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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?

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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$.

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  • $\begingroup$ But in the first example, it is only the middle column of $A - 2I$ that is an eigenvector of $A$ for the eigenvalue $1$, not the other columns. $\endgroup$ – Brahadeesh Jan 3 at 14:19
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    $\begingroup$ If you take $\lambda_1=1$ and $\lambda_2=\lambda_3=2$, then the result is $(A-1)((A-2)^2)=0$. The columns of matrix of $(A-2)^2$ do. $\endgroup$ – W. mu Jan 3 at 14:23
  • $\begingroup$ Oh! I see what you mean now. This is lovely :) $\endgroup$ – Brahadeesh Jan 3 at 14:25
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    $\begingroup$ I think one should use the minimal polynomial of $A$ in this argument, not the characteristic polynomial. For Example 3, by the given argument we will say that the columns of the matrix $(A-I)(A-2I)$ are eigenvectors of $A$ with eigenvalue $2$ because the characteristic polynomial of $A$ is $(x-1)(x-2)^2$. However, $(A-I)(A-2I) = 0$, so this does not really say anything. If we instead argue with the minimal polynomial in place of the characteristic polynomial then the argument is correct. $\endgroup$ – Brahadeesh Jan 3 at 16:24
  • $\begingroup$ @Brahadeesh You are right. $\endgroup$ – W. mu Jan 4 at 1:34

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