relation between rank of power of a singular matrix with the algebraic multiplicity of zero

Question

Could you please help me to show for an $n$ by $n$ singular matrix $A$, $n- \operatorname{rank} (A^n) =$ algebraic multiplicity of $0$ as an eigenvalue of $A$.

Answer 1

By rank-nullity theorem, you want to show that nullity of $A^n$ is equal to the algebraic multiplicity of $0$ as an eigenvalue of $A$. That is we want to show that the geometric multiplicity of $0$ as an eigenvalue of $A^n$ is equal to the algebraic multiplicity of $0$ as an eigenvalue of $A$.

Let the Jordan normal form of $A$ be

$$A=P^{-1}JP$$

where $$J=\operatorname{diag}(J_{m_1}(\lambda_1), \ldots,J_{m_k}(\lambda_k), J_{m_{k+1}}(0), \ldots, J_{m_{k+p}}(0)).$$

Here, I use the notation of $J_a(b)$ to represent Jordan block where $a$ is the size of the Jordan block with eigenvalue $b$ on the diagonal. $\lambda_1, \ldots, \lambda_k$ are non-zero.

We have the algebraic multiplicity of $0$ as an eigenvalue of $A= \sum_{j=1}^p m_{k+j}.$

We have $$A^n=P^{-1}J^nP$$

where $$J^n=\operatorname{diag}(J_{m_1}(\lambda_1)^n, \ldots,J_{m_k}(\lambda_k)^n, J_{m_{k+1}}(0)^n, \ldots, J_{m_{k+p}}(0)^n).$$

Since $J_{m_i}(\lambda_i)^n$ is nonsingular for $i \in \{ 1, \ldots, k\}$ and $J_{m_{k+j}}(0)^n=0$ for $j \in \{1, \ldots, p \}$.

The geometric multiplicty of $0$ as an eigenvalue of $A^n$ is indeed equal to $\sum_{j=1}^p m_{k+j}$ which is equal to the algebraic multiplicity of $0$ as an eigenvalue of $A$.