Characterization of Linear Operators by Spectrum

Question

For a (finite dimensional) complex vector space $V$, the Spectral Theorem tells us that for every normal operator $T$ there is an orthonormal basis of $V$ consisting of eigenvectors of $T.$ Similarly, for a real vector space $V$, the Spectral Theorem tells us that for every self adjoint operator $T$ there is an orthonormal basis of $V$ consisting of eigenvectors of $T.$

In other words;

Complex vector space: Normal operator $\implies$ existence of an orthonormal eigenvector basis.

Real vector space: Self adjoint operator $\implies$ existence of an orthonormal eigenvector basis.

What if we drop the restriction that the basis needs to be orthonormal?

Is it the case that for every operator $T$ there is a basis of of $V$ consisting of eigenvectors of $T$, though now this basis need not be orthonormal? Or is there a more general class of operators for which this holds, but it does not necessarily hold for every operator?

In other words;

Complex vector space: Property A $\implies$ existence of an eigenvector basis.

Real vector space: Property B $\implies$ existence of an eigenvector basis.

What are properties A and B?

Answer 1

Look at $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $ and try to convince yourself that it is not diagonalizable.

You might also be interested in the Jordan normal form. https://en.wikipedia.org/wiki/Jordan_normal_form