Simple exercise regarding change of basis with polynomials and derivatives

I have a couple of doubts regarding the following exercise:

Let $V = P^{\le n}[x] $ be the vector space of real polynomials with degree less or equal to $n$. If $p(x) = a_0 + a_1x + \ldots + a_nx^n$ is an n-th degree polynomial,

(i) Prove that $B_{p(x)} = \{p(x),p'(x),p''(x),\ldots,p^{(n)}(x)\}$ is a basis of $V$.

(ii) Find the change of basis matrix from $B_{p(x)}$ to $B = \{1,x,x^2,\ldots,x^n \}$.

(iii) For $n = 3$ and $p(x) = 1 + x + x^2 + x^3$, find the change of basis matrix from $B$ to $B_{p(x)}$.

In part (i), I can prove without any problem that $B_{p(x)}$ is linearly independent but I'm struggling with proving that the span of $B_{p(x)}$ is equal to V. We've seen in class a lemma which says that if $|B| = n$ and $B$ is linearly independent, then $B$ is base of $V$ ($n$ is the dimension of V), so my question here is if the second part of the demonstration would be as simple as applying this lemma.

I also have some troubles with (ii), as I feel I might be a bit confused about changes of basis. What I've done is to write vectors of $B_{p(x)}$ using the base $B$ as it follows:

• $p(x) = a_0·1 + a_1·x + \ldots + a_n·x^n$
• $p'(x) = a_1·1 + 2a_2·x + \ldots + na_n·x^{n-1} + 0·x^n$

$\ldots$

• $p^{(n-1)}(x) = (n-1)!a_{n-1}·1 + n!a_n·x + 0·x^2 + \ldots + 0·x^n$
• $p^{(n)}(x) = n!a_n·1 + 0·x + 0·x^2 + \ldots + 0·x^n$

Then the change of basis matrix has as columns the coordenates relative to $B$ of $B_{p(x)}$ vectors, so it should be something like $$ \begin{pmatrix} a_0 & a_1 & \ldots & (n-1)!a_{n-1} & n!a_n \\ a_1 & 2a_2 & \ldots & n!a_n & 0\\ a_2 & 3a_3 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{n-1} & na_n & \ldots & 0 & 0 \\ a_n & 0 & \ldots & 0 & 0 \end{pmatrix} $$

Finally, (iii) can be easily solved after doing (ii) correctly.

Solution 1:

For your doubt in (i): Yes, you can cite that a set $S$ of $n$ linearly independent vectors in an $n$-dimensional vector space $V$ is necessarily a basis. (You could show the linear span of $S$ is $n$-dimensional, therefore it's the whole space $V$.)

For your doubt in (ii): The change of basis matrix $A_{U -> W}$ from an ordered basis $U = \{\mathbf{u}_i\}$ to an ordered basis $W = \{\mathbf{w}_j\}$ has as its first column the coefficients of $\mathbf{u}_1$ when expressed in the basis $W$ (and so on for the other columns). That is, if we (uniquely) write $\mathbf{u}_1 = a_1\mathbf{w}_1 + a_2 \mathbf{w}_2 + \dots a_n \mathbf{w}_n$, then the first column on $A_{U \to W}$ is $\begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}^T.$ We do this so that the product $A_{U \to W} \mathbf{e}_1$, where $\mathbf{e}_1$ is the standard basis vector representing the coordinates of $\mathbf{u}_1$ with respect to the basis $U$, equals $\begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}^T,$ the coordinates of $\mathbf{u}_1$ with respect to the basis $W$.

So in your case, the first column gives the coordinates of $p(x)$ with respect the standard basis $\{1, x, \dots, x^n\}$ of $P^{\le n}[x]$, and so on.