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Let $A$ be the matrix $$ \begin{pmatrix} -1 & -2 & 1 & 2\\ -1 & 2 & 1 & 0\\ -4 & -3 & 3 & 3\\ -4 & -2 & 2 & 4 \end{pmatrix} $$

How can I show that $A$ has a single eigenvalue with geometric multiplicity 2 using the trace of $A$?

I know that $A$ has the eigenvalue $2$ with algebraic multiplicity 4, and I also know that $$tr(A) = \sum_i \lambda_i$$ where $\lambda_i$ is the $i$'th eigenvalue.

Edit: The whole exercise is as follows:

Consider $$ A = \begin{pmatrix} -1 & -2 & 1 & 2\\ -1 & 2 & 1 & 0\\ -4 & -3 & 3 & 3\\ -4 & -2 & 2 & 4 \end{pmatrix} ,\quad \begin{pmatrix} {\bf v}_2 & {\bf v}_1 & {\bf v}_0 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0\\ 1 & 0 & 1\\ 1 & 3 & 0\\ 2 & 2 & 2\\ \end{pmatrix} $$

Show that $\begin{pmatrix}{\bf v}_2 & {\bf v}_1 & {\bf v}_0\end{pmatrix}$ is a Jordan chain for $A$ and compute the corresponding eigenvalue $\lambda$. Show using the trace or otherwise that $A$ has only one eigenvalue and show it has geometric multiplicity 2.

share|improve this question
    
You could use trace as a check to your answer, but it isn't going to give you geometric multiplicity. I think you just have to calculate it. –  Stuart Sep 19 '12 at 2:31
    
The exercise I have implies that it should be possible. –  utdiscant Sep 19 '12 at 2:50
    
There really should be more information. Does the exercise tell you that theres only eigenvalue $2$ with multiplicity $4$? If not then how did you find out? –  EuYu Sep 19 '12 at 2:57
    
I have copied the whole exercise to the question. –  utdiscant Sep 19 '12 at 3:31
1  
So you know (from your Jordan chain), that the algebraic multiplicity of $\lambda = 2$ is $\ge 3$, right? As the trace is $\ldots$, the missing zero is... –  martini Sep 19 '12 at 8:18

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