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 be orthogonally diagonalizable over $\mathbb{K}$ 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
$$A=QDQ^{\top}\,.$$
Definition 2. A matrix $A\in\text{Mat}_{n\times n}(\mathbb{K})$ is said to be seminormal if
$$AA^\top=A^\top A\,.$$
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 $\text{char}(\mathbb{K})\neq 2$).
Question2. 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 $\mathbb{K}$?
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).
Question3. 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?
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.
Best Answer
Here are some worked examples that provide an answer to Question 2. The seminormal matrices in $\text{Mat}_{2\times 2}(\mathbb{K})$ are the symmetric matrices and matrices of the form $$T(a,b):=\begin{bmatrix}a&b\\-b&a \end{bmatrix}\,,$$ where $a$ and $b$ are elements of $\mathbb{K}$. For a symmetric matrix $$S(a,b,d):=\begin{bmatrix}a&b\\b&d\end{bmatrix}\,,$$ it is diagonalizable over $\mathbb{K}$ if and only if $a=d$ and $b=0$, or the quadratic polynomial $$x^2+(a+d)\,x+(ad-b^2)\in\mathbb{K}[x]$$ has two distinct roots in $\mathbb{K}$ (if $\text{char}(K)\neq 2$, the second condition is equivalent to stating that $$\Delta(a,b,d):=\sqrt{\left(\dfrac{a-d}{2}\right)^2+b^2}\in\bar{\mathbb{K}}$$ is a nonzero element of $\mathbb{K}$). It turns out that, if $S(a,b,d)$ is diagonalizable over $\mathbb{K}$, then
This provides a counterexample to Question 2. For example, when $\mathbb{K}$ is the field of rational numbers $\mathbb{Q}$, we can take $(a,b,d):=(6,4,0)$, so that $\Delta(6,4,0)=5$ and $\Xi(6,4,0)=2\sqrt{5}\notin\mathbb{Q}$. Therefore, $$S(6,4,0)=\begin{bmatrix}6&4\\4&0\end{bmatrix}$$ is not orthogonally diagonalizable over $\mathbb{Q}$. However, $S(6,4,0)$ is diagonalizable over $\mathbb{Q}$ because $\Delta(6,4,0)=5\in\mathbb{Q}_{\neq 0}$.
The smallest subfield $\mathbb{K}$ of $\mathbb{R}$ such that any matrix $S(a,b,d)$, with $a,b,d\in\mathbb{K}$, which is diagonalizable over $\mathbb{K}$, is always also orthogonally diagonalizable over $\mathbb{K}$ is the field of constructible real numbers. Over this field, $S(6,4,0)$ is no longer a counterexample. The same can be said for any field $\mathbb{K}$ that contains all of its square roots (that is, if $S(a,b,d)$ is diagonalizable over $\mathbb{K}$, then it is also orthogonally diagonalizable).
Now, we analyze $T(a,b)$. If $\text{char}(\mathbb{K})=2$, then $T(a,b)$ is diagonalizable over $\mathbb{K}$ if and only if $b=0$, in which case $T(a,b)$ is also orthogonally diagonalizable. If $\text{char}(\mathbb{K})\neq 2$, then $T(a,b)$ is diagonalizable over $\mathbb{K}$ if and only if $b=0$ or $\sqrt{-1}\in\mathbb{K}$; however, when $b\neq 0$, $T(a,b)$ is never orthogonally diagonalizable over $\mathbb{K}$, even when $\mathbb{K}$ contains $\sqrt{-1}$, because it is not symmetric. Unfortunately, the eigenspaces of both $T(a,b)$ and $\big(T(a,b)\big)^\top$ are identical: $$\mathbb{K}\,\begin{bmatrix}1\\+\sqrt{-1}\end{bmatrix}\text{ and }\mathbb{K}\,\begin{bmatrix}1\\-\sqrt{-1}\end{bmatrix}\,.$$
The counterexamples for $\text{Mat}_{2\times 2}(\mathbb{K})$ (for Question 2) above can be extended to counterexamples for $\text{Mat}_{n\times n}(\mathbb{K})$ whenever $n>2$. Both $S(a,b,d)$ and $T(a,b)$ so far, even when they are diagonalizable over $\mathbb{K}$ but not orthogonally diagonalizable over $\mathbb{K}$, do not provide a counterexample for Question 3.