# Homework Help: Linear Algebra: does it form a basis?

1. Dec 10, 2011

### brushman

1. The problem statement, all variables and given/known data
See attachment.

3. The attempt at a solution

I already did parts i and ii (correctly, I hope). On part iii I found 2 linearly independent elements to be: t+1, t^2 - 1.

However, I don't understand how to show that these form a basis of W. Because W is a subspace of P2, and P2 has dimension 3, we know W has at most dimension 3. But I only have 2 elements, not 3, so I can't use this to show that I have a basis. Perhaps the constraint that p(-1) = 0 reduces the dimension, but I don't know how to prove this.

Thanks.

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2. Dec 11, 2011

### ehild

A polynomial in W is can be represented by an ordered pair of numbers (a0, a1): p(t)=a0+a1t+(a1-a0)t^2. So the set W is isomorphic with the set of ordered pair of numbers ...

ehild

3. Dec 11, 2011

### genericusrnme

the constraint that -1 is a root gives
$a_0-a_1+a_2=0$
$a_0=a_1-a_2$

so then

$a(t^2-1)+b(t+1)=at^2-a+bt+a=at^2+bt+(b-a)$

since a is the coefficient on t^2 and b is the coefficient on t we can let
$a=a_2$ and $b=a_1$
to find
$a_2(t^2-1)+a_1(b+1)=a_2 t^2+a_1 t +(a_1-a_2)$
from the constraint we found $a_0=a_1-a_2$
and so
$a_2(t^2-1)+a_1(b+1)=a_2t^2+b_1t+a_0$

you follow?

I should also add, every constraint you add to a system reduces the dimension of it by 1, adding another root would make it a 1 dimensional problem

4. Dec 11, 2011

### ehild

Not really... I would do it just on the opposite way.

Those polynomials which have root -1, are of form pa,b(t)=at^2+bt+(b-a), that is pa,b(t)=a(t^2-1)+b(t+1), which means all such polynomials are linear combination of the basic "vectors" p1=t^2-1 and p2=t+1. Two basic vectors define a two-dimensional subspace.

That is right.
With one more constraint, there would be only one free parameter, so the space would become one-dimensional.

ehild