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!
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).