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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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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?
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.
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.

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I just don't understand why they knew that the eigenvalue was 0 off the bat
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.
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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