# Calculate the eigenvectors

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?

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.

• Thank you. It is a good way to check them. Jan 9, 2014 at 10:41
• 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? Jan 9, 2014 at 10:42
• 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. Jan 9, 2014 at 10:48
• That is not even correct but $2$ has to be (at least) one eigenvalue in this case Jan 9, 2014 at 10:57

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 $\operatorname{span}\left(\left( \begin{array}{ccc} 1 \\ 1 \\ 0 \\ \end{array} \right),\left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right)\right)$

• 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? Jan 9, 2014 at 10:45
• 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. Jan 9, 2014 at 10:50