Linear combinations, span কী?

Example 18.18 (Testing a candidate basis, start to finish)

Is $\mathcal{F} = (1 + X,\ 1 + X^2,\ X + X^2)$ a basis of $\R_2[X]$? Write $u_1, u_2, u_3$ for the three polynomials. Freeness: a null combination $a\,u_1 + b\,u_2 + c\,u_3 = 0$ gives, coefficient by coefficient,

subtracting the first two, $b = c$, then the third gives $2b = 0$: $a = b = c = 0$, free. Generating: instead of solving three systems, notice the symmetric combination

so $1 = \frac12(u_1 + u_2 - u_3)$; then

The monomials lie in the span, so everything does: $\mathcal{F}$ is a basis. As a bonus, assembling the three displays gives the coordinates of any $P = \alpha + \beta X + \gamma X^2$:

(Sanity check with $P = X$: coordinates $\bigl(\frac12, -\frac12, \frac12\bigr)$, as found above.) Two lessons: symmetry in the family usually hides a shortcut combination; and once dimension is available (Chapter 19), the whole generating half of this work will come free of charge — three free vectors of a $3$-dimensional space always form a basis.

Example 18.22 (A hidden relation shrinks a span)

In $\mathcal{F}(\R, \R)$, what is $\operatorname{Vect}(1,\ \cos^2,\ \sin^2)$? The identity $\cos^2 + \sin^2 = 1$ is a nontrivial null combination

the family is linked, and the span is already generated by $(1, \cos^2)$ alone ($\sin^2 = 1 - \cos^2$). That smaller family is free: $a + b\cos^2 x = 0$ for all $x$ gives, at $x = 0$ and $x = \frac\pi2$: $a + b = 0$ and $a = 0$. So the span is a plane inside the function space — and it also contains $\cos 2x = 2\cos^2 x - 1$: linear-looking families of trigonometric functions routinely collapse under identities, which is why freeness must be proved, never assumed from the list’s length.