Find the eigenvalues of this matrix

AI Thread Summary
The discussion revolves around finding the eigenvalues of the matrix C using the determinant equation det(C - λI) = 0. The user initially struggles with simplifying the matrix and contemplates using elementary row operations but is unsure if they will help. They attempt to simplify the matrix by subtracting the first column from the second, leading to a new determinant equation. The user questions the validity of this column operation, as they are more familiar with row operations. Ultimately, they recognize that the eigenvalues can be derived from their manipulation, identifying λ = 1 and λ = 0.97.
Benny
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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.

<br /> C = \left[ {\begin{array}{*{20}c}<br /> {0.98} &amp; {0.01} \\<br /> {0.02} &amp; {0.99} \\<br /> \end{array}} \right]<br />

<br /> \det \left( {C - \lambda I} \right) = 0 \Rightarrow \left| {\begin{array}{*{20}c}<br /> {0.98 - \lambda } &amp; {0.01} \\<br /> {0.02} &amp; {0.99 - \lambda } \\<br /> \end{array}} \right| = 0<br />

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.

<br /> \left| {\begin{array}{*{20}c}<br /> {0.98 - \lambda } &amp; { - 0.97 + \lambda } \\<br /> {0.02} &amp; {0.97 - \lambda } \\<br /> \end{array}} \right| = 0<br />

<br /> \left| {\begin{array}{*{20}c}<br /> {1 - \lambda } &amp; 0 \\<br /> {0.02} &amp; {0.97 - \lambda } \\<br /> \end{array}} \right| = 0<br />

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