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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 \begin{equation}A= \begin{bmatrix} 1 & 0 & 0 &0\\ 0 & 1 & 0 &0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 &0 \end{bmatrix} \end{equation} 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?

Thanks in advance.

Maths
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    This question may help: What is the relation between rank of a matrix, its eigenvalues and eigenvectors? https://math.stackexchange.com/q/1349907/1215020 – César VB Oct 13 '23 at 07:39
  • @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. – Maths Oct 13 '23 at 08:08

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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 $0$

Digitallis
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  • @Maths I don't understand what you are saying. Could you rephrase that and maybe use punctuation to clarify what you mean ? – Digitallis Oct 13 '23 at 10:28