Is this operator diagonalizable？

[Jack3]

Let M be the space of all 2 × 2 complex matrices,
satisfying 〖(X)bar〗^t = -X (skew-hermitian).
Consider M as a vector space over R.
Define a bilinear form B on M by B(X,Y) = -tr(XY)

(1) Show that B takes real values, is symmetric and positive definite.

(2) For any A ∈ M , define the operator ad_A: M → M by ad_A(X) = AX – XA.
Show that operator ad_A is diagonalizable.

(3) Let the matrix
A =
( i 1)
(-1 i) .
Compute the eigenvalues of operator ad_A.(For part (2), Maybe we should show there is a basis of M consisting of eigenvectors of ad_A?)Thanks.

 

[Fernando Revilla]

Hint: If $X,Y\in M$ then,

$X=\begin{bmatrix}{x_1i}&{x_2+x_3i}\\{-x_2+x_3i}&{x_4i}\end{bmatrix},\;Y=\begin{bmatrix}{y_1i}&{y_2+y_3i}\\{-y_2+y_3i}&{y_4i}\end{bmatrix}\quad(x_i,y_j\in \mathbb{R})$.

Now, verify: $B(X,Y)=-\mbox{trace }(XY)=\ldots=x_1y_1+x_2y_2+x_3y_3+x_4y_4$