Finding the eigenvectors of a matrix. Find the normalized eigenvectors of this matrix:
$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$
My work:
$$\det(A-\lambda I) =(\lambda-1)^2(\lambda-4)$$
$$\lambda_1=1, \lambda_2=1,\lambda_3=4$$
$$\lambda_1\iff x + y +z=0$$
$$\lambda_2\iff x + y +z=0$$
$$\lambda_3\iff -2x + y +z=0$$
$$\lambda_3\iff x + -2y +z=0$$
$$\lambda_3\iff x + y +-2z=0$$
When my lecturer did this the equations were easy to solve... How do I find the normalized eigevectors from these equations?
Please Help :)
 A: For the matrix:
$$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$$
The CP is given by:
$$|A - \lambda I| = 0$$
For the CP, we get:
$$-\lambda^3+6 \lambda^2-9 \lambda+4 = -(\lambda-4) (\lambda-1)^2 = 0$$
This leads to three eigenvalues as $\lambda_1 = 4, \lambda_2 = 1, \lambda_3 = 1$.
We have a repeated eigenvalue.
To find an eigenvector, for each eigenvalue, we solve:
$$[A - \lambda_i I]v_i = 0$$
When we have repeated eigenvalues, we may need to resort to generalized eigenvectors, which I assumed you learned in class.
Lets find an eigenvector as an example for $\lambda_2 = 1$:
$$[A - \lambda_2 I]v_2 = 0$$
$$[A - \lambda_2 I]v_2 = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v_2 = 0$$
The row-reduced-echelon-form (RREF) is given by:
$$A=\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}v_2 = 0$$
So, we can choose an eigenvector to satisfy the RREF for:
$$x + y + z = 0 \rightarrow (-1, 0 , 1)$$
Since we have a repeated eigenvalue, we can choose a second, linearly independent, eigenvector as:
$$(-1, 1, 0)$$
Notice that both of these satisfy the RREF equation.
Repeat the above process for the other eigenvector so you can practice.
This leads to the eigenvalues and eigenvectors as:
$$\lambda_1 = 4, v_1 = (1, 1, 1)$$
$$\lambda_2 = 1, v_2 = (-1, 0, 1)$$
$$\lambda_3 = 1, v_3 = (-1, 1, 0)$$
However, we are being asked to find the normalized eigenvectors, so we have to do an additional step.
The way we normalize is to divide each eigenvector by it's length.


*

*$v_2 = (-1,0, 1)$

*$|v_2| = \sqrt{-1^2 + 0^2 + 1^2} = \sqrt{2}$

*Thus, the normalized eigenvector is:


$$\tilde v_2 = \frac{v_2}{|v_2|} = \frac{1}{\sqrt{2}}(-1,0, 1)$$
You should verify that the normalized eigenvector has length one.
Repeat this for the other two eigenvectors.
A: To find the eigenvectors of the matrix $A$ you find the null space of $(A-\lambda_1 I)$, $(A-\lambda_2 I)$ and $(A-\lambda_3 I)$. Since $\lambda_1=\lambda_2$, to find the eigenvectors corresponding to those eigenvalues it suffices to find the null space of $(A-\lambda_1 I)x=0$.
Thus
$$(A-\lambda_1 I)x=0$$
$$\longrightarrow\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\\\end{bmatrix}x=0$$
$$\longrightarrow x_1=\begin{bmatrix}-1\\0\\1\\\end{bmatrix} \text{ and } x_2=\begin{bmatrix}-1\\1\\0\\\end{bmatrix}$$
Similarly
$$(A-\lambda_3 I)x=0$$
$$\longrightarrow\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\\\end{bmatrix}x=0$$
$$\longrightarrow x_3=\begin{bmatrix}1\\-1\\1\\\end{bmatrix}$$
Finally, to answer the question you just normalize them to get
$$x_1=\begin{bmatrix}-1/\sqrt{2}\\0\\1/\sqrt{2}\end{bmatrix} \text{, } x_2=\begin{bmatrix}-1/\sqrt{2}\\1/\sqrt{2}\\0\end{bmatrix}\text{, } x_3=\begin{bmatrix}1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{bmatrix}$$
