# Vector Spaces

• Jul 5th 2011, 09:11 AM
Jav
Vector Spaces
Let W be the subspace of P3(all third degree polynomials) such that p(0)=0, and let U be the subspace of all polynomials such that p(1)=0. Find a basis for W, a basis for U, and a basis for their intersection W and U.
• Jul 5th 2011, 10:00 AM
FernandoRevilla
Re: Vector Spaces
I suppose you mean by $P_3$ the vector space of all polynomials of degree $\leq 3$. Choose a generic polynomial $p(x)=ax^3+bx^2+cx+d$ then, $p\in W\Leftrightarrow p(0)=0\Leftrightarrow d=0$ and $p\in U\Leftrightarrow p(1)=0\Leftrightarrow a+b+c+d=0$ . Could you continue?
• Jul 5th 2011, 03:33 PM
Deveno
Re: Vector Spaces
hint #2. suppose a+b+c+d = 0. if we choose a,b and c, does that tell us what d has to be? what does that tell you about dim(U)?

for the intersection, we have to have a+b+c = 0 and d = 0 (why?).
• Jul 5th 2011, 04:02 PM
Jav
Re: Vector Spaces
To Fernando:
I'm not sure, how to continue. Should I set this arbitrary polynomial in a matrix?
If "d" is always zero for W, then the basis has to include only the variables x, x^2, x^3?

As for U, I have no clue, how to set a basis. There can be a large amount of combinations of a+b+c+d that yield 0.
• Jul 5th 2011, 04:26 PM
Also sprach Zarathustra
Re: Vector Spaces
Quote:

Originally Posted by Jav
To Fernando:
I'm not sure, how to continue. Should I set this arbitrary polynomial in a matrix?
If "d" is always zero for W, then the basis has to include only the variables x, x^2, x^3?

As for U, I have no clue, how to set a basis. There can be a large amount of combinations of a+b+c+d that yield 0.

But not every $(a,b,c,d)\in\mathbb{R}^4$ satisfies $a+b+c+d=0$
• Jul 5th 2011, 04:50 PM
Jav
Re: Vector Spaces
Quote:

Originally Posted by Also sprach Zarathustra
But not every $(a,b,c,d)\in\mathbb{R}^4$ satisfies $a+b+c+d=0$

Yeah, if "a" is the coefficient of the variable with the leading exponent, it can happen that by adding two vectors the result is not in the space R^4.

-ax^3+bx^2+cx+d +
ax^3+ex^2+gx+d= (b+e)x^2...

How would I set up a basis then?
• Jul 5th 2011, 06:44 PM
Deveno
Re: Vector Spaces
here is my claim: dim(U) = 3. how can i prove this? i shall exhibit a basis. so i need to find 3 linearly independent polynomials p,q,r such that p(1) = q(1) = r(1) = 0.

for p, i choose p(x) = x-1. for q, i choose q(x) = (x-1)^2. for r, i choose (x-1)^3. are these linearly independent?

suppose ap + bq + cr = 0. that means: a(x-1) + b(x^2 - 2x + 1) + c(x^3 - 3x^2 + 3x - 1) is the 0-polynomial. now,

a(x-1) + b(x^2 - 2x + 1) + c(x^3 - 3x^2 + 3x - 1) = ax - a + bx^2 - 2bx + b + cx^3 - 3cx^2 + 3cx - c

= cx^3 + (b - 3c)x^2 + (a - 2bc + 3c)x - (a - b + c). if this is the 0-polynomial, we have to have c = 0. so

bx^2 + (a - 2b)x - (a - b) = 0. from this, we see that b = 0, as well. so:

ax - a = 0, which is only identically the 0-polynomial when a = 0.

how does this relate to my hint? if a+b+c+d = 0, if we choose any 3 numbers for a,b and c, then we are forced to set d = -a - b - c.

this suggests that picking 3 coefficients determines an element of U. the three basis vectors i chose correspond the the following choices for a,b,c,d:

x - 1 (a = 0, b = 0, c = 1, d = -1)
(x-1)^2 = x^2 - 2x + 1 (a = 0, b = 1, c = -2, d = 1).
(x-1)^3 = x^3 - 3x^2 + 3x - 1 (a = 1, b = -3, c = 3, d = -1).

i could have chosen p,q, and r quite differently. suppose i just worked from d = -a - b - c.

one "natural" choice might be: a = 1, b = c = 0, d = -1 ---> r(x) = x^3 - 1; a = c = 0, b = 1, d = -1 --> q(x) = x^2 - 1;

a = b = 0, c = 1, d = -1 --> p(x) = x - 1. you may verify for yourself that {x - 1, x^2 - 1, x^3 - 1} is also a basis for U.

now for W, it is clear that {x,x^2,x^3} is a linearly independent set, and that span({x, x^2, x^3}) is clearly a subspace of W.

suppose p(x) is in W. since p(0) = 0, p has a 0 constant term. hence p(x) = ax^3 + bx^2 + cx, which is in span({x, x^2, x^3}),

so {x, x^2, x^3} is a linearly independent spanning set for W. what do we call a linearly independent spanning set?

it would be nice if the bases we had exhibited so far had a nice intersection. but, oh well. to find a basis for U∩W, my advice is

to think about how you can factor polynomials for which p(0) = 0, and p(1) = 0. another approach:

dim(U) = dim(W) = 3. since neither U nor W is contained in the other, dim(U∩W) < 3, which leaves 0,1, or 2.

x^2 - x is in U∩W, so dim(U∩W) > 0. can you find another element of U∩W linearly independent from x^2 - x?
• Jul 5th 2011, 10:19 PM
FernandoRevilla
Re: Vector Spaces
Quote:

Originally Posted by Jav
To Fernando: I'm not sure, how to continue.

When we have the implicit equations of a subspace, it is just routine to find a basis of the subspace, you only need to solve the system . Working in coordinates with respect to $B=\{x^3,x^2,x,1\}$ we have

Basis of $W$

$W \equiv (a,b,c,d)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)$

so, $\{(1,0,0,0),(0,1,0,0),(0,0,1,0)\}$ are linearly independent an span $W$ . As a consequence a basis of $W$ is $B_W=\{x^3,x^2,x\}$ .

Basis of $U$

$U \equiv (a,b,c,d)=a(1,0,0,-1)+b(0,1,0,-1)+c(0,0,1,-1)$

so, $\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}$ are linearly independent an span $U$ . As a consequence a basis of $U$ is $B_U=\{x^3-1,x^2-1,x-1\}$

Basis of $U\cap W$

$U\cap W \equiv \begin{Bmatrix} a +b+c+d=0\\d=0\end{matrix}\Leftrightarrow \ldots$
$\Leftrightarrow (a,b,c,d)=b(-1,1,0,0)+c(-1,0,1,0)$

so, $\{(-1,1,0,0),(-1,0,1,0)\}$ are linearly independent an span $U\cap W$ . As a consequence a basis of $U\cap W$ is $B_{U\cap W}=\{-x^3+x^2,-x^2+x\}$
• Jul 6th 2011, 04:41 AM
Jav
Re: Vector Spaces
Quote:

Originally Posted by Deveno
here is my claim: dim(U) = 3. how can i prove this? ?

Thank you. It's a good insight, but I'm not well rounded in Linear Algebra to come up with claims yet. Perhaps in the future I can come up with this on my own.
• Jul 6th 2011, 04:47 AM
Jav
Re: Vector Spaces
Quote:

Originally Posted by FernandoRevilla
When we have the implicit equations of a subspace, it is just routine to find a basis of the subspace, you only need to solve the system . Working in coordinates with respect to $B=\{x^3,x^2,x,1\}$ we have

Basis of $U$

$U \equiv (a,b,c,d)=a(1,0,0,-1)+b(0,1,0,-1)+c(0,0,1,-1)$

so, $\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}$ are linearly independent an span $U$ . As a consequence a basis of $U$ is $B_U=\{x^3-1,x^2-1,x-1\}$

I understand that U can be expressed as {(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)} and this vectors are linearly independent and span the polynomial.
I've been taught to put this vectors in a matrix and solve in row echelon or reduced row-echelon to find a basis, but I don't understand your methods to get the basis of U.
Could you develop further?
• Jul 6th 2011, 05:52 AM
HallsofIvy
Re: Vector Spaces
You have a+ b+ c+ d= 0 and can solve that for any one component- for example, d= -a- b- c. Then any vector is of the form <a, b, c, d>= <a, b, c, -a- b- c>= <a, 0, 0, -a>+ <0, b, 0, -b>+ <0, 0, c, -c>= a<1, 0, 0, -1>+ b<0, 1, 0, 1>+ c<0, 0, 1, -1>.
• Jul 6th 2011, 06:05 AM
Jav
Re: Vector Spaces
Quote:

Originally Posted by HallsofIvy
You have a+ b+ c+ d= 0 and can solve that for any one component- for example, d= -a- b- c. Then any vector is of the form <a, b, c, d>= <a, b, c, -a- b- c>= <a, 0, 0, -a>+ <0, b, 0, -b>+ <0, 0, c, -c>= a<1, 0, 0, -1>+ b<0, 1, 0, 1>+ c<0, 0, 1, -1>.

I understand, but why is the basis x^(3) -1, x^(2) -1, x-1 ?
• Jul 6th 2011, 06:56 AM
Deveno
Re: Vector Spaces
in P3, x^3 - 1 is the vector (1,0,0,-1) relative to the basis {x^3,x^2,x,1}: x^3 - 1 = (1)x^3 + (0)x^2 + (0)x + (-1)1.

similarly x^2 - 1 is the vector (0,1,0,-1) relative to that same basis, and x - 1 is (0,0,1,-1).

ax^3 + bx^2 + cx + d <--> (a,b,c,d).
• Jul 6th 2011, 07:03 AM
Jav
Re: Vector Spaces
Thank you Deveno and most of you. Particularly, to Fernando. You are very clear.
:)
• Jul 6th 2011, 09:58 AM
Isomorphism
Re: Vector Spaces
The new question has been moved here. The posts here must be related to the original problem only.

Jav,
Two suggestions for a healthy stay at MHF:

1) Start a new thread for a new question.