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anonymous

  • 4 years ago

For a Vandermonde Matrix (a) Explain why showing that det(A) 6= 0 is the same as showing that det(A^t) 6= 0, where A^t means the transpose of A. [This is a very short answer]. (b) Explain why showing that det(A^t) 6= 0 is the same as showing that A^t has no nonzero vector in the kernel. (c) if ~v = (c0, c1, . . . , cn−1) is a vector in the kernel of At, and ~v is not the zero vector, explain how this would give you a polynomial of degree ≤ n − 1 with more than n−1 roots, which would be a contradiction. [Hint: consider what At~v = 0 means.]

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  1. anonymous
    • 4 years ago
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    that 6 shouldn't be there, it should be (a) Explain why showing that det(A) cant equal 0 is the same as showing that det(A^t) cant equal 0, where A^t means the transpose of A. [This is a very short answer]. (b) Explain why showing that det(A^t) cant equal 0 is the same as showing that A^t has no nonzero vector in the kernel.

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