Eventually a null column of the matrix $(A-\lambda I)^{\alpha}$

Question

I was doing some calculations with some matrices and I came across an interesting pattern, and I would like to know if what is happening is some fact or if it is coincidence with the matrices I am working on.

I will try to formalize my question in general, but I am working with $3\times 3$ and $4\times 4$ matrices, maybe the result is true in some specific dimensions, I don't know....

Suppose we have an $n\times n$ matrix $A$, I think we can first restrict our case to the real case, with real eigenvalues. Assuming the factored polynomial is given by,

$$p_c(t)=(t-\lambda_1)^{k_1}\cdot(t-\lambda_2)^{k_2}\cdots(t-\lambda_n)^{k_n}$$

Let some eigenvalue $\lambda_j$ with $k_j\ge 2$, and proceed with the calculation of $A-\lambda_j I$, where $I$ is the is the identity matrix.

What I noticed was that for some $\alpha\le k_j$, the matrix $(A-\lambda_j I)^{\alpha}$ has at least one null column.

So the question is this, if I have a eigenvalue with a multiplicity greater or equal than $2$, then there will be an exponent, less than the multiplicity, that the matrix $(A-\lambda_j I)^{\alpha}$ will have at least one null column.

I would like to know if there is any result that supports this observation that I made, or if it really was a coincidence given the matrices I am working on.

Answer 1

1. Let $A$ be an $n \times n$ matrix with eigenvalue $\lambda$ of multiplicity $n,$ then $A-\lambda I$ has eigenvalue zero with multiplicity $n$, prove that $A^{n} =0.$ (Use Cayley-Hamilton)
2. Let $A$ be a matrix, then the Jordan form is given by $J= PAP^{-1}$ for some matrix $P$. Now assume $\lambda$ is an eigenvalue of $A$ of multiplicity $r,$ then \begin{align} P (A-\lambda I)^rP^{-1} &= P(A-\lambda I )P^{-1} P (A-\lambda I )P^{-1} \cdots P(A-\lambda I)P^{-1}\\ &=(P(A-\lambda I )P^{-1})^r = (J- \lambda I)^r. \end{align} Note that one block of $(J-\lambda)$ has $0$ on the diagonal(this block has eigenvalues 0), and multiplication is done block-wise, so this block ends up being $0$ in $(J-\lambda I)^r$, and as a consequence the rows and columns of this block in $(J-\lambda I)^r$ are zero, and the other blocks are of nonzero determinant.
3. $ (A-\lambda I)^r = P^{-1}(J-\lambda I)^rP.$ Conclude