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L|ama
Registered User regular

Studying for an exam coming up and this sort of question has been in there every time for the past few years:

It doesn't really seem to be covered in the notes though. I can do the first part easily, but I've got no idea where to go with B.

It doesn't really seem to be covered in the notes though. I can do the first part easily, but I've got no idea where to go with B.

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## Posts

Since this doesn't sound familiar to you, here is a handy wikipedia link: http://en.wikipedia.org/wiki/Diagonalizable_matrix

FuzzywhaleonOh, 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!

FuzzywhaleonA^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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