Benny
Oct4-05, 11:30 AM
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.
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.