# Determining which sets are spanning sets for P3?

I have a list of sets:

(a): {$1, x^2, x^2 - 2$}

(b): {$2, x^2, x, 2x + 3$}

(c): {$x + 2, x + 1, x^2 - 1$}

(d): {$x + 2, x^2 -1$}

I am supposed to determine which of the following are spanning sets in $P_3$.

My attempted solution:

(A): Ax^2 + bx + c = A(1) + B(x^2) +C(x^2-2)

Ax^2 + bx + c = (A-2C) + Bx^2 + C(x^2)


This is pretty much as far as I've got and I am not exactly sure what I'm doing or solving for. Any clarification would be wonderful.

• it might be obvious but you might want to say what $P_3$ is – user87543 Feb 21 '14 at 6:52
• P3 are polynomials of degree less than 3. – KnowledgeGeek Feb 21 '14 at 6:52
• mention that in the question.... Do you think you can get $x$ as linear combination of elements of $\{1, x^2, x^2 - 2\}$ or $\{x + 2, x^2 -1\}$ – user87543 Feb 21 '14 at 6:56
• If I am understanding correctly, no. – KnowledgeGeek Feb 21 '14 at 6:59
• Keep in mind in dealing with problems like this one that you must be able to construct, by linear combination, polynomials consisting only of $\ Ax^2 \ , Bx ,$ or constants $\ C \$ , including zero. – colormegone Feb 21 '14 at 7:20

A spanning set for this set of polynomials, $\ ax^2 \ + \ bx \ + \ c \ ,$ must be a collection of polynomials, $\ p_i(x) \ ,$ that can be combined by linear combinations to produce all possible polynomials of second-degree or lower, including one-term polynomials, such as $\ ax^2 \ , \ bx \ ,$ and $\ c \ ;$ the "constant polynomials" must include the "zero polynomial" , $\ y = 0 \ .$

What we are doing is constructing a "vector space" of polynomials*, so it must be closed under addition and scalar multiplication (thus, under linear combination) and it must include a "zero element". There is no limit to the number of "vectors" (in this case, polynomials) that can appear in a spanning set, as long as all the polynomials in $\ P_3 \$ can be produced. (For esthetic reasons -- or reasons of sanity -- we usually prefer to find "minimal" spanning sets, ones with the fewest possible elements.)

$*$ We don't ordinarily think of polynomials as "vectors", but in this context, the arrangement of the polynomials under linear combinations has the "algebraic structure" of a vector space. So, in recent times, we talk about vector spaces of lots of mathematical objects that don't seem like vectors...

So we need to be able to produce $\ ax^2 \ + \ bx \ + \ c \$ from $\ C_1 \cdot p_1(x) \ + \ C_2 \cdot p_2(x) \ + \ \ldots \ + \ C_n \cdot p_n(x) .$ There are two choices we can eliminate immediately. Choice (D) only permits

$$ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( x + 2 ) \ + \ C_2 \ (x^2 - 1) \ = \ C_2x^2 \ + \ C_1 x + ( 2C_1 - C_2 ) \ ,$$

which "forces" the choices $\ C_1 = b \ \ \text{and} \ \ C_2 = a \ ,$ which permits us no control at all as to what value of $\ c = 2C_1 - C_2 \$ we'll end up with. Further, we can't produce a polynomial $\ ax^2 \$ or $\ bx \$ without having to have a non-zero constant term, nor can we get a "constant polynomial $\ c \$ without having other terms along with it. So (D) is definitely not a spanning set. Choice (A) is even worse because it doesn't even give us a way to produce a polynomial $\ bx \$ through linear combination, as none of the polynomials in the set have a linear term.

This leaves choices (B) and (C). It turns out both of these are spanning sets, as E-theory indicates. For (C), we have

$$ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( x + 2 ) \ + \ \ C_2 \ ( x + 1 ) \ + \ C_3 \ (x^2 - 1)$$

$$= \ C_3x^2 \ + \ (C_1 + C_2) x + ( 2C_1 + C_2 - C_3) \ ,$$

which corrects the deficiency of choice (D) by leaving us a "measure of control" over each coefficient: $\ C_3 \$ must equal $\ a \ ,$ but then we can adjust $\ C_2 \$ in the sum $\ C_1 + C_2 \$ in order to match $\ b \ ,$ and in turn adjust $\ C_1 \$ in $\ 2C_1 + C_2 - C_3 \$ to give us $\ c \ .$ This will mean that it is possible to calculate just what each unknown $\ C_i \$ must be in order to make any of the polynomials in $\ P_3 \ .$ Note also that we must match three coefficients in the polynomial and we have three "basis polynomials" to manage it with; this is a "minimal spanning set".

By the same token, choice (B) gives

$$ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( 2 ) \ + \ \ C_2 \ ( x^2 ) \ + \ C_3 \ (x) \ + \ C_4 \ (2x+3) .$$

We can see pretty readily that every power of $\ x \$ needed to "build" second-degree polynomials is present in the first three elements of the set, so we can just use $\ C_1 = \frac{c}{2} \ , \ C_2 = a \ , \ \ \text{and} \ \ C_3 = b \ .$ What about $\ (2x+3) \$ ? Well, it can certainly be incorporated into the calculation of the values for obtaining $\ b \ \ \text{and} \ \ c \ ,$ but we've also just seen that it isn't needful. So we can also choose to just "turn off" its contribution by always setting $\ C_4 = 0 \ .$ So choice (B) also gives us a spanning set for $\ P_3 \$ ; it just isn't minimal.

I found b) and c) are the spanning sets. For b) You only need to show that you can find reals m, n, p such that: ax^2 + bx + c = 2m + nx^2 + px. You don't need 2x + 3 since 2x + 3 = 2x + (3/2)*2 is in the span{2,x}. So choose n = a, p = b, and m = c/2. For c) again ax^2 + bx + c = m(x+2) + n(x+1) + p(x^2 - 1). Choose p = a, and : m = b - a - c, and n = a + c. For a) x^2 + 2x + 1 is not in the span{1, x^2, x^2 - 2}. For d) ax^2 + bx + c = m(x+2) + n(x^2 -1) ==> n = a , m = b, c = 2m - n = 2b - a. So you only need to find a quadratic polynomial like x^2 + x - 5, then it is not in the span{x+2, x^2 - 1}.

• I would greatly appreciate the details. I am lost in this topic. – KnowledgeGeek Feb 21 '14 at 7:18
• maybe drink a cup of chocolate....and get back to do homework... – DeepSea Feb 21 '14 at 7:46

$P_3$ has the basis $\{1,x,x^2\}$. Since each set $S$ is a subset of $P_3$, we have that $S$ spans $P_3$ if and only if $1$, $x$ and $x^2$ can be formed by linear combinations of the polynomials in $S$.

(a): $S=\{1, x^2, x^2 - 2\}$: can't form $x$ as a linear combination. So $S$ does not span $P^3$.

(b): $S=\{2, x^2, x, 2x + 3\}$: We have the linear combinations $$1=\tfrac{1}{2} 2+0x^2+0x+0(2x+3);$$ $$x=0 \times 2+0x^2+1x+0(2x+3);$$ $$x^2=0 \times 2+1x^2+0x+0(2x+3),$$ so $S$ spans $P^3$.

The other two are of the same flavour: either show (a) the polynomials $1$, $x$ and $x^2$ are linear combinations of the polynomials in $S$, or (b) that one of them cannot be formed as a linear combination for some reason.

• The third word in the above answer should better be "a" and not "the" – DonAntonio Feb 22 '14 at 14:34