# Proof that a Hermitian Matrix is not defective?

I am taking an introductory course into Quantum Mechanics.

To me to seems pretty simple to prove most properties of Hermitian operators. However, I am stuck at an edge case, proving that if an eigenvalue has multiplicity $n>1$, it will have $n$ linearly independent eigenvectors. This is equivalent to proving that a Hermitian matrix cannot be defective. Can anyone give me an outline or some pointers for such a proof?

• This is pretty much the content of the spectral theorem: every Hermitian matrix is diagonalizable by a unitary matrix. Feb 1, 2016 at 20:35

## 1 Answer

Here is a highlight of the reasoning line:

1. A theorem: The whole vector space is the direct sum of the generalized eigenspaces, where each generalized eigenspace is associated with an eigenvalue. The algebraic multiplicity $$\mu$$ is equal to the dimension of the generalized eigenspace associated. (See https://math.stackexchange.com/questions/2917617/proving-there-are-as-many-generalized-eigenvectors-as-algebraic-multiplicity-eig and http://www.math.byu.edu/~grant/courses/m634/f99/lec9.pdf)

2. A generalized eigenspace associated with an eigenvalue $$\lambda$$ of a matrix $$A$$ can be defined as $$G_{\lambda}\hat{=}\{v|\exists k\geq 0, \; \text{s.t.}\; (\lambda I-A)^k v =0\}$$

3. The eigenspace associated with $$\lambda$$ is a subspace of the generalized eigenspace. Thus the geometrical multiplicity $$\gamma$$ is not larger than the algebraic multiplicity, i.e. $$\gamma\leq\mu$$. The matrix $$A$$ being defective, i.e. $$\gamma<\mu$$, implies the existence of generalized eigenvectors $$u\in G_{\lambda}$$ such that $$(\lambda I-A)u\neq 0$$

4. If $$A=A^{\dagger}$$ is hermitian, there does not exist a such $$u$$ stated above. Assuming there is a $$u$$ such that $$(\lambda I-A)u\neq 0$$ and $$(\lambda I-A)^2u = 0$$, $$A$$ being hermitian implies $$(\lambda I-A)u = 0$$, thus self-contradiction.