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sh3lsh

  • one year ago

More Linear Algebra. Latexing it up below. Why doesn't OpenStudy have the option to have Latex it in the question portion?

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  1. sh3lsh
    • one year ago
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    How does one find the eigenvalues through inspection or by quick algebra? Say given matrix A = \[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix}\] How does one find the eigenvalues of this?

  2. Empty
    • one year ago
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    Well one quick way is that the trace of a matrix is the sum of its eigenvalues and the determinant of a matrix is the product of its eigenvalues. It's really helpful when dealing with 2x2 matrices, and kinda with 3x3, but in general I don't know. I'm sure there's a trick for this one since the columns are all linearly dependent.

  3. sh3lsh
    • one year ago
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    The hints were: Note that the algebraic multiplicity agrees with the geometric multiplicity. (Why?) Hint: What is the kernel of A? The answer is that ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remaining eigenvalue is tr(A) = 5.

  4. sh3lsh
    • one year ago
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    I just don't understand why they knew that the eigenvalue was 0 off the bat

  5. Empty
    • one year ago
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    In this case, because the entire matrix is linearly dependent, you know that the matrix's characteristic will have a 0 solution. Think back to what an eigenvalue really represents. It's a number we could multiply by an eigenvector instead of the matrix. Since the columns are linearly dependent, we can always find a way to combine all the columns to get 0, in 4 different ways without the vectors being 0 to begin with. \[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{bmatrix}\] Here's one of 4 ways, here's another way: \[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \\0 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \\ 0 \end{bmatrix}\] So maybe it becomes more transparent now how these are eigenvalues satisfying \[A x= \lambda x\] Really what it comes down to is identifying the linearly dependent columns, that's what's going to give you the power to easily say this kinda thing.

  6. sh3lsh
    • one year ago
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    So, using this to check a hypothesis: http://www.mathportal.org/calculators/matrices-calculators/matrix-calculator.php It seems as if per every redundant column, we get a 0 eigenvector. So in this case, given 4 redundant columns. We get a 0 with an almu of 4 (# of redundant columns). Is this true?

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