Let $A$ be a real symmetric matrix of order $n$. Show that if $A$ has $m$ identical rows then $A$ has a zero eigenvalue of multiplicity at least $m-1$.
My try:
Since $A$ has $m$ identical rows by elementary row operations we have $m-1$ rows of $A$ equal to $0$.
Hence $A$ has geometric multiplicity of $0$ to be equal to atleast $m-1$. Since $A$ is symmetric so the algebraic multiplicity of $0$ equals its geometric multiplicity and so it is atleast $m-1$.
But how to show that it is equal to exactly $m-1$?
Will you please help?