# A matrix is similar to its transpose [duplicate]

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

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

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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Doesn't this require that $K$ is algebraically closed? To get around this technicality, one needs to prove that $A$ and $B$ are similar over $K$ if and only if $A$ and $B$ are similar over $K'$ (some field extension of $K$). But essentially, all we need is that $A$ and $B$ are similar if and only if they have the same rational canonical form, and that relation will not change when we consider it over a field extension. – Nicholas Stull Mar 31 '14 at 19:51
Yes. One possible (unsatisfying) repair to extend to nonclosed fields is to notice that since they are similar in an extension field, their invariant factors match. Thus they share the same rational canonical form and thus are similar over the base field. – Bill Cook Mar 31 '14 at 22:50

note that reversing the basis order (conjugating by the matrix with ones from bottom left to upper right and zeros elsewhere) takes a jordan block to its transpose, eg

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

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