The relationship between matrix rank and its characteristic polynomial coefficients

Question

Given the matrix characteristic polynomial coefficients. Is there a quick way to determine the rank of the matrix?

Answer 1

In general nothing can be said about rank of the matrix by merely looking at char polynomial. Take $$A=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$$ Rank$A=1$, though char polynomial is $x^2=0$. But if your matrix is diagonalizable, ''effective degree'' of the characteristic polynomial is equal to the matrix rank, since for a diagonalizable matrix, rank turns out to be number of non zero eigenvalues.

Answer 2

Let $p_A(X)=\sum c_i X^i$ be the characteristic polynomial of $A$. If rank$(A)<r$ then $c_r=0$: nul$(A)\geq n-r+1$, which is the geometric multiplicity of $0$, so the algebraic multiplicity of $0$ is greater or equal to $n-r+1$, hence $X^{n-r+1}$ divides $p_A$, so $c_n,\ldots,c_r=0$.

So rank$(A)\geq r$ if $c_r\neq 0$.

But the converse is not true: a nilpotent matrix has characteristic polynomial $X^n$ and may have rank up to $n-1$.

The closest satisfactory thing that can be said is the following: Let $A\in$Mat$_n(\mathbb{C})$, let $A^*$ be its transpose conjugate, and consider the characteristic polynomial of $AA^*$, $$p_{AA^*}(X)=\det(XI_n-AA^*)=X^n-a_1X^{n-1}+\ldots+a_n.$$

Then rank$(A)\leq r$ iff $a_{r+1}=0$ and consequently rank$(A)=r$ iff $a_{r+1}=0$ and $a_r\neq0$.

In particular, the rank of $A$ is $n-k$, where $X^k$ is the highest power of $X$ dividing $p_{AA^*}$.

This happens because $a_r$ is the sum of the squares of the absolute values of all the minors of $A$ of order $r$.

Answer 3

For a $n\times n$ matrix, if the characteristic polynomial is $X^mP(X)$ where $P$ is not divided by $X$, the rank is $n-m$.