# a self adjoint in complex vector space

Let $$V$$ be a complex vector space, with Hermitian inner product $$\langle z,w\rangle$$. Let $$T : V → V$$ be a linear transformation. Show that $$T$$ is self adjoint if and only if $$\langle Tz,z\rangle$$ is real for every $$z ∈ V$$.

My solution is:

In the left side: $$T$$ is self-adjoint $$\Leftrightarrow$$ $$T=T^*$$ $$\Leftrightarrow$$ $$T=UAU^*$$ where $$UU^*=I$$ and A is a diagonal matrix.

In the right side:$$\langle Tz,z\rangle$$ is real for every $$z ∈ V$$ $$\Leftrightarrow$$ $$z'T'\overline{z}$$ is real $$\Leftrightarrow$$ $$T=UU^*$$. So there is some discrepency between the two sides.

Can you tell me which step is wrong?

$\Rightarrow)\quad$ We have for all $z$
$$\langle Tz,z\rangle=\langle z,T^*z\rangle=\langle z,Tz\rangle=\overline{\langle Tz,z\rangle}$$ so $\langle Tz,z\rangle$ is real.
$\Leftarrow)\quad$ Since $\langle Tz,z\rangle$ is real for all $z$ then we get $\langle Sz,z\rangle=0$ where $S=T-T^*$. Moreover, since $S$ is skew-hermitian then it's diagonalizable and from the equality $\langle Sz,z\rangle=0$ we see that their eigenvalues are $0$ so $S=0$. Finally we get $T=T^*$.