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.
Answer
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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