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 :)

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
$$x_1=\begin{bmatrix}-1/\sqrt{2}\\0\\1/\sqrt{2}\end{bmatrix} \text{, } x_2=\begin{bmatrix}-1/\sqrt{2}\\1/\sqrt{2}\\0\end{bmatrix}\text{, } x_3=\begin{bmatrix}1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{bmatrix}$$
