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Exercise 1.6.8

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The quadratic formula gives the eigenvalues $λ_{1} = \frac{a + d + \sqrt{{(a + d)}^{2} - 4 (ad - bc)}}{2}$ and $λ_{2} = \frac{a + d - \sqrt{{(a + d)}^{2} - 4 (ad - bc)}}{2}$ . Their sum is $a + d$ .

If $A$ has $λ_{1} = 3$ and $λ_{2} = 4$ , then $\det (A - λI) = (λ - 3) (λ - 4)$

niuers

2020-03-20 00:00

Exercise 1.6.8

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