For what values of paramater form a spanning set for P3

[FrogPad]

I have this question, and I'm not really sure how to go about it. Any help would be appreciated:

* Here is the question that is asked. (It is supposed to be a general question, and the question will change for the test. Thus, there may be 4 poly's to work with, or 2... etc.) *

$$P_1(x)=x^2+\alpha$$
$$P_2(x)=x-\alpha$$
$$P_3(x)=x^2+x+1$$
For what values of parameter $$\alpha$$ form a spanning set for $$P_3$$.


* This is what I have so far. I'm not sure if I'm going about it right. So this is where I need help :) *
Ok, so I know that a spanning set of $$P_3$$ must contain at least 3 vectors that are linearly independent.
IF the system is true:
$$\left( \begin{array}{ccc}<br /> P_1(x) & 0 \\<br /> P_2(x) & 0 \\<br /> P_3(x) & 0 \\<br /> \end{array} \right)$$
THEN the vectors are linearly dependent.

So if we setup the system:
$$\left( \begin{array}{cccc}<br /> 1 & 0 & 1 & 0 \\<br /> 0 & 1 & 1 & 0 \\<br /> \alpha & -\alpha & 1 & 0 <br /> \end{array} \right)$$

wherever $$\alpha$$ causes the system to not equal 0 would be when the poly's span $$P3$$ right?

if this is right, then how do I show this?

 

[matt grime]

Forget the last column of 0s there is no need for it. You just need to show the row rank of the remaining 3x3 matrix is 3, so do it... (reduced elechon form, will be in your notes)

 

[FrogPad]

For this example:

ok after reducing the matrix I get:
$$\left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 <br /> \end{array} \right)$$

Therefore, does this mean that for any value of $$\alpha$$ the poly's form a basis for $$P_3$$ ?

 

[matt grime]

It may do - i can't say i bothered to do the question. To verify it, did you at any point do some thing that you aren't allowed to do? like divide by alpha, which isn't valid if alpha is zero...

 

[Data]

Well, if $\alpha$ is zero, then it's trivially a basis anyways for this example.

I don't know what else he could have done that would be illegal, unless it were something really silly. Then again I haven't done the question either!

Edit: I was bored and checked it. Indeed your row reduction is correct.

 

[matt grime]

he could have divided by a-1 (use a for alpha), or a+1, or any number of similar things depending on the question, which is the way to get exceptions in these kinds of questions, for example suppose in the row reduced form the final row is 0, 0, a+1 reading left to right, then it will be a basis if and only if a=/=-1.

 

[Data]

yeah, good point. I'm tired

 

[Hurkyl]

As it turns out, though, no division is necessary. But, more importantly, does the original poster know why he should be row reducing this matrix?

 

[FrogPad]

By placing P1(x), P2(x), and P3(x) in a matrix representation, then the rank of this matrix must equal 3 to span P3. So row reducing the matrix, ensuring that it is consistent, and then setting the system to 0 and solving for the unknowns allows one to deduce where the vectors are dependent/independent and then it can be shown where/if/ and when it spans P3.

I hope that's right :)

*by the way. thanks for the help. it's truly appreciated.