Is the field characteristic necessary for this diagonalization question? I was looking into this question and I stumbled upon a similar problem, but with a slightly different hypothesis:

Let $A:M_{n\times n }\left(\mathbb{F}\right)\mapsto M_{n\times n}\left(\mathbb{F}\right)$ be the operator defined by $A(X) =X^T$, with the characteristic of the field $\mathbb{F}$ not equal to 2.
Find the eigenvalues and eigenspaces of the transformation and argue if $A$ is diagonalizable or not.

In the same question I mentioned earlier, there is this answer to the first part of the problem. As far as I could tell, this proof is legitimate without the hypothesis of the characteristic of a field $\neq2$, and I think the rest of the problem can be argued without the use of it as well since they follow from seeing that the eigenvalues are $1$ and $-1$.
Is there a detail that is being glossed over where we need to use this hypothesis? Or can the problem be generalized to cases without the extra condition? Thank you!
 A: The eigenvalues are still $\pm1$ when $\operatorname{char}(\mathbb F)=2$. Of course, since $-1=1$ in this case, all eigenvalues of $A$ are equal to $1$.
The characteristic matters when eigenspaces or diagonalisation are concerned. When $\operatorname{char}(\mathbb F)\ne2$, since the annihilating polynomial $x^2-1=(x-1)(x+1)$ is a product of distinct linear factors, $A$ must be diagonalisable.
However, when $\operatorname{char}(\mathbb F)=2$, the annihilating polynomial $x^2-1=(x-1)^2$ has repeated factors. Therefore we cannot infer that $A$ is diagonalisable. In fact, since all eigenvalues of $A$ are equal to $1$ but $A$ is not the identity map, it cannot possibly be diagonalisable.
Put it another way, when $\operatorname{char}(\mathbb F)\ne2$, the eigen "vectors" of $A$ corresponding to the eigenvalue $-1$ are are the skew-symmetric matrices in $M_{n\times n}(\mathbb F)$, while the eigen "vectors " corresponding to the eigenvalue $1$ are are the symmetric matrices. Since every square matrix is the sum of a skew-symmetric matrix and a symmetric matrix, you can construct an eigenbasis of $M_{n\times n}(\mathbb F)$ from the eigenvectors of $A$.
The situation is different when $\operatorname{char}(\mathbb F)=2$. In this case, since all eigenvalues are equal to $1$, the only eigen "vectors" of $A$ are the symmetric matrices. You cannot build an eigenbasis from them because there are matrices in $M_{n\times n}(\mathbb F)$ that are not symmetric (such as most upper triangular matrices).
