Spanning sets and polynomials.

[-Dragoon-]

Homework Statement

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

Homework Equations

N/A

The Attempt at a Solution

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

$c_{1} + c_{3} - c_{4} = a$
$c_{2} + c_{4} = b$
$2c_{1} - c_{2} - 5c_{4} = c$
$c_{1} + 2c_{2} + 2c_{3} + 2c_{4} = d$
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:
$\left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&1&1&d-a-2b\\0&0&0&0&c-4a-3b+2d\end{array}\right]$

Although the book did a different last row operation in the solution, I came to the same conclusion as they did except they had -2 in the 3x3 and 3x4 position rather than 1 as in my solution. Anyways, the book concludes the system is inconsistent regardless of specific values chosen for a, b, c, and d. But, wouldn't the system be consistent if a = b = c = d = 0? Then the fourth row would be [ 0 0 0 0 | 0], correct?

 

[Essnov]

The point is that for any vector you should be able to come up with a linear combination of the coefficients to generate it.

Here, your coefficient matrix reduces to a rank 3 matrix, so your 4 vectors span a 3 dimensional space in P3. P3 is a 4 dimensional vector space and so there are definitely some vectors which you won't be able to generate.

Using an augmented matrix to figure out which vectors are spanned by your set is fine, but given the question "do the vectors span P3" you just need to see that the coefficient matrix is not rank 4 and therefore the set doesn't span P3.