Finding perpendicular eigenvectors (secondary school level)

Question

I'm working through a set of exercises on finding eigenvectors and eigenvalues of square matrices and have come to this question:

(i) Show that $\mathbf r = \left( \begin{matrix} -2 \\ 0 \\ 1 \end{matrix} \right) $ is an eigenvector of $\mathbf A = \left( \begin{matrix} 7 & 4 & -4 \\ 4 & 1 & 8 \\ -4 & 8 & 1 \end{matrix} \right)$ and determine the corresponding eigenvalue.

(ii) State two other eigenvectors of $\mathbf A$ that, together with $\mathbf r$, give three mutually perpendicular eigenvectors and state the corresponding eigenvalues.

(iii) What is the value of $\det \mathbf A$?

Part (i) is not a problem (the eigenvalue works out to be $9$), but I get the impression that the question is set up so that Part (i) naturally leads to solving Part (ii) in some efficient way.

What I've done is to work out the characteristic polynomial of $\mathbf A$ and solve it to find the other eigenvalues ($9$ again and $-9$), and I saved a little time by dividing through by $(\lambda - 9)$ in that process.

But I don't see how to use Part (i) to efficiently find orthogonal eigenvectors and I feel I'm probably missing a trick.

Answer 1

If $x$ is an eigenvector corresponding to $\lambda$ then we have,

$(A- \lambda I )x = O$

Now for $\lambda = -9$

$$\begin{pmatrix}16&4&-4\\4 & 10&8 \\-4 &8&10\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} $$

$$ (R_1\to R_1-4R_2, R_3\to R_3 +R_2)$$

$$\begin{pmatrix}0&-36&-36\\4 &10 &8 \\0 &18&18\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} $$

$$\implies b+c = 0 \text{ and } 2a+5b+4c=0$$

Now $r$ is orthogonal to $x$

$$\implies rx^T = 0 \implies -2a +c = 0 \implies c = 2a$$

Let $a = 1 \implies c = 2 \text{ and } b = -c = -2$

So,

$$x = \begin{pmatrix}1 \\-2\\2\end{pmatrix}$$

Also $A$ is symmetrical. So automatically the eigenvectors are orthogonal. Let $$y = \begin{pmatrix}u\\v\\w\end{pmatrix}$$ be the 3rd eigen vector.

$$ry^T = 0 \implies-2u+w = 0\implies w = 2u$$

$$xy^T = 0\implies u-2v+2w = 0 \implies 5u+2v = 0 \implies v = -5/2u$$

Let $$u = 2\implies v = 5\text{ and } w = 4$$

So,

$$ y = \begin{pmatrix}2\\5\\4\end{pmatrix}$$

And $\det(A) = 9\times9\times-9 = -729$