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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?

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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.

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  • $\begingroup$ oh ok but for $v_3$ the equation you put up is exactly what I did, I got $\begin{pmatrix} 1&0&-1\\ 0&1&-1\\ 0&0&0 \end{pmatrix} \begin{pmatrix} v_1\\ v_2\\ v_3 \end{pmatrix}= \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} $. From there I got $v_1= v_3+1$, $v_2=v_3 + 1$, and $v_3= 1$. So that gave me $v_3 = \begin{pmatrix} 2 \\ 2\\ 1 \end{pmatrix} $. Where did I go wrong there? $\endgroup$ – Ruth Gutierrez May 7 '14 at 6:17
  • $\begingroup$ The matrix you just posted in your comment is not $A-2I$. $\endgroup$ – David May 7 '14 at 6:22
  • $\begingroup$ Well yeah it is just reduced. Its originally $\begin{pmatrix} -2&1&1\\ -2&1&1\\ -3&1&2 \end{pmatrix} \begin{pmatrix} v_1\\ v_2\\ v_3 \end{pmatrix}= \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}$ and then reduces to what I wrote on the above comment. Or is my set up wrong? $\endgroup$ – Ruth Gutierrez May 7 '14 at 6:29
  • $\begingroup$ The setup is right but you need to do the solution more carefully. The system in your first comment has no solution because the third row says $0=1$. $\endgroup$ – David May 7 '14 at 6:35
  • $\begingroup$ Oh yeah that's right. I forgot about that. Yeah that can't happen. So would I say there is no solution? $\endgroup$ – Ruth Gutierrez May 7 '14 at 6:50
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For part B you can set $v = \left(v_1, v_2, v_3\right)$ and then simply write out what $Av$ and $A^2v$ would be in terms of $v1, v2, v3$. Then put these 3 columns into a matrix, and choose $v1, v2, v3$ such that the determinant is not 0.

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  • $\begingroup$ I'm not sure I grasped what you said. Is there something you could direct me to to see an example? $\endgroup$ – Ruth Gutierrez May 7 '14 at 5:26

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