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Given the endomorphism $F: (M^{2 \times2 }) \rightarrow (M^{2 \times2 }), T \mapsto T^t$ between $K-$vector spaces of $2\times 2$ matrices, I need to find the eigenvalues, eigenvectors, the characteristic polynomial, the minimal polynomial and decide if the endomorphism is diagonalizable. We know that a matrix in $M^{2\times2}$ has the form A \begin{pmatrix} a & b\\ c & d \end{pmatrix} with the transpose \begin{pmatrix} a & c\\ b & d \end{pmatrix}

I assume the matrix I need to work on is the transposed one. Since the scalars $a,b,c,d$ are meant general, as far as I can see, one has to provide an expression for the characteristic polynomial, $p_F (t) = \det \begin{pmatrix} a-t & c\\ b & d-t \end{pmatrix}.$ Finding the roots of the characteristic polynomial and for each such eigenvalue finding the eigenspaces can be done in usual way. One then can discuss if $F$ is diagonalizable depending on the relationship between the geometric and algebraic multiplicities and considering different cases. Without writing all these details, I was evaluated my answer as wrong. Can somebody say or provide some suggestion what would be the correct answer ?

user996159
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    As long as $K$ is a PID, any matrix is similar to its transpose. https://math.stackexchange.com/questions/62497/matrix-is-conjugate-to-its-own-transpose – pancini Jul 28 '23 at 22:02
  • I'm afraid that you're working in the wrong space. You're expected to work with $F$ in the space of $2\times 2$ matrices. Since it is a $4$ dimensional space you'll need a $4\times 4$ matrix to represent $F$, the characteristic polynomial will be of degree $4$, it has $4$ eigenvalues (although one of them has multiplicity $3$), etc. – jjagmath Jul 28 '23 at 22:32
  • Thanks. Now I see the error. – user996159 Jul 28 '23 at 22:38
  • note that $F$ is an involution, so an immediate question should be whether or not $\text{char }\mathbb K=2$ – user8675309 Jul 29 '23 at 00:37

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