Quiz11-Solution

Quiz11-Solution -...

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Unformatted text preview: Name:.......................................................................................... Math 331 - Quiz 11 Thursday, December 8 1. Let A= 05 −2 2 find an invertible matrix P and a rotation plus scaling matrix of the form C= a −b ba such that A = P CP −1 . Solution: First we find the characteristic polynomial det (A − λI ) = det −λ 5 −2 2 − λ = (−λ)(2 − λ) + 10 = λ2 − 2λ + 10 To find the roots, we use the quadratic formula √ √ 2 ± −36 2 ± 6i 2 ± 4 − 40 = = = 1 ± 3i λ= 2 2 2 We choose one of the roots, for example 1 − 3i. (Choosing 1 + 3i would give a different, but also correct answer.) Then 1 −3 C= 31 To find P we must find an eigenvector. To do so we consider −(1 − 3i) 5 |0 −2 2 − (1 − 3i) | 0 −1 + 3i 5 |0 −2 1 + 3i | 0 = Now we can choose either equation to find a complex eigenvector. I’ll use the first. (Choosing the second would give a different, but also correct answer.) Then we have (−1 + 3i)x1 + 5x2 = 0 1 − 3i −1 + 3i x1 = − x1 x2 = − 5 5 So we can choose x1 = 5 and then x2 = 1 − 3i. Then our complex eigenvector is x= 5 1 − 3i = 5 1 + 0 −3 So P= Re(x) Im(x) = 50 1 −3 i Remark: As mentioned above, there are a lot of possible answers to this question, one could choose 1 + 3i instead of 1 − 3i, this would change C to 13 −3 1 And then would also change our P , which must now come from a eigenvector for λ = 1 + 3i. Another place where a choice is made is in deciding which equation to use to find the complex eigenvector. Using the second equation will give a different answer which is also correct. (This is a somewhat subtle point that has to do with the fact that the two equations are equivalent in terms of complex numbers, but not in terms of real numbers). Of course, we can always check whether we have obtained a correct answer. Check: P CP −1 = 50 1 −3 1 −3 31 1 5 1 15 0 −1 3 = 05 −2 2 =A ...
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