# For what values of paramater form a spanning set for P3

1. Mar 24, 2005

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:
$$\left( \begin{array}{ccc} P_1(x) & 0 \\ P_2(x) & 0 \\ P_3(x) & 0 \\ \end{array} \right)$$
THEN the vectors are linearly dependent.

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

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?

Last edited: Mar 24, 2005
2. Mar 24, 2005

### 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)

3. Mar 24, 2005

For this example:

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

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

4. Mar 24, 2005

### 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...

5. Mar 24, 2005

### 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.

6. Mar 24, 2005

### 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.

7. Mar 24, 2005

### Data

yeah, good point. I'm tired :grumpy:

8. Mar 24, 2005

### Hurkyl

Staff Emeritus
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?

Last edited: Mar 24, 2005
9. Mar 29, 2005