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The statement is :- number of non zero eigen values of a matrix A is atmost rank(A).

Can someone help me to prove this?

I read some of the answers to the question similar to this here but I am not able to understand it fully.

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  • $\begingroup$ you have read some answers that you are not able to understand, would you like to tell us what are those to avoid people posting those answers? $\endgroup$ – Siong Thye Goh Jul 28 '17 at 4:20
  • $\begingroup$ I read one one question about number of distinct eigen values of a matrix with all one's and the answer was 1 because the rank of matrix is 1. I am not able to understand that. $\endgroup$ – Zephyr Jul 28 '17 at 10:08
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If a matrix $A$ is diagonalizable, then the algebraic multiplicity of any eigenvalue of $A$ is equal to its geometric multiplicity. In particular, if $A$ is symmetric (Hermitian) or all the eigenvalues of $A$ are distinct, then $A$ is diagonalizable. From rank-nullity theorem, it is known that $$\text{rank}(A)+\text{nullity}(A)=\text{dim} (A).$$ Nullity is the dimension of the kernel space of $A$. Meaning the number of linearly independent eigenvectors x for which $Ax=0\cdot x$. So nullity in this case implies the multiplicity of $0$ as an eigenvalue of $A$ and hence rank implies the number of nonzero eigenvalues of $A$.

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  • $\begingroup$ Here algebraic multiplicity of non zero eigen values + algebraic multiplicities of zero eigen values = n right? $\endgroup$ – Zephyr Jul 29 '17 at 4:28
  • $\begingroup$ Considered the cases where algebraic multiplicity=geometric multiplicity of each eigenvalue. $\endgroup$ – G_0_pi_i_e Jul 29 '17 at 4:47
  • $\begingroup$ @Zephyr Number of nonzero eigenvalues including their multiplicities+algebraic/geometric multiplicity of zero eigenvalue = n. $\endgroup$ – G_0_pi_i_e Jul 29 '17 at 4:50
  • $\begingroup$ Can I prove like this ? Let AM and GM denote algebraic and geometric complexities of eigen values. Now AM of non zero eigen values + AM of zero eigen values = n. As AM >= GM, for zero eigen values, AM >= n-r and hence AM of non zero eigen values <=r . $\endgroup$ – Zephyr Jul 29 '17 at 4:53
  • $\begingroup$ @Zephyr 'AM of nonzero eigenvalues'! instead use sum of algebraic multiplicities of all nonzero distinct eigenvalues. $\endgroup$ – G_0_pi_i_e Jul 29 '17 at 5:02

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