Verify the results for F in the discussion of (11.34).

$\frac{1}{7}\left(\begin{array}{ccc}2& 6& 3\\ 6& -3& 2\\ 3& 2& -6\end{array}\right)$

If the transpose of a square matrix containing real numbers or components equals the inverse matrix, the matrix is said to be orthogonal.

$F=\frac{1}{7}\left(\begin{array}{ccc}2& 6& 3\\ 6& -3& 2\\ 3& 2& -6\end{array}\right)$

The determinant of F will be

$$\begin{gathered} detF=\frac{1}{7^3}\left(2\left(\left(-3\right)\left(-6\right)-2\times 2\right)-6\left(-6\right)-2\times 3\right)+3\left(6\times 2-\left(-3\right)\times 3\right)) \\ =1 \end{gathered}$$

The eigen Values of F

$\frac{1}{7}\left|\begin{array}{ccc}2-7\lambda & 6& 3\\ 6& -3-7\lambda & 2\\ 3& 2& -6-7\lambda \end{array}\right|=0$

That gives

$$\begin{gathered} =\left(2-7\lambda \right)\left(3+7\lambda \right)\left(6+7\lambda \right)+36+36+9\left(3+7\lambda \right)+36\left(6+7\lambda \right)-4\left(2-7\lambda \right) \\ =343+343\lambda -343{\lambda }^2-343{\lambda }^3 \\ ={\lambda }^3+{\lambda }^2-\lambda -1 \\ =\left(\lambda +1\right)\left({\lambda }^2-1\right)=0 \end{gathered}$$

This provides the eigenvalues $1,-1,-1.$

The eigenvector for$\lambda =1$can be obtained as

role="math" localid="1658824586021" $\frac{1}{7}\left(\begin{array}{ccc}2& 6& 3\\ 6& -3& 2\\ 3& 2& -6\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$

Thus,

$\frac{1}{7}\left(\begin{array}{ccc}-5& 6& 3\\ 6& -10& 2\\ 3& 2& -13\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=0$

When everything multiplied by 7 , we obtain the equations

$-5x+6y+3z=0$

And

$6x-10y+2z=0$

And

$3x+2y-13z=0$

When last equation to be multiplied by 2 and subtracted from the second, we have $y=2z$. When into the last equation, we have $x=3z$.

Therefore,

The eigenvector can be written as

$3i+2j+k$