Determine an Eigenbasis given a characteristic polynomial with only two different roots

Question

Let $f: \mathbb{R}^3 \longrightarrow \mathbb{R}^3$ be an endomorphism given by $f(x)=Ax$, in which $$ A=\left( \begin{array}{rrr} {-6} & {2} & {-1} \\ {2} & {-3} & {-2} \\ {-1} & {-2} & {-6} \end{array} \right) $$ The characteristic polynomial is $$ \chi_{f}(t)=(-7-t)^{2}(-1-t) $$ Determine an Eigenbasis for $\mathbb{R}^3$ for $f: \mathbb{R}^3 \longrightarrow \mathbb{R}^3$.

How is this possible when there is only two different roots $\lambda_1 = -7$ and $\lambda_2=-1$ ? I could not find any useful theorem or corollary.

Answer 1

Did you try to find the eigenvectors? Here it's $-7$ that is the multiple root. If you solve the system $(A+7I)x=0$, you get the matrix $$ \begin{bmatrix} 1&2&-1\\ 2&4&-2\\ -1&-2&1\end{bmatrix} $$ Row reduction quickly gives you $$ \begin{bmatrix} 1&2&-1\\ 0&0&0\\ 0&0&0\end{bmatrix}. $$ So $ x_1+2x_2-x_3=0$. If you take $x_2,x_3$ as a parameters $s,t$, you get $x_1=-2s+t$. Thus the solutions are $$ \begin{bmatrix} -2s+t \\ s\\ t \end{bmatrix} =s\begin{bmatrix} -2 \\ 1\\ 0 \end{bmatrix} +t\begin{bmatrix} 1 \\ 0\\ 1 \end{bmatrix} $$ That gives you two linearly independent eigenvectors for $-7$.

Answer 2

If you try to determine the eigenvectors corresponding to the eigenvalue $-7$, you will get that they are the linear combinations of $(1,0,1)$ and $(-2,1,0)$. Since $(-1,-2,1)$ is an eigenvector corresponding to the eigenvalue $-1$,$$\bigl\{(1,0,1),(-2,1,0),(-1,-2,1)\bigr\}$$is an eigenbasis.