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Is there some kind of relation between the rank of the matrix and its characteristic polynomial?After searching through various posts

Say if $A \in M_{5}(\Bbb{R})$ and its characteristic polynomial is $\alpha x^5 + \beta x^4 + \gamma x^3 =0 $,then the rank of matrix $A$ = ?

I am unable to estalish the relation ,like I know that from characteristic polynomial i can obtain the eigenvalues and hence the trace and determinant of the matrix and now the question is if i know the trace and determinat of the matrix can i obtain some information about the rank of the matrix(the number of linearly independent rows in the rref).

I was looking at this question but still i am not aware of any trick or relation.

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  • $\begingroup$ Any reference\notes to these kind of problems is great ? $\endgroup$ – BAYMAX Sep 28 '17 at 9:49
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If the matrix is diagonalizable, rank = degree of the characteristic polynomial minus the order of multiplicity of root 0 (in the example, the rank of the matrix is 5 - 3 = 2).

In fact, in this case, writing $M=PDP^{-1}$ with $D$ diagonal matrix with $n-r$ zeros, and transforming it into $MP=PD$, it means that if the columns of $P$ are denoted $P_k$, we have $MP_k=\lambda_k P_k$ with, say, the last $n-r$ vectors associated with eigenvalue $0$, (and only them) i.e. we have exactly $n-r$ independant vectors belonging to the kernel.

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  • $\begingroup$ @ Lord Shark the Unknown Thank you for your remark. I was forgetting the condition of diagonalizibility. $\endgroup$ – Jean Marie Sep 28 '17 at 6:08
  • $\begingroup$ Why is that formula so,any reference you want to give regarding rhis?What happens if the matrix is not diagonalizable? Then is there any relation? $\endgroup$ – BAYMAX Sep 28 '17 at 6:33
  • $\begingroup$ See what I have added to my answer. $\endgroup$ – Jean Marie Sep 28 '17 at 6:58
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Let $A$ be an $n$-by-$n$ matrix, and suppose that $0$ is a zero of the characteristic polynomial with multiplicity $m\ge1$. Then the rank of $A$ can be any number between $n-m$ and $n-1$ inclusive.

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  • $\begingroup$ For actually any other eigenvalue. The Jordan blocks form shows it need not be diagonizable (necessarily) $\endgroup$ – dEmigOd Sep 28 '17 at 6:33
  • $\begingroup$ Jean Marie has included the condition of diagonalizability of the matrix? $\endgroup$ – BAYMAX Sep 28 '17 at 6:34
  • $\begingroup$ @BAYMAX See also this answer math.stackexchange.com/questions/2340541/… for supporting Lord Shark the Unknown's answer $\endgroup$ – Widawensen Sep 28 '17 at 8:46

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