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I am given a square matrix $A$, and I need to prove that if c is its eigenvalue, then it is also an eigenvalue of its transpose. How should I approach this? Clearly $Av$=$cv$, but I am not sure how to bring transpose into the equation.

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If $M$ is diagonalizable it seems easy enough since $M = Q^{-1}\Lambda Q$ implies $M^t = Q^t \Lambda^t (Q^{t})^{-1}$, so $M$ and $M^t$ share the same multiset of eigenvalues. Maybe some well chosen theorem or observation extend this for other matrices? – Myself Mar 4 '11 at 2:26
@Myself: What you want is Jordan canonical form. – Soarer Mar 4 '11 at 3:55
up vote 11 down vote accepted

A simple way would be to look at $\left | A^T-cI \right |$. $$\left | A^T-cI \right | = \left| (A-cI)^T \right| = \left| (A-cI) \right|$$


$\left| B \right|$ denotes the determinant of $B$. First note that for any matrix $B$, $\left| B \right| = \left| B^T \right|$.

(This is true since you get the same determinant if you find the determinant along the row or column.)

$c$ is an eigenvalue of $A$ iff $\left| A - cI\right| = 0$.

Note that $\left| A^T - cI \right| = \left| A^T - cI^T \right| = \left| (A - cI)^T \right| = \left| A - cI\right| = 0$.

Hence, $c$ is an eigenvalue of $A^T$.

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Wow, this makes perfect sense -- how did I not realize that... Thanks a bunch. – LinAlgStudent Mar 4 '11 at 2:24
@LinAlgStudent: Right. However you could look for a more direct proof. What I have done above is a bit round about. – user17762 Mar 4 '11 at 2:28
Thanks. Going your way, I assume that by || you mean determinant. I am actually not quite clear how to prove that the determinants of those two matrices will be equal to each other (and equal to zero). – LinAlgStudent Mar 4 '11 at 2:34

Another way to look at it is to see that it's equivalent to the following statement: $A$ is regular iff $A^T$ is regular. That follows from the fact that the row rank is equal to the column rank. But I'm not sure how easy it is to show that directly. Anyone?

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It is actually interesting to think why intuitively the left and right eigenvalues must be the same. – user17762 Mar 4 '11 at 2:44
because they are similar. – Soarer Mar 4 '11 at 3:52
@Soarer: How does that help with the intuition? – user17762 Mar 4 '11 at 3:58
Sorry for the imprecision, I mean that $A$ and $A^T$ are similar matrices. – Soarer Mar 4 '11 at 4:04
@Soarer: What is the intuition for them to be similar? – user17762 Mar 4 '11 at 4:36

Here is another way, that is a bit overboard: Suppose $A:V\rightarrow V$ is a linear operator. Then by definition, $A^t$ is the operator on the dual space $V^*$ given by $$\left(A^t(w)\right)(v)=w\left(A(v)\right)$$ where $w\in V^*$ and $v\in V$.

Suppose $A$ is invertible and $Ax=\lambda x$. Let $W=\text{span}(x)$ and then decompose $V$ into $V=W\oplus W^c$ where $W^c$ is also invariant under $A$. Let $w$ be any non-zero functional in the annihilator of $W^c$. Then lets look at $A^t(w)$. We have $$(A^t(w))(v)=w(A(v))$$ for every vector $v$, but it suffices to only consider $v\in W$ or $v\in W^c$. In the previous case, $w(A(v))=w(\lambda v)=\lambda w(v)$, and in the second case $w(A(v))=0$ since $W^c$ is invariant under $A$. Thus $$A^t(w)=\lambda w$$ so we see the transpose will have the same eigenvalues.

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Let $\lambda_1,v_1$ and $\lambda_2,v_2$ be eigenpairs of $A$ and $A^T$ respectively.

$$Av_1 = \lambda_1 v_1$$

$$A^Tv_2 = \lambda_2 v_2$$


$$v_2^TAv_1 = \lambda_2 v_2^T v_1$$

$$\lambda_1 v_2^Tv_1 = \lambda_2 v_2^T v_1$$


  1. $v_2^T v_1 = 0$ and $\lambda_1 \neq \lambda_2$.
  2. $\lambda_1 = \lambda_2$
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So what have you shown in the case where $v_2^T v_1=0$? – Myself Mar 4 '11 at 4:07

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