# Math Help - Span and polynomials.

1. ## Span and polynomials.

Question: Do the polynomials $t^{3} + 2t + 1$, $t^{2} - t + 2$, $t^{3} +2$, $-t^{3} + t^{2} - 5t + 2$ span $P_{3}$?

My attempt: Let $at^{3} + bt^{2} + ct + d$ be an arbitrary vector in $P_{3}$, then:
$c_{1}(t^{3} + 2t + 1) + c_{2}(t^{2} - t + 2) + c_{3}(t^{3} +2) + c_{4}(-t^{3} + t^{2} - 5t + 2) = at^{3} + bt^{2} + ct + d$

Combining the terms on the left side and equating corresponding coefficients gives the linear system:
$c_{1} + c_{3} - c_{4} = a$
$c_{2} + c_{4} = b$
$2c_{1} - c_{2} - 5c_{4} = c$
$c_{1} + 2c_{2} + 2c_{3} + 2c_{4} = d$

Which becomes the augmented matrix:
$\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\2&-1&0&-5&c\\1&2&2&2&d\end{array}\right]$
Row reducing it, I come to the matrix:
$\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&0&0&-\frac{1}{2}(c-2a+b)(d-a-2b)\end{array}\right]$

Since there are a row of zeros, then $-\frac{1}{2}(c-2a+b)(d-a-2b)$ must be equal to zero for the system to be consistent, and hence, the polynomials do not span $P_{3}$. However, my book concluded that the reason that the polynomials do not span $P_{3}$ is because it is an inconsistent system, that no value for a, b, c, and d would yield 0 for the 4th row? How did they come to this conclusion? Thanks in advance.

2. ## Re: Span and polynomials.

my first instinct is to say check your arithmetic for your row-reduction

EDIT: also, note that you are choosing a,b,c,d arbitrarily, you are not solving for them.

3. ## Re: Span and polynomials.

Originally Posted by Deveno
my first instinct is to say check your arithmetic for your row-reduction
My row reductions:
$\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\2&-1&0&-5&c\\1&2&2&2&d\end{array}\right]$ $\rightarrow$ $\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&-1&-2&-3&c-2a\\0&2&1&3&d-a\end{array}\right]$ $\rightarrow$ $\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&1&1&d-a-2b\end{array}\right]$ $\rightarrow$ $\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&0&0&-\frac{1}{2}(c-2a+b)(d-a-2b)\end{array}\right]$

First row operations:
$r_{3} - 2r_{1} \rightarrow r_{3}$
$r_{4} - r_{1} \rightarrow r_{4}$
Second row operations:
$r_{3} + r_{2} \rightarrow r_{3}$
$r_{4} - 2r_{2} \rightarrow r_{4}$
Third row operation:
$r_{4} - \frac{1}{2}r_{3} \rightarrow r_{4}$

4. ## Re: Span and polynomials.

Originally Posted by Deveno
EDIT: also, note that you are choosing a,b,c,d arbitrarily, you are not solving for them.
Yes, that is why if I were to set each one of a, b, c, and d to 0, the system should be consistent. No?

5. ## Re: Span and polynomials.

the product on the last row in the last column should be a sum.

if your four polynomials span P3, then one linear combination (at least) must be t^3 (which is a = 1, b = c = d = 0).

this leads to 0 = -2 (from the 4th row, where your augmented column should read (d-a-2b)+(1/2)(c-2a+b)), which is inconsistent.

the zero-vector (the polynomial 0t^3 + 0t^2 + 0t + 0 in this case) is ALWAYS in the span of ANY subset.

thus setting a = b = c = d = 0 is a spectacularly bad choice for a test polynomial, because it tells us nothing (you can

always make a linear combination of 0's be 0).