Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

There's this... – J. M. Dec 28 '11 at 3:26
This question has already been asked and answered here… . – 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


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

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.

share|cite|improve this answer
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

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

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.