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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?

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$.