Compute a transition matrix

Let F be a field and let n be a positive natural number. Denote by V \subseteq F[x] the set of all polynomials having degree at most n. Fix a nonzero \lambda \in F. Let B = \{ x^k \ |\ 0 \leq k \leq n \} and let E = \{ (x-\lambda)^k \ |\ 0 \leq k \leq n \}. Compute the transition matrix M^B_E(1). Conclude that E is a basis for V.


Using the binomial theorem, we have (x-\lambda)^k = \sum_{t=0}^k {k \choose t} (-\lambda)^{k-t} x^t. Letting these coordinates (in B) be the entries of the kth column of M^B_E(1), we see that M^B_E(1) = [a_{i,j}] where a_{i,j} = {j \choose i} (-\lambda)^{j-i} if j \geq i and 0 otherwise. In particular, M^B_E(1) is upper triangular and has all 1s on the main diagonal- hence this matrix is invertible. Thus E is a basis for V.

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