Mathematics, 15.01.2020 06:31 hosteenimport21
In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. the same is true for a lower triangular matrix. verify this for 2 × 2 and 3 × 3 matrices.
Answers: 3
![A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0](/tpl/images/0455/6968/cd3cf.png)
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![A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0](/tpl/images/0455/6968/c6415.png)
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In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal....
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