# A curious observation regarding eigenvectors of $3 \times 3$ matrices - Hoffman and Kunze's *Linear Algebra*

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?

## 1 Answer

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 $$\lambda_1$$.

• 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. – Brahadeesh Jan 3 at 14:19
• 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. – W. mu Jan 3 at 14:23
• Oh! I see what you mean now. This is lovely :) – Brahadeesh Jan 3 at 14:25
• 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. – Brahadeesh Jan 3 at 16:24
• @Brahadeesh You are right. – W. mu Jan 4 at 1:34