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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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This document was uploaded on 04/02/2012.
 Fall '09
 Math

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