Relation between the rank of the matrix and its characteristic polynomial?

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.

• Any reference\notes to these kind of problems is great ? – BAYMAX Sep 28 '17 at 9:49

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.

• @ Lord Shark the Unknown Thank you for your remark. I was forgetting the condition of diagonalizibility. – Jean Marie Sep 28 '17 at 6:08
• 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? – BAYMAX Sep 28 '17 at 6:33
• See what I have added to my answer. – Jean Marie Sep 28 '17 at 6:58

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.

• For actually any other eigenvalue. The Jordan blocks form shows it need not be diagonizable (necessarily) – dEmigOd Sep 28 '17 at 6:33
• Jean Marie has included the condition of diagonalizability of the matrix? – BAYMAX Sep 28 '17 at 6:34