The task is: Determine eigenvalues and corresponding eigenvectors for the following matrix:
$A = \begin{pmatrix}
1 & -3 & 3 \newline
3 & -5 & 3 \newline
6 & -6 & 4
\end{pmatrix} \in \mathbb{R}^3_3$
I know how to calculate the eigenvalues and eigenvectors for a 2×2 matrix, but all the tutorials I found online use factorization to get the eigenvalues for a 3×3 that my brain cannot really transfer onto this problem.
When calculating $det(A – \lambda I)$ I am left with $-\lambda^3 + 66\lambda + 16 = 0$. How do I go from there?
Help would greatly be appreciated!
Best regards
Best Answer
As mentionned twice by Will Jagy in the comments, your 66 is wrong. The characteristic polynomial is $$P(\lambda):=\det(A-\lambda I)=-\lambda^3+12\lambda+16.$$ By the rational root theorem, one easily finds that $-2$ and $4$ are roots and finally, $$P(\lambda)=-(\lambda-4)(\lambda+2)^2.$$ An eigenvector for $4$ is e.g. $$u:=\begin{pmatrix}1\\1\\2\end{pmatrix}.$$ At this stage, we still don't know if $A$ is diagonalizable. The eigenspace for $-2$ could be of dimension only $1.$ Actually, the calculation shows it is of dimension $2:$ two independent eigenvectors for $-2$ are (for instance) $$v:=\begin{pmatrix}1\\0\\-1\end{pmatrix},\quad w:=\begin{pmatrix}1\\1\\0\end{pmatrix}.$$ So, $A$ is diagonalizable and $(u,v,w)$ is an eigenbasis for it (your question "basis for $A$" has been meanwhile corrected).