linear algebra - find eigenvalues and eigenvectors

Question: A).Consider a matrix $A= \begin{pmatrix}0&1&1\\ -2&3&1\\-3&1&4 \end{pmatrix}$ Find all eigenvalues and the corresponding eigenvecors(and generalized eigenvectors) of A.

B). Find a vector $v$ such that $v$,$Av$, and $A^2v$ are linearly independent.

Proof: A) Ok so for this one I got my eigenvalues to be $\lambda= 2,3$ and my vectors to be $v_1= \begin{pmatrix} 1\\1\\2 \end{pmatrix}$ such that $\lambda=3$, $v_2= \begin{pmatrix} 1\\1\\1 \end{pmatrix}$ such that $\lambda=2$, and because we have a generalized eigenvector we get $v_3= \begin{pmatrix} 2\\2\\1 \end{pmatrix}$. Are these results right?

and for part B) I'm not quite sure how to show that. Where can I go to see how to show linear independence?

Answer

Your ${\bf v_3}$ is wrong. An easy way to see this is that the generalised eigenvectors should form a basis for $\Bbb R^3$, but for your vectors
$${\bf v}_1+{\bf v}_3=3{\bf v}_2\ ,$$
so they are not independent.

You can find ${\bf v}_3$ in various ways, for example by solving
$$(A-2I){\bf v}_3={\bf v}_2\ ,$$
which gives the useful relations
$$A{\bf v}_1=3{\bf v}_1\ ,\quad A{\bf v}_2=2{\bf v}_2\ ,\quad
A{\bf v}_3={\bf v}_2+2{\bf v}_3\ .\tag{$*$}$$

These also give a (relatively) simple way to do the second question. Since $\{{\bf v}_1,{\bf v}_2,{\bf v}_3\}$ is a basis for $\Bbb R^3$, we can take
$${\bf v}=\alpha{\bf v}_1+\beta{\bf v}_2+\gamma{\bf v}_3\ .\tag{${*}{*}$}$$
Using $(*)$, we have
$$\eqalign{
A{\bf v}&=3\alpha{\bf v}_1+2\beta{\bf v}_2+\gamma({\bf v}_2+2{\bf v}_3)\cr
A^2{\bf v}&=9\alpha{\bf v}_1+4\beta{\bf v}_2+\gamma(4{\bf v}_2+4{\bf v}_3)\ .
\cr}$$
For these vectors to be independent we need the determinant
$$D=\det\pmatrix{\alpha&3\alpha&9\alpha\cr \beta&2\beta+\gamma&4\beta+4\gamma\cr

\gamma&2\gamma&4\gamma\cr}$$
to be non-zero. But
$$\eqalign{D
&=\alpha\gamma
\det\pmatrix{1&3&9\cr \beta&2\beta+\gamma&4\beta+4\gamma\cr 1&2&4\cr}\cr
&=\alpha\gamma\det\pmatrix{1&3&9\cr 0&2\gamma&4\gamma\cr 1&2&4\cr}\cr
&=\alpha\gamma^2\det\pmatrix{1&3&9\cr 0&2&4\cr 1&2&4\cr}\cr
&=-\alpha\gamma^2\ ;\cr}$$
so you can take any ${\bf v}$ from $({*}{*})$ as long as neither $\alpha$ nor $\gamma$ is zero.