4

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?

user1551
  • 139,064
  • 1
    I've edited the problem statement, because the original one is false (counterexample: $A=\pmatrix{1&1&0\ 1&1&0\ 0&0&0}$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$). – user1551 Aug 26 '18 at 09:17
  • @user1551; thanks but will u say what to assume so that the problem is true –  Aug 26 '18 at 09:18
  • @user1551;i guess here the problem is itself wrong –  Aug 26 '18 at 09:18

0 Answers0