# Transpose Operator is diagonalizable?

Let $T \colon \mathbb{M}_{n\times n}(\mathbb{R}) \to \mathbb{M}_{n\times n}(\mathbb{R})$ the linear operator such that $T(M)=M^t$, where $M^t$ is the transpose of the matrix $M$. Prove that $T$ is diagonalizable.

I thought I'd follow the path that $T^3=T$ but I have not resolved the problem.

You can do it with the elementary characterization for diagonalizable operators:

The operator $T$ is diagonalizable if and only if, for any eigenvalue, the algebraic multiplicity is equal to the geometric multiplicity.

Since $T^2$ is the identity, the operator can only have the eigenvalues $1$ and $-1$, which are indeed eigenvalues as soon as $n>1$, because symmetric matrices are the eigenspace for $1$ and antisymmetric matrices are the eigenspace for $-1$.

Since the subspace of $\mathbb{M}_{n\times n}(\mathbb{R})$ consisting of symmetric matrices has dimension $n(n+1)/2$, the algebraic multiplicity of $1$ is at least $n(n+1)/2$.

Similarly, the algebraic multiplicity of $-1$ is at least $n(n-1)/2$.

Since $$\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=n^2= \dim\mathbb{M}_{n\times n}(\mathbb{R})$$ we have that the algebraic and geometric multiplicities must agree, so the operator is diagonalizable and the characteristic polynomial is $$(1-X)^{n(n+1)/2}\,(-1-X)^{n(n-1)/2}$$

Note that this holds over any field, provided it has characteristic $\ne2$. In the case of characteristic $2$, the operator is not diagonalizable, because there's only the eigenvalue $1$, and the operator is not the identity (provided $n>1$).

Notice that $T^2=id$ then the polynomial with simple roots $x^2-1$ annihilates $T$ so it's diagonalizable.

A linear operator $T:V\to V$ is diagonalizable if and only if all the vectors can be written as linear combinations of eigenvectors.

If $M$ is any matrix, then $M+M^t$ is an eigenvector of $T$ belonging to eigenvalue $+1$ (or it is equal to zero), and $M-M^t$ is an eigenvector of $T$ belonging to eigenvalue $-1$ (or zero). Furthermore, $$M=\frac12(M+M^t)+\frac12(M-M^t)$$ shows $M$ as a sum of eigenvectors. The claim follows from this.

The proof works whenever $+1\neq-1$ and we are able to divide by $2$. In other words the proof works whenever the characteristic of the base field is $\neq2$.

$$T^3=T\iff T^3-T=0\iff T(T^2-id)=0\iff T(T-id)(T+id)=0$$ Then, $p(X)=X(X-1)(X+1)$ is the minimal polynomial, and $p(X)$ has only simple roots, then $T$ is diagonalizable.

In super operator formalism, transposition of matrix $A$ also has a matrix representation $\hat T$: $$vec(A^T)=\hat T vec(A)$$ Since transposition only flips positions of elements $\hat T$ is a permutation matrix, which is invertible. Even self-invertible because $(A^T)^T=A$. In other words $\hat T^2=1$, and therfore eigenvalues are $\pm1$. Matrices invariant under this permutation are the diagonal matrices, which form an eigenspace with eigenvalue $+1$...