A matrix is similar to its transpose 
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 $A^T=PAP^{-1}$.
 A: 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
$$
\left(
\begin{array}{ccc}
0&0&1\\
0&1&0\\
1&0&0\\
\end{array}
\right)
\left(
\begin{array}{ccc}
a&1&0\\
0&a&1\\
0&0&a\\
\end{array}
\right)
\left(
\begin{array}{ccc}
0&0&1\\
0&1&0\\
1&0&0\\
\end{array}
\right)=
\left(
\begin{array}{ccc}
a&0&0\\
1&a&0\\
0&1&a\\
\end{array}
\right)
$$
A: 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.
Edit: What if our field is not algebraically closed? [From the comments.]
Of course, we only have a Jordan form when our characteristic polynomial splits over our field. Thus if we are working over a field such as $\mathbb{R}$ or $\mathbb{Q}$, my quick argument does not apply. Here is a fix:
Say we are working over a field $\mathbb{F}$. First, tack on all of the eigenvalues and call that field $\mathbb{K}$ (i.e., $\mathbb{K}$ is the splitting field for our characteristic polynomial).
By the above answer, using Jordan form, $A$ is similar to $A^T$ (working over $\mathbb{K}$). Therefore, $A$ and $A^T$ share the same rational canonical form. Since a matrix is similar to its rational canonical form over the field generated by its entries, $A$ and $A^T$ are similar to this common form working over $\mathbb{F}$ (or possibly even a smaller field). Therefore, $A$ and $A^T$ are similar over $\mathbb{F}$.
