & \dots & (A')^{n-1}c \end{bmatrix}[/tex]

[tex]P= O C ^{-1}[/tex]

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