91Ó°ÊÓ

First, we need to find the characteristic polynomial of matrix \( \mathbf{A} \). This requires calculating \( \det(\mathbf{A} - \lambda \mathbf{I}) \). So we have, \( \mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} 7 - \lambda & \sqrt{6} \ \sqrt{6} & 2 - \lambda \end{bmatrix} \).The determinant is calculated by:\[\det(\mathbf{A} - \lambda \mathbf{I}) = (7-\lambda)(2-\lambda) - (\sqrt{6})^2 = 14 - 9\lambda + \lambda^2 - 6 \]This simplifies to:\[\lambda^2 - 9\lambda + 8\]

The characteristic polynomial \( \lambda^2 - 9\lambda + 8 \) needs to be solved to find the eigenvalues. Using the quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -9 \), and \( c = 8 \).\[\lambda = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2}\]This gives us eigenvalues \( \lambda_1 = 8 \) and \( \lambda_2 = 1 \).

For each eigenvalue, find the eigenvector. Start with \( \lambda_1 = 8 \).Substitute \( \lambda_1 \) into \( (\mathbf{A} - \lambda_1 \mathbf{I})\mathbf{x} = \mathbf{0} \) results in:\[\begin{bmatrix} -1 & \sqrt{6} \ \sqrt{6} & -6 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Using the first equation: \( -x_1 + \sqrt{6}x_2 = 0 \), solving it gives \( x_1 = \sqrt{6}x_2 \). Choose \( x_2 = 1 \), then \( x_1 = \sqrt{6} \). The eigenvector associated with \( \lambda_1 = 8 \) is \( \begin{bmatrix} \sqrt{6} \ 1 \end{bmatrix} \).Now, consider \( \lambda_2 = 1 \).\[\begin{bmatrix} 6 & \sqrt{6} \ \sqrt{6} & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]The first equation gives \( 6x_1 + \sqrt{6}x_2 = 0 \), and solving gives \( 6x_1 = -\sqrt{6}x_2 \), so \( x_1 = -\frac{\sqrt{6}}{6}x_2 \). Choose \( x_2 = 6 \), then \( x_1 = -\sqrt{6} \). The eigenvector is \( \begin{bmatrix} -\sqrt{6} \ 6 \end{bmatrix} \).

It's common to express eigenvectors in unit form. Normalize each vector.For eigenvector \( \begin{bmatrix} \sqrt{6} \ 1 \end{bmatrix} \), the magnitude is \( \sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{7} \). The normalized vector is \( \begin{bmatrix} \frac{\sqrt{6}}{\sqrt{7}} \ \frac{1}{\sqrt{7}} \end{bmatrix} \).For eigenvector \( \begin{bmatrix} -\sqrt{6} \ 6 \end{bmatrix} \), the magnitude is \( \sqrt{6 + 36} = \sqrt{42} \). The normalized vector is \( \begin{bmatrix} -\frac{\sqrt{6}}{\sqrt{42}} \ \frac{6}{\sqrt{42}} \end{bmatrix} \).