Hmm, diagonalization wasn't even mentioned in our lectures, and I don't see how that would help. I think it just wants it rewritten some other way in terms of the eigenvectors, not explicitly calculated. Trying to follow the method on wikipedia gets me with rows 1 and 2 both being (1,0), so unless I'm doing something wrong (very possible) I don't think it's even diagonalizable.
Oh, there's some basic answers to the exams up:
Q9 (a) Characteristic equation is L(L − 1) = 0, e’values are L = 0 with corresponding e’vector (2, 3),
and L = 1 with corresponding e’vector (1, 1).
(b) A^4(v1 + v2) = v1.
uh... huh. It works out and makes sense, but seems like an almost trivial question.
And for the similar questions for the other two sets of answers given, one of them is 256v1+v2, the other is v1+243v2, so there must be some other method than just multiplying it out and doing it by inspection.
Isn't the characteristic equation det(A-lambda*I)=0?
If it is, then for this matrix the equation is x^2-6x+5=0, which has roots 1 and 5. Plugging those back into the matrix and then finding the null space should give you the eigenvectors.
I obviously don't want to contradict your exam solutions, so I guess i'm wrong. At any rate I kind of want to know what else it could be...
edit: yeah i missed the minus on the bottom right 2. thanks!
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Since this doesn't sound familiar to you, here is a handy wikipedia link: http://en.wikipedia.org/wiki/Diagonalizable_matrix
Oh, there's some basic answers to the exams up:
uh... huh. It works out and makes sense, but seems like an almost trivial question.
And for the similar questions for the other two sets of answers given, one of them is 256v1+v2, the other is v1+243v2, so there must be some other method than just multiplying it out and doing it by inspection.
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If it is, then for this matrix the equation is x^2-6x+5=0, which has roots 1 and 5. Plugging those back into the matrix and then finding the null space should give you the eigenvectors.
I obviously don't want to contradict your exam solutions, so I guess i'm wrong. At any rate I kind of want to know what else it could be...
edit: yeah i missed the minus on the bottom right 2. thanks!
A^4 * (v1 + v2) = A^4 * v1 + A^4 * v2, of course. Let e1 and e2 be the corresponding eigenvalues, then we have:
A^4 * v1 = A^3 * A * v1 = A^3 * e1 * v1.
Continuing to reduce the exponent on A in that fashion gives the final result
A^4 * v1 = e1^4 * v1, and likewise with A^4 * v2. So
A^4 * (v1 + v2) = e1^4 * v1 + e2^4 * v2.
The eigenvalues of this matrix are 0 and 1, so declaring that e1 = 1 and e2 = 0 gives
A^4 * (v1 + v2) = v1.
(bolded line is the important one, should explain the question with the more complicated answer as well.)
Fuzzywhale: Det(A - x*I) = x^2 - x. I think you lost a minus sign or something to get that result.
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