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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?

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    $\begingroup$ This is pretty much the content of the spectral theorem: every Hermitian matrix is diagonalizable by a unitary matrix. $\endgroup$
    – Javier
    Feb 1, 2016 at 20:35

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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.

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