Given $q(x)=a_0+a_1 x+a_2 x^2+ ... + a_n x^n$, prove $B$ and $B'$ are bases of $V$ and obtain coordinates of $q(x)$ with respect to each of them

Question

Given $V=\{p(x)\in \mathbb{R}[x]:grad(p(x))\leq n \}$ and the polynomial $q(x)=a_0+a_1 x+a_2 x^2+ ... + a_n x^n$ in V.

a) If $c\in \mathbb{R}$, prove $B = \{1,x-c,(x-c)^2,...,(x-c)^n\}$ is a basis of $V$ and find the coordinates of $q(x)$ relative to $B$.

b) Prove $B' = \{1,x,x (x-1),x (x-1) (x-2),...,x (x-1) (x-2) ... (x-n+1)\}$ is a basis of $V$ and find the coordinates of $q(x)$ relative to $B$.

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My attempt at a solution:

a) Given that every element of $B$ is in $V$, $B$ is a basis of $V$ iff the elements in $B$ are linearly independent and they are in the same number as the dimension of $V$.

$$P = \begin{pmatrix} 1 &0 &0 &... &0 \\ -c &1 &0 &... &0 \\ c^2 & &... & &0 \\ & &... & & \\ (-c)^n& &... & &1 \end{pmatrix}$$

We can guess (I don't really know how to prove this, nor do I think is necessary) that this is a triangular matrix, so the condition given in the previous statement holds, thus $B$ is a basis of $V$. The smartest way I can give the coordinates of $q(x)$ in this basis is $$q_B =P \begin{pmatrix} a_0 \\ a_1 \\ ... \\ a_n \\ \end{pmatrix}$$ I don't know how to give a more elaborate and forthright formula for each coordinate or if there's an easy way to get it.

b) In the same fashion, I can calculate some coordinates of the polynomials in $B'$, $$P'=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & ... &0 \\ 0 & 1 & 0 & 0 & 0 & ... &0 \\ 0 & -1 & 1 &0 & 0 & ... &0 \\ 0 &2 &-3 & 1 & 0 & ... &0 \\ 0 &-6 &11 & -6 & 1 & ... &0 \\ & & & ... & & & \\ 0 & & & ... & & &1 \end{pmatrix}$$ and guessing it is a triangular matrix, if we follow the same reasoning as above we get to a similar conclusion.

I didn't know how to get a formula for $P$ and I'm even more perplexed with $P'$. So the only correct way I can find to give the coordinates is as timid as before $$q_{B'} =P' \begin{pmatrix} a_0 \\ a_1 \\ ... \\ a_n \\ \end{pmatrix}$$

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Any help and advice is welcome. Thank you.

Answer 1

a) The coordinates of $q(x)$ in the canonical basis are $(a_0,a_1,...,a_n)$. Now, with respect to B, we got to solve $$q(x)=b_0+b_1(x-c)+b_2(x-c)^2+...+b_n(x-c)^n$$ Now, $q(c)=b_0$

If we differentiate, $$q'(x)=b_1+2b_2(x-c)+...+nb_n(x-c)^{n-1}$$ And so, $q'(c)=b_1$

If we keep going, $$q''(c)=2b_2$$ $$q'''(c)=3\ 2b_3$$ $$...$$

And we can guess that

$$q_{B}=(q(c),q'(c),\frac{q''(c)}{2},...,\frac{q^{(n)}(c)}{n!})$$

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b) In a similar fashion, $$q(x)=b_0+b_1x+b_2x(x-1)+...+b_nx(x-1)(x-2)...(x-n+1)$$ $$q(0)=b_0$$ $$q(1)=b_1$$ $$...$$ $$q(n)=n!b_n$$ And we can guess: $$q_{B'}=(q(0),q(1),\frac{q(2)}{2},...,\frac{q(n)}{n!})$$

Answer 2

Note that, $dim(V)=n+1$ and $|B|=n+1$.

So to show that $B$ is a basis for $V$, it is enought to show that $B$ is linearly independent, because any set of $n+1$ linearly independent vectors must be a basis for $V$.

Consider, $a_0+a_1(x-c)+a_2(x-c)^2...+a_n(x-c)^n=0$.

On using binomial expansion and clubbing the terms according to the powers of $x$ we get,

$a_nx^n+(a_{n-1}-na_n)x^{n-1}+...+(\text{some combination of} \;a_i's)x+(\text{some combination of} \;a_i's).1=0$

Since $\{1,x,x^2,...,x^n\}$ is a basis for $V$ $\implies$ All coefficients in the above expression must be $0$

Thus, $a_n=0$, $a_{n-1}-na_n=0$,...

$\implies a_i=0 \;\; \forall i=0,1,...,n$

So $B$ is Linearly Independent and hence it forms a Basis for $V$

Now to find the coordinates of $q(x)$ relative to $B$, first you must find the Change of Basis matrix from $B'$ to $B$, here $B'$ dentes the standard basis for $V$

If $Q$ is a change of basis matrix then, $[q(x)]_{\displaystyle B}=Q[q(x)]_{\displaystyle B'}$

Answer 3

The coefficients of the direct matrices are given by the Binomial numbers (Pascal's triangle) times the powers of $-c$, and the Stirling numbers respectively.

The matrices are readily inverted by the resolution of the linear system which is in echelon form:

$$\begin{align}&ax&=A\\&bx+cy&=B\\&dx+ey+fz&=C\\&gx+hy+iz+jt&=D,\end{align}$$ then $$\begin{align}ax&=A\\cy&=B-bx\\fz&=C-dx-ey\\jt&=D-gx-hy-iz.\end{align}$$