If a matrix and its transpose both have the same eigenvectors, is it necessarily symmetric? 
If a matrix and its transpose both have the same eigenvectors, is it necessarily symmetric?

It's clear to see how if $A$ = $A^T$, they would have the same eigenvectors, but is it the only way? And how would you show it?
 A: Suppose $A$ is a nonzero square matrix such that $A^T=-A$.

Then $A$ and $A^T$ have the same eigenvectors, but $A$ is not symmetric.

As an example, let 
$A=\pmatrix{
0&1\\
-1&0\\
}
$.
A: 
Proposition.  Let $A\in\text{Mat}_{n\times n}(\mathbb{R})$ be such that $A$ is diagonalizable over $\mathbb{R}$.  Then, $A$ and $A^{\top}$ have the same set of $\mathbb{R}$-eigenspaces if and only if $A$ is a symmetric matrix.

Proof.  One direction is trivial, so we prove the more difficult direction.  Let $v_1,v_2,\ldots,v_n\in\mathbb{R}^n$ be linearly independent eigenvectors of $A$, with the corresponding eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n\in\mathbb{R}$, respectively.  Then, there exists a permutation $\sigma$ on the set of indices $\{1,2,\ldots,n\}$ such that $v_1,v_2,\ldots,v_n$ are eigenvectors of $A^\top$ associated to the eigenvalues $\lambda_{\sigma(1)},\lambda_{\sigma(2)},\ldots,\lambda_{\sigma(n)}$, respectively.  Thus,
$$AA^\top\,v_k=A\,(A^\top\,v_k)=A\,(\lambda_{\sigma(k)}\,v_k)=\lambda_{\sigma(k)}\,(A\,v_k)=\lambda_{\sigma(k)}\,(\lambda_k\,v_k)\,.$$
Similarly,
$$A^\top A\,v_k=A^\top\,(A\,v_k)=A^\top\,(\lambda_kv_k)=\lambda_{k}\,(A^\top\,v_k)=\lambda_{k}\,(\lambda_{\sigma(k)}\,v_k)\,.$$
Therefore,
$$AA^\top\,v_k=\lambda_k\lambda_{\sigma(k)}\,v_k=A^\top A\,v_k$$
for $k=1,2,\ldots,n$.  Since $\mathbb{R}^n$ is spanned by $\{v_1,v_2,\ldots,v_n\}$, this proves that $AA^\top=A^\top A$, whence $A$ is normal.  Therefore, $A$ can be diagonalized using an orthogonal matrix.
Let now $A=Q\Lambda Q^\top$ be a diagonalization of $A$ via an orthogonal matrix $Q\in\text{Mat}_{n\times n}(\mathbb{R})$, where $\Lambda$ is the diagonal matrix
$\text{diag}_n\left(\lambda_1,\lambda_2,\ldots,\lambda_n\right)$.  This means $$A^\top =(Q\Lambda Q^\top)^\top=(Q^\top)^\top\Lambda^\top Q^\top =Q\Lambda Q^\top=A\,,$$
as $\Lambda^\top=\Lambda$.  Therefore, $A$ is symmetric.



Corollary 1. Let $A\in\text{Mat}_{n\times n}(\mathbb{R})$ be such that $A$ is diagonalizable over $\mathbb{C}$.  Then, $A$ and $A^{\top}$ have the same set of $\mathbb{C}$-eigenspaces if and only if $A$ is a normal matrix.



Corollary 2. Let $A\in\text{Mat}_{n\times n}(\mathbb{C})$ be such that $A$ is diagonalizable over $\mathbb{C}$.  Then, $A$ and $A^{\dagger}$ have the same set of $\mathbb{C}$-eigenspaces if and only if $A$ is a normal matrix.  Here, $(\_)^\dagger$ represents the Hermitian conjugate operator.


Remark.  Let $\mathbb{K}$ be a field and suppose that $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is diagonalizable over $\mathbb{K}$.  It is only known that, if $A$ and $A^\top$ have the same set of $\mathbb{K}$-eigenspaces, then $AA^\top=A^\top A$.  I do not think that the converse holds for all $\mathbb{K}$.  See also my question here.
Update.  I forgot that diagonalizable matrices commute if and only if they can be simultaneously diagonalized.  Since $A$ is diagonalizable over $\mathbb{K}$, $A^\top$ is also diagonalizable over $\mathbb{K}$.  Therefore, $A$ and $A^\top$ commute (i.e., $AA^\top=A^\top A$) if and only if $A$ and $A^\top$ can be simultaneously diagonalized, which is equivalent to the condition that $A$ and $A^\top$ have the same $\mathbb{K}$-eigenspaces.  Therefore, we have the following theorem.

Theorem.  Let $\mathbb{K}$ be a field and $n$ a positive integer.  Suppose that a matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is diagonalizable over $\mathbb{K}$.  Then, $A$ and $A^\top$ have the same $\mathbb{K}$-eigenspaces if and only if $$AA^\top=A^\top A\,.$$

A: If you mean that $A\in M_n(\mathbb R)$ and its transpose share a common eigenbasis $\mathcal B\subset\mathbb R^n$, then $A$ must be symmetric.
By definition, for any $v\in\mathcal B$, we have $Av=\lambda v$ and $A^Tv=\mu v$ for some $\lambda,\mu\in\mathbb R$. Since $\langle Av,v\rangle=\langle v,A^Tv\rangle$, we get $\lambda=\mu$. Therefore $A=A^T$ on $\mathcal B$ and in turn, $A=A^T$ on $\mathbb R^n$. Hence $A$ is symmetric.
