What is the proof for the formula used in calculating generalized eigenvectors?

Question

Definitionally, a generalized eigenvector for matrix $A$ is a vector $\textbf{x}$ such that

\begin{align} (A - \lambda I)\textbf{x} \neq \textbf{0} \\ (A - \lambda I)^m\textbf{x} = \textbf{0} \end{align}

where $\lambda$ is some eigenvalue of $A$, and $m$ is some integer greater than one.

However, when I search for resources on calculating $\textbf{x}$ in textbooks and online, I always find this:

\begin{align} (A - \lambda I)\textbf{x} = \textbf{v} \end{align}

where, I believe, $\textbf{v}$ is a (possibly generalized) eigenvector for $m-1$.

My question is: what is the proof for this equation? I can only assume that I'm missing some painfully obvious algebra trick that would show its validity, but I have been unable to figure it out.

Answer 1

Hint:

Observe that if $(A-\lambda1\!\!1)\mathbf v=\mathbf 0$ and $(A-\lambda1\!\!1)\mathbf x=\mathbf v$ then \begin{eqnarray*} (A-\lambda 1\!\!1)^2\mathbf x&=&(A-\lambda 1\!\!1)(A-\lambda1\!\!1)\mathbf x,\\ &=&(A-\lambda1\!\!1)\mathbf v,\\ &=&\mathbf 0. \end{eqnarray*}.

Answer 2

The Cayley-Hamilton Theorem is where these notions originated. For any $n\times n$ complex matrix $A$, the characteristic polynomial $p(\lambda)=(\lambda-\lambda_1)^{n_1}(\lambda-\lambda_2)^{n_2}\cdots(\lambda-\lambda_k)^{n_k}$ annihilates $A$. So, for example, if you look at all vectors $$ x\in\mathcal{R}((A-\lambda_2I)^{n_2}(\cdots)(A-\lambda_k)^{n_k}) $$ you know that $(A-\lambda_1)^{n_1}x=0$. If $m_1$ is the smallest positive integer such that $(A-\lambda_1)^{m_1}=0$ on this subspace, then you end up with a linearly independent set of vectors $$ \{ x,(A-\lambda_1I)x,\cdots,(A-\lambda_1I)^{m_1-1}\} $$ On the subspace spanned by these vectors, you can see that $A$ has the following matrix representation with $\lambda_1$'s on the main diagonal and $1$'s on the diagonal just above the main one. $$ \left[\begin{array}{4} \lambda_1 & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_1 & 1 & 0 & \cdots & 0 \\ 0 & 0 & \lambda_1 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \lambda_1 \end{array}\right] $$