# A matrix is similar to its transpose [duplicate]

Possible Duplicate:
Matrix is conjugate to its own transpose

How can I prove that a matrix is similar to its transpose?

My approach is: if $A$ is the matrix then $f$ is the associated application from $K^n\rightarrow K^n$. Define $g:K^n\rightarrow (K^n)^*$ by $g(e_i)=e_i^*$, and define $f^T$ to be the transpose application of $f$. I proved that $f^T=gfg^{-1}$. What I don't understand is, what is the matrix associated to $g$, so I can write $A^T=PAP^{-1}$.

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There's this... –  Ｊ. Ｍ. Dec 28 '11 at 3:26
This question has already been asked and answered here math.stackexchange.com/questions/62497/… . –  Mariano Suárez-Alvarez Dec 28 '11 at 4:58
could you help me to complete my argument? –  Alex M Dec 28 '11 at 5:03
@Ｊ.Ｍ.: The article you linked to shows that $A$ and $A^T$ are always conjugate by a nonsingular symmetric matrix, and uses the "well known" fact that $A$ and $A^T$ are always conjugate. An interesting precision, but it does not really answer this question. –  Marc van Leeuwen Dec 9 '13 at 8:27

## marked as duplicate by Hans Lundmark, t.b., Jonas Meyer, Henning Makholm, Zev ChonolesDec 31 '11 at 4:58

Consider...

$$B^{-1} = B = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad J = \begin{bmatrix} \lambda & 1 & & \\ & \ddots & \ddots & & \\ & & \lambda & 1 \\ & & & \lambda \end{bmatrix}$$

Then $B^{-1}J^TB = J$. Thus a Jordan block $J$ and its transpose $J^T$ are similar. So using $B_1,\dots B_\ell$ for each Jordan block $J_1,\dots,J_\ell$ and letting $$B = \begin{bmatrix} B_1 & & & \\ & B_2 & & \\ & & \ddots & \\ & & & B_\ell \end{bmatrix} \qquad \mathrm{and} \qquad J = \begin{bmatrix} J_1 & & & \\ & J_2 & & \\ & & \ddots & \\ & & & J_\ell \end{bmatrix}$$ Then $B^{-1}J^TB=J$. Therefore, a Jordan form and its transpose are similar.

Finally, put $A$ into its Jordan form: $P^{-1}AP=J$ then $J^T = (P^{-1}AP)^T=P^TA^T(P^T)^{-1}$ so thus $A$ is similar to $J$. $J$ is similar to $J^T$ and $J^T$ is similar to $A^T$. Hence by transitivity $A$ and $A^T$ are similar.

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$$\left( \begin{array}{ccc} 0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array} \right) \left( \begin{array}{ccc} a&1&0\\ 0&a&1\\ 0&0&a\\ \end{array} \right) \left( \begin{array}{ccc} 0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array} \right)= \left( \begin{array}{ccc} a&0&0\\ 1&a&0\\ 0&1&a\\ \end{array} \right)$$