Assume $T$ has an eigenvalue $k$.
Then if $T$ is hermitian, then eigenvalue is real.
I don't understand this statement.
If $T$ is hermitian, does that mean that inner product is complex number?
In my book it says in the complex case $T$ is called hermitian and skew hermitian.
So inner product is complex but how can eigenvalue be real?
Assume $T$ has some eigenvalue $k$ for some eigenvector $x$, then we know $$ Tx = k x$$ In terms of inner products, we call $T$ hermitian if $$ \langle Tx , y \rangle = \langle x , T^* y \rangle = \langle x , T y \rangle $$ Hence $T^* = T$. So what can we say about eigenvalues now? Well, we have $$ k \langle x,x \rangle = \langle k x,x \rangle =\langle T x ,x \rangle = \langle x , Tx \rangle = \langle x, k x \rangle = \bar k \langle x,x \rangle $$ This implies that (where $\bar k$ is the complex conjugate of $k$) $$ k = \bar k$$ which means that $k$ must be real.
Edit: If we're working in a nice vectorspace, we have $$ \langle x,y \rangle = x^T \overline{y} $$ So if everything is real, we have $\overline{y} = y$ and then $\langle x,y \rangle$ must be real. In the case of complex $x,y$, this need not be the case ( but the norm is induced from it, and is real valued)
In the context of the above, we have Hermitian $A=\overline A ^{T} = A^* $. Thus if $A$ is real, we have that $A = \overline A$ so we obtain $A = A^T$(i.e. symmetric ). If $A$ is skew-Hermitian, this means $A = -A^*$, i.e. in the case that $A$ is real, we have $A = -A^T$ (skew-symmetric)