\(|A - \lambda I| = 0\)
⇒ \(\begin{vmatrix}2-\lambda&2&1\\1&3-\lambda&1\\0&2&2-\lambda\end{vmatrix}=0\)
⇒ \((2 - \lambda) (\lambda^2 - 5\lambda + 6-2) -1 (4 - 2\lambda -2) = 0\)
⇒ \(\lambda ^3 - 5\lambda^2 + 4\lambda - 2\lambda^2+10\lambda - 8 + 2 - 2\lambda = 0\)
⇒ \(\lambda ^3 - 7\lambda^2 + 12 \lambda - 6 = 0\)
⇒ \((\lambda - 1) (\lambda ^2 - 6\lambda + 6) = 0\)
⇒ \(\lambda = 1, \frac{6 \pm \sqrt{36 - 24}}2\)
⇒ \(\lambda = 1 , 3 \pm \sqrt 3\)
\(\therefore \) Eigen values are \(1, 3-\sqrt 3\) and \(3+\sqrt 3\).
\((A - \lambda _1I) \vec V = \vec 0\)
⇒ \(\begin{bmatrix}1&2&1\\1&2&1\\0&2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
⇒ \(2x_2 + x_3 = 0\)
⇒ \(x_2 = \frac{-k}2\)
Let \(x_3 = k\)
Also
\(x_1 +2x_2 + x_3 = 0\)
⇒ \(x_1 = 0\)
Hence, \((0, \frac{-1}2, 1) \) is eigen vector of matrix A w.r.t. eigen value 1.
\((A - \lambda _2I)V = 0\)
⇒ \(\begin{bmatrix}-1+\sqrt 3&2&1\\1&\sqrt 3&1\\0&2&-1 + \sqrt 3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
Let \(x_3 = k\)
\(2x_2 + (-1 + \sqrt 3)x_3 = 0\)
⇒ \(x_2 = \frac{(1 - \sqrt 3)k}2\)
Also
\(x_1 + \sqrt 3x_2 + x_3 = 0\)
⇒ \(x_1 + \frac{\sqrt 3 - 3}2 k + k = 0\)
⇒ \(x_1 = \left(\frac{3-\sqrt 3}2 - 1\right)k = \left(\frac{1-\sqrt 3}2\right)k\)
\(\therefore \) \(\left(\frac {1 - \sqrt 3}2, \frac{1-\sqrt 3}2, 1\right)\) is eigen vector of matrix A w.r.t eigen value \(3 -\sqrt 3\).
\((A - \lambda _3I)V = 0\)
⇒ \(\begin{bmatrix}-1-\sqrt 3&2&1\\1&-\sqrt 3&1\\0&2&-1 - \sqrt 3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
Let \(x_3 = k\)
\(2x_2 - (1 + \sqrt 3)x_3 = 0\)
⇒ \(x_2 = \frac{(1 + \sqrt 3)k}2\)
Also
\(x_1 - \sqrt 3x_2 + x_3 = 0\)
⇒ \(x_1 = \sqrt 3 x_2 - x_3 = \left(\frac{\sqrt 3 + 3}2 - 1\right)k\)
⇒ \(x_1 = \left(\frac{1 +\sqrt 3}2\right)k\)
\(\therefore \left(\frac {1 + \sqrt 3}2, \frac{1+\sqrt 3}2, 1\right)\) is eigen vector of matrix A w.r.t. eigen value \(3 +\sqrt 3\).
