Inner-product on skew-hermitian matrices

Let $$\mathfrak{u}(n)=\{X\in M(n,\Bbb C):X+X^*=0\}$$ where $X^*$ is the conjugate transpose. Then, $\mathfrak{u}(n)$ is a real vector space.

Problem. Show that $\langle X,Y\rangle=\DeclareMathOperator{\Tr}{Tr}\Tr(XY^*)$ is a real inner-product on $\mathfrak{u}(n)$ that is invariant under conjugation by $U(n)$.

Attempts.

The part about conjugation by $U(n)$ is easy: $$\Tr((UXU^{-1})(UYU^{-1})^*)=\Tr(UXU^{-1}UY^*U^*)=\Tr(UXY^*U^{-1})=\Tr(XY^*)$$ since the trace is invariant under conjugation.

But now, I don't know how to prove that $\Tr(XY^*)$ is an inner-product.

It is clear every matrix in $\mathfrak{u}(n)$ is normal and hence unitarily diagonalizable. So $X\sim \DeclareMathOperator{\diag}{diag}\diag(a_1,\ldots,a_n)$ and $Y\sim \diag(b_1,\ldots,b_n)$ where $a_i,b_i\in\Bbb R$. My guess is that $$\Tr(XY^*)=a_1b_1+\cdots+a_nb_n,$$ and hence it is a real inner-product. This is true if $X$ and $Y$ are diagonal. But I can't prove it in general.

• Wouldn't it just be a matter of checking the four properties of an inner product, of which three of them are almost trivial? – user305860 Jun 29 '16 at 13:25
• I am curious to understand why you would like to restrict this product to $u(n)$. This product is the most natural inner product on $M(n,C)$ whit associated quadratic form the sum of square of module of entries of the matrix. – Thomas Jun 29 '16 at 13:35
• Note that this inner product is simply the "dot-product" using the matrix entries. – Omnomnomnom Jun 29 '16 at 15:35
• @Thomas That's not a real inner-product, but a hermitian one. – user350977 Jun 29 '16 at 15:46
• @PatrickAbraham The entries are not necessarily real, but the set $\mathfrak{u}(n)$ has the structure of a real vector space: it is closed under addition and multiplication by real numbers. Note that it is not closed under multiplication by complex numbers and hence is not a complex vector space. – user350977 Jun 29 '16 at 16:13