# number of non-zero eigen values of a matrix (counting multiplicities) is equal to rank of that matrix.

I am having problem in understanding the fact the

number of non-zero eigen values of a matrix (counting multiplicities) is equal to rank of that matrix.

Where I got stuck:

Let $k$ be the rank of the matrix $A_{n\times n}$ .Then the nullity of $A$ is $n-k$.So the eigen space corresponding to the eigen value $0$ has dimension $n-k$.Since algebraic multiplicity $\geq$ geometric multiplicity ,so the characteristic polynomial of $A$ will have the form $g(x)=x^pf(x);p\geq n-k$ and $\deg f(x)\leq k$.

So the roots of $f(x)$ will give non-zero eigen values of $A$.

But how it is that the number of non-zero eigen values of $A$(counting multiplicities)=rank of $A$?

• what can you say if $p>n-k$? – uniquesolution Oct 7 '15 at 11:28
• Do you know about Jordan normal form? If so, think about what the Jordan form of the matrix looks like, and the rank of the Jordan form. – Gerry Myerson Oct 7 '15 at 11:29
• If possible can you please give me some more hints how to complete this problem @GerryMyerson;I really cant proceed anymore – Learnmore Oct 7 '15 at 13:14
• You have to meet me halfway. I asked you a question: do you know about Jordan form? You didn't answer me. – Gerry Myerson Oct 7 '15 at 21:54
• yes ;@GerryMyerson; I read it now – Learnmore Oct 8 '15 at 2:07