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We calculate the eigenvectors for the matrix $$ \begin{equation*} \mathbf{A} = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & -1 & 3 \\ \end{array} \right) \end{equation*} $$

First, I calculted the eigenvalue polynomial $\det(\mathbf{A}-\lambda \mathbf{I})=0$, and got triple overlapping eigenvalues $\lambda=2$ $$\begin{equation*} \mathbf{A} - \lambda \mathbf{I} = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & -1 & 1 \\ 1 & -1 & 1 \\ \end{array} \right) \end{equation*}$$

So, I got only two eigenvectors $x_1=\left( \begin{array}{ccc} 1 \\ 1 \\ 0 \\ \end{array} \right)$ and $x_2=\left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right)$. So the matrix $\mathbf{A}$ is a degenerate matrix.

Did I calculate it correctly?

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2 Answers 2

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You can check the correctness of your eigenvectors if you apply the matrix on them.

By the way: Since there is only one not-zero entry (2) in the first row of the matrix and this is the first entry, this is the only candidate for an eigenwert.

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  • $\begingroup$ Thank you. It is a good way to check them. $\endgroup$
    – John-Annual
    Jan 9, 2014 at 10:41
  • $\begingroup$ This is my first time to see this method(only one not-zero entry (2) in the first row). Is it a usual method for calculating the eigenvalues? How does it come? What's the principle? $\endgroup$
    – John-Annual
    Jan 9, 2014 at 10:42
  • $\begingroup$ Yes very unusual (I just edited it: it only works if the entry is also on the diagonal). You should always do it your way - that's standard. $\endgroup$
    – Bernd
    Jan 9, 2014 at 10:48
  • $\begingroup$ That is not even correct but $2$ has to be (at least) one eigenvalue in this case $\endgroup$
    – user127.0.0.1
    Jan 9, 2014 at 10:57
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Your results are correct, but that does not mean that $A$ is a degenerate matrix, just that $A$ is not diagonalizable.

Just one little thing: $x_1=\left( \begin{array}{ccc} 1 \\ 1 \\ 0 \\ \end{array} \right)$ and $x_2=\left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right)$ are not the only eigenvectors but every vector in the whole subspace $\operatorname{span}\left(\left( \begin{array}{ccc} 1 \\ 1 \\ 0 \\ \end{array} \right),\left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right)\right)$

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  • $\begingroup$ Thank you for pointing out my blunder. In this case, A is not degenerate. Is a degenerate matrix always undiagonalizable? No? There is not relation between degenerate and diagonalizable, right? $\endgroup$
    – John-Annual
    Jan 9, 2014 at 10:45
  • $\begingroup$ The last sentence is correct, there is no relation. A diagonalizable matrix might be degenerated or not, and a degenerate matrix might be diagonalizable or not. $\endgroup$
    – user127.0.0.1
    Jan 9, 2014 at 10:50

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