What are the Eigenvectors in the following matrix?

Question

I have the matrix A:

\begin{bmatrix} 4 & 2 & 2\\ 2 & 4 & 2\\ 2 & 2 & 4\\ \end{bmatrix}

I found $\lambda I_n - A$ to be:

\begin{bmatrix} (\lambda -4) & -2 & -2\\ -2 & (\lambda -4) & -2\\ -2 & -2 & (\lambda -4)\\ \end{bmatrix}

Applying Sarrus' Rule, and with help in a previous question, I determined the Eigenvalues to be $\lambda = 8$ or $2$ from the polynomial $\lambda^3 -12\lambda^2 +36\lambda -32 = 0$

If I now get the matrix

\begin{bmatrix} -2 & -2 & -2\\ -2 & -2 & -2\\ -2 & -2 & -2\\ \end{bmatrix} when verifying $\lambda = 2$, does this mean that $\lambda = 2$ is not a valid Eigenvalue? Or does it mean that $(\lambda I_n -A)v = 0$ Vector v = 1? And creates an Eigenspace from this?

Checking for $\lambda = 8$ gives me: \begin{bmatrix} 6 & -2 & -2\\ -2 & 6 & -2\\ -2 & -2 & 6\\ \end{bmatrix}

How do I get the Eigenvectors for this?

Answer 1

An easier way to check for eigenvalues is to factorize $$ x^3 - 12x^2 + 36x -32 = \left(x-8\right)\left(x-2\right)^2 $$ Then to get the eigenvectors, input the eigenvalues one by one to $$ (\lambda I - A) \cdot \begin{bmatrix}S_1\\S_2\\S_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} $$ and solve for $S_1$, $S_2$ and $S_3$

Answer 2

Bear in mind that your characteristic equation has double roots, so you can find infinitely many (coplanar) eigenvectors corresponding to $\lambda=2.$

You need to find any two non-zero and non-parallel vectors satisfying the single equation $-2x-2y-2z=0$