Adaptation of this proof of spectral theorem to the complex case

Question

My question is quite simple, I would like to know why we can't use this proof to the complex case, i.e., the operator $T$ is self adjoint on a complex n-dimensional inner product space $V$.

Can we adapt this proof to the complex case?

Thanks in advance

Answer 1

I can't see why not. The proof should be the same...the only words that must change is symmetric to self-adjoint, and real to complex in the statement of the theorem, but the proof should be the same using the standard inner product for complex vector spaces.

The matrix version you probably know is

Every Hermitian matrix is unitarily similar to a real diagonal matrix.

and it's easy to prove using Schur's theorem.

Now just an example of a matrix that is complex and symmetric, but not diagonalizable (just for kicks): \begin{equation} \begin{bmatrix}1&i\\i&-1 \end{bmatrix}. \end{equation} It has characteristic polynomial $x^2$, and so it's minimum polynomial must also be $x^2$. This means that this matrix is not diagonalizable, and has Jordan normal form \begin{equation} \begin{bmatrix}0&0\\1&0 \end{bmatrix}. \end{equation} So clearly the theorem is not directly applicable to operators over the complex field.