What is the proof for the formula used in calculating generalized eigenvectors? 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.
 A: 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*}.
A: 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.
$$
            \left[\begin{array}{4}
                 \lambda_1 & 1 & 0 & 0 & \cdots & 0 \\
                    0 & \lambda_1 & 1 & 0 & \cdots & 0 \\
                    0 & 0 & \lambda_1 & 1 & \cdots & 0 \\
                    \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
                     0 & 0 & 0 & 0 & \cdots & \lambda_1
                  \end{array}\right]
$$
