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Thread: Find the characteristic polynomial of an arbitrary-sized matrix

  1. #1
    Member Last_Singularity's Avatar
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    Find the characteristic polynomial of an arbitrary-sized matrix

    Question: Let A \in M_{n \times n}(R) be the matrix defined by A_{ij} = 1 for all i,j. Find the characteristic polynomial of A.

    My attempt: Call A_k to be the k \times k matrix of all 1's. Then its characteristic polynomial f(t)_k = (-1)^k [t^k - k t^{k-1}]. Use proof by induction.

    Base case: A_1 = [1]. Then its characteristic polynomial f(t)_1 = 1-t = -t-(-1) = (-1)[t^1 - 1 t^0]. This checks.

    Then for induction step, assume that it holds for k and attempt to prove it to be true for k+1. My problem with this was that as I expanded out the determinant along the top row of A_{k+1} - t I_{k+1}, things got very messy and I couldn't see a pattern.

    Could you give me a hand, please? Thanks!
    Last edited by Last_Singularity; Nov 5th 2009 at 06:55 PM.
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  2. #2
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    Opalg's Avatar
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    One way of doing this is to think in terms of eigenvalues and eigenvectors. It's easy to see that v = (1,1,...,1)^{\textsc t} is an eigenvector of A with eigenvalue n. Also, if w is any vector orthogonal to v then Aw=0. So 0 is an eigenvalue whose eigenspace has dimension n1. Therefore the characteristic polynomial of A must be a scalar multiple of (t-1)t^{n-1}. But the coefficient of t^n has to be (-1)^n. That gives f_n(t) = (-1)^n(t^n-nt^{n-1}), as you wanted.
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