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missylulu

  • 2 years ago

how do you find a matrix A given its eigenvectors and eigenvalues?? please helpp Dx

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  1. JamesJ
    • 2 years ago
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    By definition, a vector v is an eigenvector of matrix A with eigenvalue \( \lambda \) if \[ Av = \lambda v \] Hence \[ (A - \lambda I)v = 0 \] and this is equivalent to saying \[ \det(A - \lambda I) = 0 \] This is therefore the equation you solve to find the eigenvalues. Once you have found the eigenvalues, \( \lambda_i \), you want to find the null space of the operator \[ A - \lambda_i I \] for each \( i \). This corresponds to the eigenspace for \( \lambda_i \). This all probably seems a bit abstract, but is absolutely correct. I STRONGLY recommend you go through you lecture notes and/or text book to find some worked examples. Once you understand them, attempt your particular problem.

  2. missylulu
    • 2 years ago
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    ok thx for the advice !

  3. missylulu
    • 2 years ago
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    @JamesJ i can get to the part where you use matrix multiplication to get like 4 separate equations..2 with a, b and 2 with c, d...do you just solve the 4 unknowns and put it in as the matrix??sorry

  4. JamesJ
    • 2 years ago
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    You want to solve for lambda

  5. JamesJ
    • 2 years ago
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    If you look at an example, you would see that

  6. missylulu
    • 2 years ago
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    yeah but in this problem its given so do you just work backwards to get the original matrix?

  7. JamesJ
    • 2 years ago
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    Then for each eigenvector Av = lambda.v Let v1, v2, ..., vn be basis vectors for each of the eigenspaces. If we treat each of those as column vectors, then you can see that A [ v1 v2 ... vn ] = [ lambda1.v1 lambda2.v2 .... lambdan.vn ] where here lambdaj is the eigenvector corresponding to vj. Call this matrix V = [ v1 v2 ... vn ] Then V^-1 A V = V^-1 [ lambda1.v1 lambda2.v2 .... lambdan.vn ] = D where D is a diagonal matrix with the lambdaj down the diagonal. Hence A = V D V^-1 So that's how you recover A from the lambda and the vj. Again, read an example!

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