# Homework Help: Find a basis

1. Feb 3, 2009

### fk378

1. The problem statement, all variables and given/known data
Let W equal the set of all polynomials in F[x] with degree less than or equal to n-1 such that the sum of the coefficients of the terms is 0. Find a basis of W over F.

3. The attempt at a solution

I don't know where to begin to find the basis. Do I use the fact that the sum of the coefficients is 0?

Last edited: Feb 4, 2009
2. Feb 4, 2009

### Office_Shredder

Staff Emeritus
Is not a vector space. Do you mean with degree less than or equal to n-1?

3. Feb 4, 2009

### fk378

yes I just fixed it.

4. Feb 4, 2009

### Office_Shredder

Staff Emeritus
You definitely need to use the fact that the sum of the coefficients is zero. Notice in particular that if f1,...,fm is a basis, then the sum of the coefficients in each fi must be zero also. Try writing down some simple polynomials that satisfy this

5. Feb 4, 2009

### fk378

can you just use
1-x^(n-1)?

The first coefficient is 1 and the (n-1)th coefficient is -1. But I don't think that is a basis.....I don't see how to find a basis for it and find coefficients at the same time. Can it be alternating positive and negative ones?

6. Feb 5, 2009

### Office_Shredder

Staff Emeritus
That's actually a fairly good start.

Here's something that can help with problems like this: The dimension of the set of all polynomials of degree less than or equal to n-1 is n. Considering the canonical basis{1,x,x2,...,xn-1} you have a single linear equation in the coefficients of linear combinations. So intuitively, the dimension of the new subspace should be n-1. So try finding n-1 polynomials that look like the one in your post

7. Feb 5, 2009

### cesc

check that

The vector space of set of the polynomials is isomorphic to the space X
a0+a1+a2=...an-1=0

now find a basis for X

8. Feb 5, 2009

### fk378

Can you explain this more? I don't understand the part where you say we have a "linear equation in the coefficients of linear combinations". Do you just mean that W is made up of linear equations?

And how did you get that the dimension of the new subspace is n-1? Is it because we're saying dimW=n so a subspace of W must be n-1? But why are we trying to find a *new* subspace when we just need a basis for W?

9. Feb 5, 2009

### Office_Shredder

Staff Emeritus
Looking back, I definitely didn't phrase it very well.

Sorry to confuse you. I meant that the condition for the subspace is that your vector has to satisfy a single linear equation. In general, that decreases the dimension of your subspace by 1 for each equation as long as they are independent from each other. So we know we should be looking for a vector subspace of dimension n-1

10. Feb 5, 2009

### fk378

By subspace do you mean basis? And why does it decrease the dimension by 1 for each equation?

11. Feb 5, 2009

### fk378

Here is what I have so far.

If the sum of the coefficients (call each coefficient w_i) = 0 then we can have a vector of all the w_i's (n-1 dimension) and an n-1 dimension unit vector. Multiplying these two vectors together we get 0.

We have w_0= -(summation w_i from 1 to n-1)
implies -(w_0)= summation w_i from 1 to n-1

We can write M= w_1 (-1,1,0,..0) + w_2 (-1,0,1,0...0) +.....+ w_n-1 (-1,0,...,1).

Then the basis for the whole space W (from the question) is made up of the vectors
(-1,1,0,..0)
(-1,0,1,0..0)
.
.
.
.
.
(-1,0,...,1)

Is this right?

12. Feb 6, 2009

### HallsofIvy

Yes, that's right. Generally in problems like this you can take whatever equations you have (here is is that $w_1+ w_2+ ...+ w_{n-1}= 0$) and solve for as many of the numbers as you can in terms of the others. Then take each of those "others" 1 in turn.

Since here you have only one equation, you can solve for one of them, say $w_1= -(w_2+ ...+ w_{n-1})$. Now let $w_2= 1$, $w_2= ...= w_{n-1}= 0$ and you get $w_1= -1$ so your first basis vector is (-1,1,0, ..., 0) as you say. Taking $w_3= 1$, $w_2= w_4= ...= w_n= 0]/itex] you get [itex]w_1= -1$ you get (-1, 0, 1, 0, ...,0), again as you have.

Because this is a set of polynomials, it would be better to write the "vectors" in that way: the basis is the set
$$\{x-1, x^2- 1, x^3- 1, \cdot\cdot\cdot, x^{n-1}-1\}$$

13. Feb 6, 2009

### fk378

How do you get from the list of vectors to that basis you just got for polynomials?
In other words, how do you get from (-1,1,0..0) to (x-1)?

Is it because x=1
so x-1=0?
what happened to the first term that equals -1?

Last edited: Feb 6, 2009