## 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 $k$th 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 $V$.