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Anonymous
Eigenvalues of a matrix with trigonometric functions 2016-03-31 22:32:21 Post No. 7970565
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Eigenvalues of a matrix with trigonometric functions 2016-03-31 22:32:21 Post No. 7970565
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Eigenvalues of a matrix with trigonometric functions
Anonymous
2016-03-31 22:32:21
Post No. 7970565
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Hope someone could help me.
How can you obtain the eigenvalues of a matrix which has trigonometric functions as elements?
For example the matrix I show in the image (matrix A)
| cos(x), sin(x), 0 |
| -sin(x), cos(x), 0 |
| 0, 0 , 1 |
According to the equation
det(A - E*y) = 0
where A is the matrix above, E is the identity matrix, and y is the eigenvalues
if I get the determinant established by that equation then I got a cubic polynomial from whose roots I can get the three eigenvalues for the matrix A.
I already get to the polynomial as:
y^3 + [-2*cos (x) -1]*y^2 + [2*cos (x) +1]*y - 1
and I find in a solution book the answer for the eigenvalues as:
y(1) = 1,
y(2) = cos (x) + i sin (x),
y(3) = cos (x) - i sin (x)
But I don't have any idea on how to get to that answer in this case of a cubic polynomial with trigonometric functions. I don't even imagine how the answer involve the imaginary "i sin (x)".
If it's not trouble and someone can at least tell me about a identity, formula, or some good book or web page where I can find and easy explanation I would be very grateful.
Thanks in advance.
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>>7970565
it is apparent from just looking at the equation that y=1 is a root because
[eqn][-2*cos (x) -1]*y^2 + [2*cos (x) +1]*y[/eqn]
clearly cancel.
I don't think there is a general rule to easily this kind of polynomial but I do think you're expected to be able to see obvious roots like this. The other two innolve solving a quadratic which is simple enough.
If worst comes to worst there is also the cubic formula for the roots of a polynomial but I'd avoid that.
Or you could just use a CAS...
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>>7970565
You should be able to read of the first eigenvalue, just from the characteristic equation
(1-y)((cos(x)-y)^2+(sin(x)-y)^2)
And from expansion note that the polynomial is equivalent to
2y^2 - 2(cos(x)+sin(x))y +1 = 0
so if we calculate the quadratic formula we see that
[math]
\frac{2cos(x) + 2sin(x) \pm \sqrt{ 4(1 + 2cos(x)sin(x)) - 4(2)(1)}{4}
[/math]
and using some trig identities you can find that this should equal the other two eigenvalues.
Alternately you could observe that for any fixed x, the 2x2 contained in the 3x3 is just a rotation matrix, and solve like that.
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>>7970602
Actually, the second method is surely the easiest. Since a rotation matrix has eigenvalue a+/-bi where a is the entry in the first column and b is the entry in the second column of the first row.
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>>7970565
This is a rotation in a plane inside of 3-space. One line will be fixed, so 1 will be an eigenvalue. The other two eigenvalues will be complex, since rotations will not take any line to a real multiple of itself. Hopefully visualising this will lead you to the actual solution.
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>>7970596
>If worst comes to worst there is also the cubic formula for the roots of a polynomial but I'd avoid that.
It's a block diagonal matrix. You only have to solve a single quadratic equation.
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>>7970565
Note [0 0 1]^T is an obvious eigenvector. It's eigenvalue is 1.
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