It is proven that $A$ and $A^T$ have the same eigenvalues. I want to study what stands for eigenvectors. Let me make a try. Given:

$$Ax=\lambda x$$ we know that $x\in C(A)$ for $\lambda \neq 0$. Suppose that for $A^T$ we have the same eigenvectors $x$:

$$A^Tx=\lambda x$$ but now we have that $x\in C(A^T)$. Based on this, eigenvector's $x$ belong both in column and row space which is impossible. So, $A$ and $A^T$ have different eigenvectors.

Am I right about this deduction? In any case, could you please suggest a different way if possible?


PS: After @G Tony Jacobs comments I made some changes hopping that I have less mistakes.

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    $\begingroup$ By $N(A)$ and $C(A)$, do you mean the null space and column space of $A$? Unless $\lambda=0$, an eigenvector isn't in the nullspace of $A$. It's in the nullspace of $A-\lambda I$. $\endgroup$ – G Tony Jacobs Aug 21 '16 at 16:25
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    $\begingroup$ Also, a diagonal matrix and its transpose are identical, so they have the same eigenvalues and eigenvectors. $\endgroup$ – G Tony Jacobs Aug 21 '16 at 16:26
  • $\begingroup$ Keep in mind that the zero vector cannot be an eigenvector. $\endgroup$ – user84413 Aug 21 '16 at 16:43
  • $\begingroup$ @G Tony Jacobs i have made some changes. Could you please place a comment for the new explanation? Thanks! $\endgroup$ – darkmoor Aug 21 '16 at 18:14
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    $\begingroup$ @darkmoor Let $A = \begin{bmatrix}1 & 4 & 0\\ 3 & 0 & 0\\ 0 & 0 & 7 \end{bmatrix}$. Then $x=\begin{bmatrix}0\\ 0\\ 1 \end{bmatrix}$ is an eigenvector for both $A$ and $A^T$ $\endgroup$ – David Peterson Aug 21 '16 at 18:34

Let $n\geq 2$ and $Z_n=\{A\in M_n(\mathbb{C}); A,A^T \text{have at least one common eigenvector}\}$.

Proposition. $M_n(\mathbb{C})\setminus Z_n$ is a Zariski open dense subset. That implies (for example) that if you randomly choose the $(a_{j,k})=(\alpha_j+i\beta_k)$ according to a normal law, then $A,A^T$ have no common eigenvector, with probability $1$.

Proof. According to Shemesh, cf. Remark 3.1 in https://ac.els-cdn.com/0024379584900855/1-s2.0-0024379584900855-main.pdf?_tid=36b4b2a4-1983-11e8-b76a-00000aacb35e&acdnat=1519491386_7b7c5a295dda427ce55e86ffc250bc15

$A\in Z_n$ IFF $(*)$ the matrix $[[A,A^T]^T,\cdots,[A^k,{A^l}^T]^T,\cdots,[A^{n-1},{A^{n-1}}^T]^T]$ has rank $<n$. Since $(*)$ can be written as a complex algebraic system of relations, $Z_n$ is a Zariski closed subset. It remains to show that, for every $n$, $Z_n$ is not $M_n(\mathbb{C})$.

Choose $a_{j,j}=j$ and if $j<k$ then $a_{j,k}=1$, the other $a_{i,j}$ being $0$.


For matrices with distinct eigenvalues, (same eigenvalues) + (same eigenvectors) = (same matrix).

Therefore any asymmetric $A$ with distinct eigenvalues is an example where $A$ and $A^T$ have different eigenvectors.

To write down such an example, take any upper triangular matrix with distinct entries on the diagonal.

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    $\begingroup$ Though one could imagine a case where two matrices have the same sets of eigenvalues and eigenvectors, but the eigenvectors are associated with different eigenvalues. Not sure if that's possible with $A$ and $A^T$ though. $\endgroup$ – Rahul Aug 21 '16 at 21:28
  • $\begingroup$ Right. I assume, though, that the OP was asking the simpler question of whether the eigenvectors of eigenvalue $x$ can be the same for all $x$. $\endgroup$ – zyx Aug 21 '16 at 22:25

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