How should I prove this bilinear form is symmetric and positive definite?

Question

I am trying to prove this problem:

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)

How should I show that B takes real values, is symmetric and positive definite?

Answer 1

$$X\in M\Longleftrightarrow X^t=-\overline X\Longrightarrow X=\begin{pmatrix}\,ia&x+iy\\-x+iy&ib\end{pmatrix}\;\;,\;\;a,b,x,y,\in\Bbb R$$

Thus, for example, taking

$$Y:=\begin{pmatrix}\,i\alpha&\xi+i\upsilon\\-\xi+i\upsilon&i\beta\end{pmatrix}\;\;,\;\;\alpha,\beta,\xi,\upsilon,\in\Bbb R$$

we get

$$B(X,Y)=-tr.(XY)=-tr\begin{pmatrix}\,-a\alpha+(x+iy)(-\xi+i\upsilon)&***\\***&-(x-iy)(\xi+i\upsilon)-b\beta\end{pmatrix}=$$

Can you take it from here?

$$=a\alpha+b\beta-2(x\xi+y\upsilon)\in\Bbb R$$