# Symmetric matrix with n rows has n linearly independent eigenvectors?

I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?

EDIT: The comments led me to a nice answer from another user which I think I understand.

Essentially the argument is:

Take any eigenvector of the symmetric matrix $$A$$, call it $$x$$. Then $$A$$ maps the set $$\{v : \langle v, x \rangle=0\}=:V$$ to itself since take $$v \in V$$ then $$(Av)^Tx=v^TA^Tx=v^TAx=\lambda v^T x=0$$ but this space V has 1 dimension less than $$V \cup \text{ span} (x)$$. Then because all endomorphisms which are linear functions have eigenvalues and $$A$$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $$V$$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.