Using the determinant and trace to find eigenvalues in FP3 edexcel Watch

gagafacea1
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Okay so I just found about how the determinant of a matrix is equal to the product of its eigenvalues and its trace is equal to the sum of the eigenvalues. My question is can I use this in the exam?
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Gome44
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(Original post by gagafacea1)
Okay so I just found about how the determinant of a matrix is equal to the product of its eigenvalues and its trace is equal to the sum of the eigenvalues. My question is can I use this in the exam?
Why would you want to? Finding eigenvalues requires finding a determinant...
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gagafacea1
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(Original post by Gome44)
Why would you want to? Finding eigenvalues requires finding a determinant...
because it's easier to just find the determinant in my calculator ( I don't use the method with the minors) and the trace is just addition.
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rayquaza17
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How exactly would you find the eigenvalues from this?
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FireGarden
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(Original post by rayquaza17)
How exactly would you find the eigenvalues from this?
For any nxn matrix you can find a polynomial with coefficients in terms of the trace and determinant. As a result, the eigenvalues of 2x2, 3x3 and 4x4 matrices all have formulae.

For a 2x2 it's very easy; \lambda^2 - \text{tr}(A)\lambda +\det{A} =0.

For 3x3 you also need a coefficient  \dfrac{1}{2}(\text{tr}(A)^2 - \text{tr}(A^2)) .
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gagafacea1
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(Original post by rayquaza17)
How exactly would you find the eigenvalues from this?
http://algebra.math.ust.hk/eigen/04_...lecture4.shtml
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rayquaza17
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(Original post by FireGarden)
For any nxn matrix you can find a polynomial with coefficients in terms of the trace and determinant. As a result, the eigenvalues of 2x2, 3x3 and 4x4 matrices all have formulae.

For a 2x2 it's very easy; \lambda^2 - \text{tr}(A)\lambda +\det{A} =0.

For 3x3 you also need a coefficient  \dfrac{1}{2}(\text{tr}(A)^2 - \text{tr}(A^2)) .
Is there an easy formula for any nxn matrix or is it only easy up to n=4?


Can you not do it both ways using the 'normal' method to get the marks, and this method as a check?
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FireGarden
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(Original post by rayquaza17)
Is there an easy formula for any nxn matrix or is it only easy up to n=4?
Well, it's just a polynomial. General formulae for 5th degree and higher don't exist. There are formulae for some 5th degree polynomials, and similarly for higher degrees, but knowing/finding out which polynomials they work for would be such a pain.

For 2x2 it might make life like 2 seconds faster, but generally it isn't that useful.
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gagafacea1
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(Original post by rayquaza17)
Can you not do it both ways using the 'normal' method to get the marks, and this method as a check?
Oh yeah, good idea, thanks!
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gagafacea1
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(Original post by FireGarden)
Well, it's just a polynomial. General formulae for 5th degree and higher don't exist. There are formulae for some 5th degree polynomials, and similarly for higher degrees, but knowing/finding out which polynomials they work for would be such a pain.

For 2x2 it might make life like 2 seconds faster, but generally it isn't that useful.
Wait, I thought it was simply det[A] = product of all eigenvalues of A , and tr[A]= sum of all eigenvalues .
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FireGarden
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(Original post by gagafacea1)
Wait, I thought it was simply det[A] = product of all eigenvalues of A , and tr[A]= sum of all eigenvalues .
Yeah, that's right.. but those facts don't immediately give you the eigenvalues. If you know a+b and ab then to determine a and b you'd have to solve x^2-(a+b)x + ab =0. Of course, this is exactly the 2x2 matrix case. The coefficient of x is (minus) the trace and the constant is the determinant.

3x3 and higher, it's not going to be fun.
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gagafacea1
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(Original post by FireGarden)
Yeah, that's right.. but those facts don't immediately give you the eigenvalues. If you know a+b and ab then to determine a and b you'd have to solve x^2-(a+b)x + ab =0. Of course, this is exactly the 2x2 matrix case. The coefficient of x is (minus) the trace and the constant is the determinant.

3x3 and higher, it's not going to be fun.
Oh yeah no that's fine. Because they usually give you one value and ask you to work out the others, and they're usually very nice numbers; so knowing this fact, they'd be easy to guess.
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