Is $1-3x$ in the span of {$x^2+x, 1-x^2,x^2-x$}?

Question

Having trouble understanding how to approach these types of problems. Here's what I have so far:

Using the definition of span, I'm looking for some scalars $s_1, s_2, s_3$ such that $s_1(x^2+x) + s_2(1-x^2) + s_3(x^2-x) = 1-3x$.

Now, as to how to actually solve this in a meaningful way - I'm stuck. Were this presented as simple matrices with simple numbers, it'd be easier for me to compute. Any tips or advice would be appreciated.

Answer 1

You are on the right track \begin{align*} s_1(x^2+x) + s_2(1-x^2) + s_3(x^2-x) & = 1-3x\\ x^2(s_1-s_2+s_3) + x(s_1-s_3) + s_2 = 1-3x. \end{align*} Now compare the coefficients of both sides to get \begin{align*} s_1-s_2+s_3 & =0\\ s_1-s_3 & =-3\\ s_2 &=1 \end{align*} See if you can solve this.

Answer 2

Presumably, we are working with the vector space $\Bbb R_2[x]$ of polynomials with degree $\leq 2$. Note that $\dim\Bbb R_2[x]=3$ since $\{1, x, x^2\}$ is a basis of $\Bbb R_2[x]$.

Our polynomials of interest are $p_1, p_2, p_3\in\Bbb R_2[x]$ given by \begin{align*} p_1(x) &= x^{2} + x & p_2 (x) &= -x^{2} + 1 & p_3(x) &= x^{2} - x \end{align*} We wish to determine if the polynomial $f(x)=-3 \, x + 1$ is in $\operatorname{Span}\{p_1, p_2, p_3\}$.

Let's take a general approach by explicitly describing all $f\in\operatorname{Span}\{p_1, p_2, p_3\}$. These are exactly the polynomials $f$ of the form $f(x)=b_{2} x^{2} + b_{1} x + b_{0}$ for which there are scalars $c_1$, $c_2$, and $c_3$ satisfying $$ c_1\cdot p_1+c_2\cdot p_2+c_3\cdot p_3 = f\tag{$\ast$} $$ Now, note that ($\ast$) is equivalent to $$ c_1\cdot(x^{2} + x)+c_2\cdot(-x^{2} + 1)+c_3\cdot(x^{2} - x)=b_{2} x^{2} + b_{1} x + b_{0} $$ Combining like terms gives $$ (c_{1} - c_{2} + c_{3})\, x^{2} + (c_{1} - c_{3})\, x + c_{2}=b_{2} x^{2} + b_{1} x + b_{0} $$ Comparing coefficients then gives the system $$ \begin{array}{rcrcrcr} c_1 &-& c_2 &+& c_3 &=& b_2 \\ c_1 & & &-& c_3 &=& b_1 \\ & & & & c_3 &=& b_0 \end{array} $$ Converting this system to augmented form and row reducing gives $$ \begin{align*} \left[\begin{array}{rrr|r} 1 & -1 & 1 & b_{2} \\ 1 & 0 & -1 & b_{1} \\ 0 & 0 & 1 & b_{0} \end{array}\right] \xrightarrow{R_1-R_2\to R_1}\left[\begin{array}{rrr|r} 1 & -1 & 1 & b_{2} \\ 0 & 1 & -2 & b_{1} - b_{2} \\ 0 & 0 & 1 & b_{0} \end{array}\right] \\ \xrightarrow{R_1+R_2\to R_1}\left[\begin{array}{rrr|r} 1 & 0 & -1 & b_{1} \\ 0 & 1 & -2 & b_{1} - b_{2} \\ 0 & 0 & 1 & b_{0} \end{array}\right] \\ \end{align*} \\ \xrightarrow{ \begin{array}{rcrcr} R_1 &+& R_3 &\to& R_1 \\ R_2 &+& 2\cdot R_3 &\to& R_1 \end{array} }\left[\begin{array}{rrr|r} 1 & 0 & 0 & b_{0} + b_{1} \\ 0 & 1 & 0 & 2 \, b_{0} + b_{1} - b_{2} \\ 0 & 0 & 1 & b_{0} \end{array}\right] $$ This shows that our original equation ($\ast$) is solved by taking \begin{align*} c_1 &= b_0+b_1 & c_2 &= 2\,b_0+b_1-b_2 & c_3 &= b_0 \end{align*} In particular, for $f(x)=-3\,x+1$, we have $$ -2\cdot p_1-p_2+p_3=f $$

Answer 3

Here is another, more tedious way that lets you compute the answer using matrices:

Let $u_1(x) = x^2+x$, and $u_2,u_3$ similarly. Let $f(x) = 1-3x$.

Note that the maximum degree is 2, and so any polynomial is completely specified by the values at 3 distinct points.

Given a polynomial $p$ of degree $2$ or less, let $\hat{p} = ( p(0) , p(1), p(2) )^T$.

Then let $\hat{U} = \begin{bmatrix} \hat{u}_1 & \hat{u}_2 & \hat{u}_3 \end{bmatrix}$. In particular, if $f = \sum_k \alpha_k u_k$ we have $\hat{f} = \hat{U} \alpha$, where $\alpha= (\alpha_1, \alpha_2, \alpha_3)^T$.

It is straightforward to check that $\hat{U}$ is invertible (since the $u_k$ are linearly independent), so we compute $\alpha = \hat{U}^{-1} \hat{f}$, and then $f = \sum_k \alpha_k u_k$ (since $f$ is uniquely specified by $\hat{f}$).