# Find the eigenvalues of this matrix

1. Oct 4, 2005

### Benny

I'm experiencing difficulties trying to find the eigenvalues of the follow matrix. The hint is to use an elementary row operation to simplify $$C - \lambda I$$ but I can't think of a suitable one to use or figure out whether a single row operation will actually make the calculations simpler.

$$C = \left[ {\begin{array}{*{20}c} {0.98} & {0.01} \\ {0.02} & {0.99} \\ \end{array}} \right]$$

$$\det \left( {C - \lambda I} \right) = 0 \Rightarrow \left| {\begin{array}{*{20}c} {0.98 - \lambda } & {0.01} \\ {0.02} & {0.99 - \lambda } \\ \end{array}} \right| = 0$$

Out of desperation, and having seen it being done once(not sure if it is correct) I decided to then subtract the first column from the second column.

$$\left| {\begin{array}{*{20}c} {0.98 - \lambda } & { - 0.97 + \lambda } \\ {0.02} & {0.97 - \lambda } \\ \end{array}} \right| = 0$$

$$\left| {\begin{array}{*{20}c} {1 - \lambda } & 0 \\ {0.02} & {0.97 - \lambda } \\ \end{array}} \right| = 0$$

I'm not even sure if subtracting columns from each other was a valid step. I know that subtracting rows is but I'm not sure about columns. I'm just wondering if that step was correct because if it is then I can get the eigenvalues from it fairly easily.

2. Oct 4, 2005

### TD

You're doing great! You can even see the eigenvalues right now, l = 1 or l = 0.97

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