Determining if a vector belongs in the linear span

Question

Fix a $\in$ $\mathbb{R}$. Consider the vectors $a + x + x^2$, $1 + x^2$ and $x$ in the $\mathbb{R}$-vector space $P(\mathbb{R})$ of polynomial functions over $\mathbb{R}$. Then $x^2$ belongs to the linear span of $a + x + x^2$, $1 + x^2$, $x$ iff $a \neq 0$.

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This was on a multi-choice practice test and it was a supposedly correct lucky guess. I understand what a linear span is (the collection of all linear combinations), but is there a more concrete way of actually determining if a vector is in a span? Our notes in class are pretty vague.

Answer 1

You want to see if there exists real numbers $p,q,r$ such that $$x^2=p(a+x+x^2)+q(1+x^2)+r(x).$$ Comparing coefficients of powers of $x$ on both sides we get \begin{align*} p+q&=1\\ p+r&=0\\ ap+q&=0. \end{align*} Now see if you can find value(s) of $a$ for which the system is consistent.

Answer 2

$x^{2}=b(a+x+x^{2})+c(1+x^{2})$ iff $ab=-c, b=0$ and $b+c=1$. This gives $0=b+c=0+0=1$ a contradiction. Hence $x^{2}$ never belongs to the linear span of $a+x+x^{2}$ and $1+x^{2}$.

Answer 3

The statement is not true. $x^2$ belongs in the span of those three vectors iff $a \neq 1$ because $$x^2 = \frac{1}{1-a}\left[1\cdot(a+x+x^2)-a\cdot(1+x^2)-1\cdot(x)\right]$$