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Math Help - Help with determinants and eigenvalues.

  1. #1
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    Help with determinants and eigenvalues.

    Help with determinants and eigenvalues.-bsmath.png

    For #1, I currently have that the two values are equal if A is either a matrix whose determinant is zero or a 1x1 matrix.

    I don't really have any direction for the others. I see that the determinant of a square matrix whose columns add to zero is 0, but I'm not sure of why. Can it somehow be reduced to the zero matrix?

    Edit: Okay, messed around with row reduction. Quickly found that all columns adding to zero forces at least one row to row-reduce to a zero row, forcing a determinant of zero.
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  2. #2
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    det(cA)=c^n*det(a) where n is the number of rows so as long as you have an odd number of rows and the scalar is -1, then det(-A)=-det(A)
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  3. #3
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    I'm not following.

    -det(A) = det(-A) and det(-A) = -1^n*det(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A).

    Oh, okay. Nevermind. Odd number of rows.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pantsaregood View Post
    I'm not following.

    -det(A) = det(-A) and det(-A) = -1^n*det(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A).

    Oh, okay. Nevermind. Odd number of rows.
    2) If c_1,\cdots,c_n are the columns it's asking for what \det\left[\begin{array}{c|c|c}& & &\\ c_1 & \cdots & c_n\\ & & &\end{array}\right] given that \displaystyle \sum_{j=1}^{n}c_j=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}. Let me ask you this: what is the determinant of a matrix whose columns are linearly dependent equal? Moreover, how does that apply to this question?

    3) This is pretty straightforward. I don't know how to give a decent hint that doesn't give it away. Merely note that (AB)\bold{x}=A\left(B\bold{x}\right) and go from there.

    4) If \lambda is an eigenvalue for A note that \bold{0}=A^k\bold{x}=\lambda A^{k-1}\bold{x}=\cdots=\lambda^k\bold{x}...so...
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