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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?

Thanks.

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

glS
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darkmoor
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    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 $A-\lambda I$. – G Tony Jacobs Aug 21 '16 at 16:25
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    Also, a diagonal matrix and its transpose are identical, so they have the same eigenvalues and eigenvectors. – G Tony Jacobs Aug 21 '16 at 16:26
  • Keep in mind that the zero vector cannot be an eigenvector. – user84413 Aug 21 '16 at 16:43
  • @G Tony Jacobs i have made some changes. Could you please place a comment for the new explanation? Thanks! – darkmoor Aug 21 '16 at 18:14
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    Why do you think a vector cannot be in both the column space and the row space? General advice: when you modify a failed conjecture, you should re-check it against the counterexamples to the old one, in this case G Tony Jacobs' suggestion of a diagonal matrix. –  Aug 21 '16 at 18:21
  • @Rahul diagonal and symmetric matrices are special cases, what if $A$ is not in these cases? – darkmoor Aug 21 '16 at 18:25
  • If $A$ has full rank then both the column space and the row space are $\mathbb R^n$. –  Aug 21 '16 at 18:26
  • @Rahul your comment intuitive:) So I have to find another way. – darkmoor Aug 21 '16 at 18:31
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    @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 $A^T$ – David P Aug 21 '16 at 18:34
  • @David thanks for you comment. Also, based on a counter example logic we may found a matrix $A$ where all eigenvectors are different: $A = \begin{bmatrix}1 & 2\ 0 & 3 \end{bmatrix}$. So, based on the comments, until now we may say that $A$ and $A^T$ may have all or some or none or the eigen vectors common. The counter examples, suggest that in not special cases $A$ and $A^T$ have different eigenvectors but returning to my question how can we prove this analytically? – darkmoor Aug 21 '16 at 19:03
  • @darkmoor You have to define what the "special cases" are. It isn't at all clear what they are. – David P Aug 21 '16 at 19:56
  • @DavidP so far I think two of them have mentioned for $A$, is diagonal or symmetric, where both eigenvalues and eigenvectors are the same. – darkmoor Aug 21 '16 at 20:42
  • Does this answer your question? A matrix and its transpose have the same set of eigenvalues/other version: $A$ and $A^T$ have the same spectrum – PinkyWay May 20 '20 at 09:36

3 Answers3

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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:

$A\in Z_n$ 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 $0$.

PinkyWay
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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.

zyx
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    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. –  Aug 21 '16 at 21:28
  • 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 $x$. – zyx Aug 21 '16 at 22:25
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A matrix A is diagonalizable into a diagonal matrix D (D's diagonal components are eigenvalues). The transformation matrix from the original basis to the diagonal basis of A is P. In other words, P is the eigenvectors of A expressed in the original basis (or the columns of P are the eigenvectors).

This is equivalent to: There exists a diagonal matrix D and an invertible matrix P such that: $D=P^{-1}AP$

Tranposing the above eqt gives: $D=P^{T}A^TP^{-T}$

We conclude that the matrix $A^T$ is diagonalizable into the same diagonal matrix D, thus its eigenvalues are identical to those of A. Additionally, the transformation matrix from the original basis to the diagonal matrix of $A^T$ is $P^{-T}$. In other words, $P^{-T}$ is the eigenvectors of $A^T$ expressed in the the original basis. Generally, $P^{-T}\ne P$.

If P is orthogonal i.e eigenvectors of A are orthonormal, we have $P^{-T}= P$, which means that eigenvectors of A and $A^T$ are identical. This result is logic because if P is orthogonal, A is symmetric hence $A=A^T$.

Minh Hưng Nguyễn
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