# Relation between eigenvalues and rank of matrix

Prove/Disprove: Let $$A$$ be a square matrix of order $$n$$ then rank of $$A$$ is atleast number of non zero eigenvalues of $$A.$$

My approach: For any matrix $$A$$, $$A^TA$$, and $$AA^T$$ are both symmetric and hence diagonalizable so the rank of $$A^TA$$ and $$AA^T$$ are the same and are equal to the number of non-zero eigenvalues.

Also, we know one result that for any real matrix $$A$$, $$\operatorname{rank}(A^TA)=\operatorname{rank}(AA^T)=\operatorname{rank}(A)=\operatorname{rank}(A^T).$$

Using the two above facts can we say that the statement is proven?

and to justify "atmost" take $$$$A= \begin{bmatrix} 1 & 0 & 0 &0\\ 0 & 1 & 0 &0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 &0 \end{bmatrix}$$$$ here rank of $$A$$ is $$3$$ which is greater than the number of non zero eigenvalues of $$A$$ that is $$2$$.

Is my approach okay or is there any nice explanation to justify the statement?

• This question may help: What is the relation between rank of a matrix, its eigenvalues and eigenvectors? math.stackexchange.com/q/1349907/1215020 Oct 13, 2023 at 7:39
• @CésarVB Thanks, that is a very interesting approach, but I need to know if my logic is right or if it has some problem if had written in some exam. Oct 13, 2023 at 8:08

You claimed the following

1. the rank of $$A^TA$$ and $$AA^T$$ are equal to the number of non-zero eigenvalues.
1. The rank of $$A$$ is the same of the rank of $$AA^T$$.

From the above we deduce that $$rank(A)$$ is the same as the number of non-zero eigenvalues. Which your chosen example proves is False.

It is in fact the first statement which is wrong.

I guess the classical approach to prove this statement is to use the rank nullity theorem and point out that the kernel is an eigenspace for the eigenvalue $$0$$

• @Maths I don't understand what you are saying. Could you rephrase that and maybe use punctuation to clarify what you mean ? Oct 13, 2023 at 10:28