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.]

Question

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.]

anonymous · Answer

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.