Does this set of polynomials span P3?

[mitch_1211]

hey i want to find out if the set
s = {t²-2t , t³+8 , t³-t² , t²-4} spans P3

For vectors, i would setup a matrix (v1 v2 v3 v4 .. vn | x) where x is a column vector (x , y ,z .. etc) and reduce the system. If a solution exists then the vectors span the space, if there are no solutions then the space spanned is either the line or plane made up of the x , y ,z = 0

My question is, how do u setup a similar system for polynomials?

i know that the colums would be
t³
t²
t
1

and then put in the polynomials like (p1 p2 p3 p4 .. ) but what do i augment this matrix with?

many thanks

mitch

 

[henry_m]

Hi,

It is an identical situation. Rather than having a vector $(w,x,y,z)=w \mathbf{i} + x\mathbf{j}+y\mathbf{k}+z\mathbf{l}$, you have a vector $(w,x,y,z)=w 1 +x t + y t^2 +z t^3$. The $\mathbf{i},\mathbf{j},\mathbf{k},\mathbf{l}$ and the $1,t,t^2,t^3$ are standard basis vectors, and the $w,x,y,z$ are components. In this notation, your basis is $\{(0,-2,1,0),(8,0,0,1),(0,0,-1,1),(-4,0,1,0)\}$, a trivial change which hopefully brings you into familiar territory!

 

[mitch_1211]

So after I've got the basis vectors you have described, I augment them in a matrix with w,x,y,z just like for usual vectors?

 

[micromass]

So after I've got the basis vectors you have described, I augment them in a matrix with w,x,y,z just like for usual vectors?

Yes!

 

[HallsofIvy]

Another way to do this, without directly using matrices, is to use the definition:
The set $\{t^2-2t , t^3+8 , t^3-t^2 , t^2-4\}$ spans P³ if and only if, for any a, b, c, d there exist $\alpha$, $\beta$, $\gamma$, $\delta$ such that
$$\alpha(t^2- 2t)+ \beta(t^3+ 8)+ \gamma(t^3- t^2)+ \delta(t^2- 4)= at^3+ bt^2+ ct+ d$$
which is the same as
$$(\beta+ \gamma)t^3+ (\alpha- \gamma+ \delta)t^2- (2\alpha+ \gamma- \delta)t+ (8\beta- 4\delta)= at^3+ bt^2+ ct+ d$$
That gives the four equations
$$\beta+ \gamma= a$$
$$\alpha- \gamma+ \delta= b$$
$$2\alpha+ \gamma- \delta= c$$
$$8\beta- 4\delta= d$$

The set spans the space if and only if it is possible to solve for $\alpha$, $\beta$, $\gamma$, and $\delta$ in terms of any numbers, a, b, c, and d.

Of course, solving that system of equations could be done in terms of the matrix of coefficients which gets right back to your method!
[tex]\beta+ \gam,

 

[mitch_1211]

Yes!

Perfect, thank you!

 

[mitch_1211]

Of course, solving that system of equations could be done in terms of the matrix of coefficients which gets right back to your method!

Thanks for explaining that, much appreciated.