TITLE: A curious observation regarding eigenvectors of $3 \times 3$ matrices - Hoffman and Kunze's *Linear Algebra*
QUESTION [5 upvotes]: I am reading Hoffman and Kunze's Linear Algebra, 2nd ed., and I made a curious observation in a couple of the examples relating to computing eigenvalues and eigenvectors in Chapter 6.
In Example 2 on pages 184-185, we have the (real) $3 \times 3$ matrix
$$
A = \begin{bmatrix}
3 & 1 & -1\\
2 & 2 & -1\\
2 & 2 & \phantom{-}0
\end{bmatrix}.
$$
The characteristic polynomial for $A$ is $(x-1)(x-2)^2$. Thus, the characteristic values of $A$ are $1$ and $2$. We have
$$
\begin{align}
A - I &= 
\begin{bmatrix}
2 & 1 & -1\\
2 & 1 & -1\\
2 & 2 & -1
\end{bmatrix}\\\\
A - 2I &= 
\begin{bmatrix}
1 & 1 & -1\\
2 & 0 & -1\\
2 & 2 & -2
\end{bmatrix}.
\end{align}
$$
The characteristic spaces associated to each characteristic value is one-dimensional in this case. The vector $\alpha_1 = (1,0,2)$ spans the null space of $T - I$ and the vector $\alpha_2 = (1,1,2)$ spans the null space of $T - 2I$.
Here, my observation is that $\alpha_1$ is the middle column vector of $A - 2I$, and $\alpha_2$ is the middle column vector of $A - I$.
A similar thing happens in Example 3 (pages 187-188):
$T$ is the linear operator on $\Bbb{R}^3$ which is represented in the standard ordered basis by the matrix
$$
A =
\begin{bmatrix}
\phantom{-}5 & -6 & -6 \\
-1 & \phantom{-}4 & \phantom{-}2 \\
\phantom{-}3 & -6 & -4
\end{bmatrix}.
$$
The characteristic polynomial is computed to be $(x-2)^2(x-1)$. Then, we have
$$
\begin{align}
A - I &=
\begin{bmatrix}
\phantom{-}4 & -6 & -6 \\
-1 & \phantom{-}3 & \phantom{-}2 \\
\phantom{-}3 & -6 & -5
\end{bmatrix}\\\\
A - 2I &=
\begin{bmatrix}
\phantom{-}3 & -6 & -6 \\
-1 & \phantom{-}2 & \phantom{-}2 \\
\phantom{-}3 & -6 & -6
\end{bmatrix}.
\end{align}
$$
The null space of $T-I$ is one-dimensional and the null space of $T-2I$ is two-dimensional. The vector $\alpha_1 = (3,-1,3)$ spans the null space of $T-I$. The null space of $T-2I$ consists of the vectors $(x_1,x_2,x_3)$ with $x_1 = 2x_2 + 2x_3$, so the authors give an example of a basis of the null space of $T-2I$ as $$\begin{align}\alpha_2 &= (2,1,0)\\ \alpha_3 &= (2,0,1).\end{align}$$ However, we can also take $$\begin{align}\alpha_2 &= (-6,3,-6)\\ \alpha_3 &= (-6,2,-5)\end{align}$$ and we see again that $\alpha_1$ is the first column of $A - 2I$ and $\alpha_2,\alpha_3$ are the second and third columns of $A - I$.
I find this quite curious, more so since the authors don't mention this observation at all. Is there a simple explanation for why this is happening, and can this observation be used to quickly find eigenvectors of linear transformations?

REPLY [4 votes]: Let the minimal polynomial of $A$ be
$$
f(x) = \prod_{i=1}^n (x-\lambda_i),
$$
where $\lambda_1,\dots,\lambda_n$ are eigenvalues of $A$ (not necessarily distinct).
Then,
$$\prod_{i=1}^n(A-\lambda_i)=0.$$
So, we have 
$$(A-\lambda_1)(\prod_{i=2}^n(A-\lambda_i))=0,$$
that is, the columns of $\prod_{i=2}^n(A-\lambda_i)$ are eigenvectors of $
\lambda_1$.