Question

If A.x=
λx
,where A=
∣∣∣∣211232−212∣∣∣∣
,determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If
λ=1
,what is a

Answer 1

We have the matrix:

$$A = \left( \begin{array}{ccc} 2 & 1 & 1 \\ 2 & 3 & 2 \\ 2 & 1 & 2 \end{array} \right)$$

Let's find eigenvalues:

$$\begin{array}{l} \det (A - \lambda I) = \det \left( \begin{array}{ccc} 2 - \lambda & 1 & 1 \\ 2 & 3 - \lambda & 2 \\ 2 & 1 & 2 - \lambda \end{array} \right) \\ = (2 - \lambda) \left( (3 - \lambda) (2 - \lambda) - 2 \right) - (2 (2 - \lambda) - 4) + (2 - 2 (3 - \lambda)) \\ = -\lambda^3 + 7\lambda^2 - 10\lambda + 4 = -(\lambda - 1)(\lambda^2 - 6\lambda + 4) = 0 \end{array}$$

The roots of this polynomial are:

$$\lambda_1 = 1$$ $$\lambda_{2,3} = 3 \pm \sqrt{5}$$

Let's find eigenvectors corresponding to each of the eigenvalues.

Eigenvectors for given $\lambda$ are the fundamental solutions of the system

$$(A - \lambda I)x = 0$$

For $\lambda_1 = 1$ we have such matrix:

$$A - \lambda I = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 2 & 1 & 1 \end{array} \right) \sim \left( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & 1 \end{array} \right) \sim \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \end{array} \right)$$

Eigenvector corresponding to $\lambda_1$ is thus $v_1 = C_1(0, -1, 1)$, where $C_1$ is a real constant, $C_1 \neq 0$.

For $\lambda_2 = 3 - \sqrt{5}$:

$$A - \lambda I = \left( \begin{array}{ccc} \sqrt{5} - 1 & 1 & 1 \\ 2 & \sqrt{5} & 2 \\ 2 & 1 & \sqrt{5} - 1 \end{array} \right) \sim \left( \begin{array}{ccc} \sqrt{5} - 3 & 0 & 2 - \sqrt{5} \\ 2 - 2\sqrt{5} & 0 & -3 + \sqrt{5} \\ 2 & 1 & \sqrt{5} + 1 \end{array} \right) \sim \left( \begin{array}{ccc} \sqrt{5} - 3 & 0 & 2 - \sqrt{5} \\ 2 & 1 & \sqrt{5} + 1 \end{array} \right)$$

Corresponding eigenvector equals to

$v_2 = C_2(1 - \sqrt{5}, 2 - 2\sqrt{5}, 4)$, where $C_2$ is a real constant, $C_2 \neq 0$.

For $\lambda_3 = 3 + \sqrt{5}$:

$$A - \lambda I = \left( \begin{array}{ccc} -\sqrt{5} - 1 & 1 & 1 \\ 2 & -\sqrt{5} & 2 \\ 2 & 1 & -\sqrt{5} - 1 \end{array} \right) \sim \left( \begin{array}{ccc} -\sqrt{5} - 3 & 0 & 2 + \sqrt{5} \\ 2 + 2\sqrt{5} & 0 & -3 - \sqrt{5} \\ 2 & 1 & \sqrt{5} + 1 \end{array} \right) \sim \left( \begin{array}{ccc} -\sqrt{5} - 3 & 0 & 2 + \sqrt{5} \\ 2 & 1 & \sqrt{5} + 1 \end{array} \right)$$

Corresponding eigenvector equals to

$v_{3} = C_{3}\big(1 + \sqrt{5},2 + 2\sqrt{5},4\big),$ where $C_3$ is a real constant, $C_3\neq 0$.