Relationship between matrix's eigenvalue and rank

Question

Can we determine the rank of a matrix from its eigenvalues ? Let's say there is a matrix with eigenvalues (0,0,1,2,3), can we determine its rank ?

Answer 1

The rank of an $m\times n$ matrix is $n-\dim\ker A$, therefore for square matrices the rank is completely determined by the geometric multiplicity of the eigenvalue $0$. Tautologically, the geometric multiplicity of $\lambda$ in the matrix $A$ is just $m_g(\lambda)=\dim\ker (A-\lambda I)$. $\lambda$ is then an eigenvalue of $A$ if and only if $m_g(\lambda)\ge1$.

There is the issue of whether you are listing the eigenvalues by geometric multiplicity or by algebraic multiplicity. The algebraic multiplicity of $\lambda$ is $$m_a(\lambda)=\max\{k\in\Bbb N\,:\, (x-\lambda)^k\mid \det(A-xI)\},$$ id est, the exponent with which $(x-\lambda)$ appears in the factorization of the characteristic polynomial of $A$. It is true that $$m_a(\lambda)\ge1\Leftrightarrow m_g(\lambda)\ge1\Leftrightarrow \lambda\text{ is an eigenvalue of }A,$$ but in general $m_g(\lambda)\le m_a(\lambda)$, and therefore $n-m_a(0)\le \operatorname{rk}A$. By the previous remark that $m_g(0)\ge 1\Leftrightarrow m_a(0)\ge1$, this becomes $n-m_a(0)\le \operatorname{rk}A\le n-1$ as soon as $m_a(0)\ne0$.

For an instance of that inequality being effective, if in your example you are indicating the algebraic multiplicity of the eigenvalues, then that piece of information alone does not distinguish the cases $$\operatorname{rk}\begin{bmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&2&0\\ 0&0&0&0&3\end{bmatrix}=4\\ \operatorname{rk}\begin{bmatrix}0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&2&0\\ 0&0&0&0&3\end{bmatrix}=3$$

Answer 2

Yes. The rank is the number of eigenvalues that are unequal to zero.

But beware, this is only true for diagonalizable matrices. For arbitrary matrices, look at Saucy O'Paths answer.