Range of vectors that turn into eigenvectors after recursive multiplication by a matrix

Question

Suppose $\mathbf{x}$ is a vector, and $\mathbf{A}$ is a square matrix.

Which $\mathbf{x}$'s will satisfy the equation $\mathbf{A}^n\mathbf{x} = \lambda\mathbf{A}^{n-1}\mathbf{x}$, where $\lambda$ is an eigenvalue of the matrix and $n \in\mathbb{N^+}$? So I want to determine what range of vectors $\mathbf{x}$ will generate an eigenvector $\mathbf{y}$ for the matrix $\mathbf{A}$ where $\mathbf{y}=\mathbf{A^{n-1}x}$.

Also, is the value of $n$ needed for a specific vector predictable?

Answer 1

1. If $\lambda=0$ then the solutions for $A^nx=0$ are the vectors of the $\ker(A^n)$.
2. If $\lambda\neq 0$ then the solutions for $A^nx=\lambda A^{n-1}x$ are the vectors of $\ker(A-\lambda Id)\oplus\ker(A^{n-1})$.

Proof of 2.:

If $A^{n-1}x=0$ then $x\in \ker(A^{n-1})$ and $x$ is a solution.

If $A^{n-1}x=y\neq 0$ then $Ay=\lambda y$ and $x=\frac{1}{\lambda^{n-1}}y+x-\frac{1}{\lambda^{n-1}}y$.

Now $\frac{1}{\lambda^{n-1}}y\in \ker(A-\lambda Id)$ and $x-\frac{1}{\lambda^{n-1}}y\in \ker(A^{n-1})$.

Thus $\{x, A^nx=\lambda A^{n-1}x\}\subset \ker(A-\lambda Id)\oplus\ker(A^{n-1})$.

Next, $\ker(A-\lambda Id)\oplus\ker(A^{n-1})\subset \{x, A^nx=\lambda A^{n-1}x\}$ is obvious.

Therefore, we have $\ker(A-\lambda Id)\oplus\ker(A^{n-1})= \{x, A^nx=\lambda A^{n-1}x\}$.