# spanning set V

• Jul 16th 2013, 07:28 PM
n22
spanning set V
Hello,
How should I solve this question.thanks.

Let X = {f,g,h} where
f(x) = 3 + x, g(x) = 1 + x^5 and h(x) = 1 + x -2x^5
Let V ={f is an element of F such that f(x)=α+βx+γx^5

a)determine whether X is a spanning set for V.

(a) Determine whether X is a spanning set for V .
• Jul 16th 2013, 09:17 PM
hollywood
Re: spanning set V
Since $h(x)+2g(x) = 1+x-2x^5+2+2x^5 = 3+x = f(x)$, it can not be a spanning set. There are only two independent functions and the subspace has 3 dimensions.

- Hollywood
• Jul 17th 2013, 06:43 AM
HallsofIvy
Re: spanning set V
Hollywood's answer is best. Being not as sharp, I would have gone directly to the definition of "span":

Given any numbers, $\alpha$, $\beta$, and $\gamma$, do there exist numbers, a, b, and c, such that
$a(3+ x)+ b(1+ x^5)+ c(1+ x- 2x^5)= \alpha+ \beta x+ \gamma x^5$
$(3a+ b+ c)+ (a+ c)x+ (b- 2c)x^5= \alpha+ \beta x+ \gama x^5$

So $3a+ b+ c= \alpha$, $a+ c= \beta$, and $b- 2c= \gamma$
From $a+ c= \beta$, $c= \beta- a$.
From $b= \gamma+ 2c= \gamma+ 2\beta- 2a$.

Putting those into the first equation, $3a+ b+ c= 3a+ \gamm+ 2\beta- 2a+ \beta- a= \gamma+ 3\beta= \alpha$.

Since the "a"s cancelled we cannot solve for a.

In fact, in order that there exist such a representation of $\alpha + \beta x+ \gamma x^5$ we must have $\alpha= \gamma+ 3\beta$ so that $\alpha+ \beta x+ \gamma x^5= \gamma+ 3\beta+ \beta x+ \gamma x^5= (3+ x)\beta+ (1+ x^5)\gamma$. That is, $3+x$, $1+ x^5$, and $1+ x- 2x^5$ span the two dimensional subspace having $\{3+ x, 1+ x^5\}$ as basis.