Complexification of a self-adjoint operator is self-adjoint on the complexified space

Question

In "Linear algebra done right" 9.b.4 :

Suppose $V$ is a real inner product space and $T \in \mathcal{L}(V)$ is self-adjoint. Show that $T_\mathbb{C}$ is a self-adjoint operator on the inner product space $V_\mathbb{C}$.

My way to do it is as follow:

Since $T$ is self-adjoint under real inner product vector space, by real spectral theorem, there exists an orthonormal basis $\mathcal{B}$ such that $\mathcal{M}(T,\mathcal{B},\mathcal{B})$ is diagonal matrix and all $\lambda_i \in \mathbb{R}$ are on the diagonal.

W.r.t the same basis $\mathcal{B}$, $\mathcal{M}(T_\mathbb{C},\mathcal{B},\mathcal{B})$ is the same as $\mathcal{M} (T,\mathcal{B},\mathcal{B})$, and because $\lambda_i$ are all real, we have $\mathcal{M}(T_\mathbb{C}^*,\mathcal{B},\mathcal{B})$=$(\overline{\mathcal{M}(T_\mathbb{C},\mathcal{B},\mathcal{B})})^t$=$\mathcal{M}(T_\mathbb{C},\mathcal{B},\mathcal{B})$, which implies $T_\mathbb{C}$ is self-adjoint.

Can someone tell me is there anything wrong in this proof?

Answer 1

Your proof is correct. However, your proof uses the real spectral theorem, which is a deep tool. There is a simpler proof that just uses the definitions. Specifically, you should verify from the definitions that $$ \langle{ T_{\mathbf{C}}(u + iv), x+iy}\rangle = \langle{u + iv, T_{\mathbf{C}}(x + iy)}\rangle, $$ for all $u, v, x, y \in V$, which implies that $T_{\mathbf{C}}$ is self-adjoint.