Find the eigenvalues of this matrix

[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 [tex]C - \lambda I[/tex] 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.

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

[tex]
\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
[/tex]

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.

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

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

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.

 

[TD]

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

 

[quicknote]

I would first like to commend you for your effort and determination in trying to find the eigenvalues of this matrix. It is not an easy task, and it requires a lot of mathematical understanding and problem-solving skills.

Now, to address your concerns, subtracting columns from each other is a valid step in elementary row operations, as it does not change the determinant of the matrix. However, in this case, it may not lead to a simpler calculation for finding the eigenvalues.

A better approach would be to use the characteristic equation, as you have correctly done, and solve for the roots. In this case, we can use the quadratic formula to find the eigenvalues:

\lambda = \frac{0.98 + 0.99 \pm \sqrt{(0.98 - 0.99)^2 - 4(0.98)(0.99 - \lambda)}}{2} = 0.985 \pm 0.014i

Therefore, the eigenvalues of this matrix are 0.985 + 0.014i and 0.985 - 0.014i.

In general, finding eigenvalues can be a challenging task, and sometimes there is no simple way to do it. As a scientist, it is important to be familiar with different methods and techniques in order to solve such problems efficiently. Keep exploring and learning, and don't be afraid to ask for help or consult with experts in the field.