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.

Re: Span and polynomials.

Quote:

Originally Posted by

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

My row reductions:

$\displaystyle \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]$$\displaystyle \rightarrow$$\displaystyle \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]$$\displaystyle \rightarrow$$\displaystyle \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]$$\displaystyle \rightarrow$$\displaystyle \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:

$\displaystyle r_{3} - 2r_{1} \rightarrow r_{3}$

$\displaystyle r_{4} - r_{1} \rightarrow r_{4}$

Second row operations:

$\displaystyle r_{3} + r_{2} \rightarrow r_{3}$

$\displaystyle r_{4} - 2r_{2} \rightarrow r_{4}$

Third row operation:

$\displaystyle r_{4} - \frac{1}{2}r_{3} \rightarrow r_{4}$

Re: Span and polynomials.

Quote:

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?

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).