**Update 1.** I still need help with Question 1, Question 2' (as well as the bonus question under Question 2'), and Question 3'.

**Update 2.** I believe that all questions have been answered if $\mathbb{K}$ is of characteristic not equal to $2$. The only thing remains to deal with is what happens when $\text{char}(\mathbb{K})=2$.

Let $\mathbb{K}$ be a field and $n$ a positive integer. The notation $\text{Mat}_{n\times n}(\mathbb{K})$ represents the set of all $n$-by-$n$ matrices with entries in $\mathbb{K}$. The subset $\text{GL}_n(\mathbb{K})$ of $\text{Mat}_{n\times n}(\mathbb{K})$ is composed by the invertible matrices. Here, $(\_)^\top$ is the usual transpose operator. Also, $\langle\_,\_\rangle$ is the standard nondegenerate bilinear form on $\mathbb{K}^n$.

Definition 1.A matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is said to beover $\mathbb{K}$ if there exist matrices $D\in\text{Mat}_{n\times n}(\mathbb{K})$ and $Q\in\text{GL}_{n}(\mathbb{K})$ where $D$ is diagonal and $Q$ is orthogonal (i.e., $Q^\top=Q^{-1}$) such that $$A=QDQ^{\top}\,.$$orthogonally diagonalizable

Definition 2.A matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is said to beif $$AA^\top=A^\top A\,.$$seminormal

For clarification, when $\mathbb{K}$ is $\mathbb{R}$, seminormal matrices are the same as normal matrices. However, when $\mathbb{K}$ is $\mathbb{C}$, the terms *seminormal* and *normal* are different. We have an obvious proposition.

Proposition.Let $A\in\text{Mat}_{n\times n}(\mathbb{K})$.(a) If $A$ is orthogonally diagonalizable over $\mathbb{K}$, then $A$ is symmetric.

(b) If $A$ is symmetric, then $A$ is seminormal.

The converse of (a) does not hold (but it does if $\mathbb{K}$ is $\mathbb{R}$). For example, when $\mathbb{K}$ is the field $\mathbb{C}$ or any field with $\sqrt{-1}$, we can take $$A:=\begin{bmatrix}1&\sqrt{-1}\\\sqrt{-1}&-1\end{bmatrix}\,.$$ Then, $A$ is symmetric, but being nilpotent, it is not diagonalizable. The converse of (b) does not hold trivially (nonzero antisymmetric matrices are seminormal, but not symmetric).

Here are my questions. Crossed-out questions already have answers.

Question 1.Is there a way to characterize all orthogonally diagonalizable matrices over an arbitrary field $\mathbb{K}$?

As in Proposition (a), these matrices must be symmetric, but the counterexample above shows that this is not a sufficient condition. Due to the answer by user277182, I believe that this is a correct statement.

Theorem.Suppose that $\text{char}(\mathbb{K})\neq 2$. A matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is orthogonally diagonalizable over $\mathbb{K}$ if and only if(a) $A$ is symmetric and diagonalizable over $\mathbb{K}$, and

(b) there exists a basis $\{v_1,v_2,\ldots,v_n\}$ of $\mathbb{K}^n$ consisting of eigenvectors of $A$ such that $\langle v_i,v_i\rangle$ is a nonzero perfect square element of $\mathbb{K}$ for each $i=1,2,\ldots,n$.

In the case where $\mathbb{K}$ contains all of its square roots (or when $\mathbb{K}$ is algebraically closed), the condition (b) in the theorem above is redundant. This theorem also answers Question 2' below (in the case $\text{char}(\mathbb{K})\neq 2$).

If a symmetric matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is already known to be diagonalizable over $\mathbb{K}$, is it also orthogonally diagonalizable over $\mathbb{K}$?~~Question~~2.

The answer of Question 2 turns out to be *no* (see a counterexample in my answer below). In light of this discovery, I propose a modified version of Question 2.

Question 2'.Let $\mathbb{K}$ be an algebraically closed field. If a symmetric matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is diagonalizable over $\mathbb{K}$, is it also orthogonally diagonalizable over $\mathbb{K}$?

**Bonus.** If $\mathbb{K}$ is not an algebraically closed field, what is a minimal requirement of $\mathbb{K}$ such that, if a symmetric matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is diagonalizable over $\mathbb{K}$, it is always also orthogonally diagonalizable over $\mathbb{K}$? This requirement may depend on $n$.

My guess for the bonus question is that, for every $x_1,x_2,\ldots,x_n\in\mathbb{K}$, $x_1^2+x_2^2+\ldots+x_n^2$ has a square root in $\mathbb{K}$. For example, a minimal subfield of $\mathbb{R}$ with this property is the field of constructible real numbers. Any field of characteristic $2$ automatically satisfies this condition.

**Edit.** According to this paper and that paper, when $\mathbb{K}=\mathbb{C}$, a symmetric matrix $A$ with an isotropic eigenvector $v$ (that is, $v^\top\,v=0$) is *nonsemisimple* (i.e., it is not diagonalizable). Therefore, at least, when $\mathbb{K}$ is a subfield of $\mathbb{C}$ such that, for every $x_1,x_2,\ldots,x_n\in\mathbb{K}$, $x_1^2+x_2^2+\ldots+x_n^2$ has a square root in $\mathbb{K}$, then a symmetric matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is orthogonally diagonalizable over $\mathbb{K}$ if and only if it is diagonalizable over $\mathbb{K}$. The result for other fields is currently unknown (to me).

As a generalization of this question, suppose that $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is diagonalizable over $\mathbb{K}$. Does it hold that $A$ and $A^\top$ have the same set of eigenspaces if and only if $A$ is seminormal?~~Question~~3.

~~Only the forward direction ($\Rightarrow$) of this biconditional statement is known to be true. It is clear, however, that when $A$ is orthogonally diagonalizable over $\mathbb{K}$, then $A$ is symmetric, whence $A$ and $A^\top$ have the same eigenspaces. As a result, the converse is true at least when $\mathbb{K}$ is a subfield of $\mathbb{R}$ because the seminormal (whence normal) matrices which is diagonalizable over $\mathbb{R}$ are the symmetric matrices.~~

The answer to Question 3 is *yes*. I forgot that diagonalizable matrices commute if and only if they can be simultaneously diagonalized. See my answer in the other thread for a more detailed proof. Therefore, I proposed a more generalized version of Question 3.

Question 3'.Let $A\in\text{Mat}_{n\times n}(\mathbb{K})$ be such that all roots of the characteristic polynomial of $A$ lie in $\mathbb{K}$. What is a necessary and sufficient condition for $A$ and $A^\top$ to have the same set of generalized eigenspaces?

Clearly, seminormality is not one such conditions. Over any field $\mathbb{K}$, the matrix $A:=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ has the same set of generalized eigenspaces as does $A^\top$. (The only eigenvalue of $A$ is $0$, and the generalized eigenspace associated to this eigenvalue is the whole $\mathbb{K}^2$. The same goes with $A^\top$.) However, $$AA^\top=\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq \begin{bmatrix}0&0\\0&1\end{bmatrix}=A^\top A\,.$$ In fact, any matrix $A\in\text{Mat}_{2\times 2}(\mathbb{K})$ which has an eigenvalue in $\mathbb{K}$ with multiplicity $2$ has $\mathbb{K}^2$ as its unique generalized eigenspace, and it follows immediately that $A$ and $A^\top$ have the same generalized eigenspace.