How to calculate the basis of the subspace 'U'

My subspace is:

U = ax^3 + bx^2 cx+ d: d + a = b + 2c

How would I calculate this basis?

Reply 1

nuodai

17

The space of polynomials of degree 3 or less has dimension 4, and your subspace is the set of polynomials in this space whose coefficients satisfy a linear equation; so the subspace has dimension 4-1=3.

The simplest way I can see to do this is to express one of the coefficients in terms of the other three. Then you have free choice over the first three coefficients (i.e. 3 degrees of freedom = 3 dimensions), and the fourth is determined by the first three.

Reply 2

claret_n_blue

OP

17

Original post by nuodai

The simplest way I can see to do this is to express one of the coefficients in terms of the other three. Then you have free choice over the first three coefficients (i.e. 3 degrees of freedom = 3 dimensions), and the fourth is determined by the first three.

Excellent. Thanks for that. Couple of questions:

1) Does it matter what I rearrange it for? Because I got it as a = and that gave me a different answer to when I put it as d =, which is what the answers said.

2) What do you do if yu have 2 equals signs? Another question gives me a subspace as:

U = ax^3 + bx^2 +cx + d : c + 2d = a - b = 2c + d

How would I do this? I'm assuming similar method but where would I rearrange everything?

3) Is there a "set" method in calculating the dimension of the intersection between two vector spaces?

Reply 3

nuodai

17

Original post by claret_n_blue

Excellent. Thanks for that. Couple of questions:

1) Does it matter what I rearrange it for? Because I got it as a = and that gave me a different answer to when I put it as d =, which is what the answers said.

2) What do you do if yu have 2 equals signs? Another question gives me a subspace as:

U = ax^3 + bx^2 +cx + d : c + 2d = a - b = 2c + d

How would I do this? I'm assuming similar method but where would I rearrange everything?

3) Is there a "set" method in calculating the dimension of the intersection between two vector spaces?

1. I imagine your answers are actually equivalent, just written differently.

2. Well what you actually have is two equations, which you could write separately as

\begin{cases} c+2d = a-b \\ c+2d = 2c+d \end{cases}

and then you could eliminate as usual. (Rearranging the second equation reveals something quite useful for when it comes to using the first equation.)

3. I dunno, maybe, but if there is it's probably not as good as common sense! Just see which elements of one space also belong to the other; this set is the intersection.

Reply 4

claret_n_blue

OP

17

Original post by nuodai

2. Well what you actually have is two equations, which you could write separately as

\begin{cases} c+2d = a-b \\ c+2d = 2c+d \end{cases}

and then you could eliminate as usual. (Rearranging the second equation reveals something quite useful for when it comes to using the first equation.)

When I rearrange, do I need to put it as one thing? Because I have got c = a - b - 2d and c = d. Then do I put these together and get a = b - 3c and then use this to work out my basis?

Original post by nuodai

3. I dunno, maybe, but if there is it's probably not as good as common sense! Just see which elements of one space also belong to the other; this set is the intersection.

How lol? I can't seem to do this as the vectors r in different forms each time. And this messes me up because the question after is to find the dimension of the sum of the spaces and I can't do that without knowing the intersection. Unless there's a way to work that out?

Reply 5

nuodai

17

Original post by claret_n_blue

When I rearrange, do I need to put it as one thing? Because I have got c = a - b - 2d and c = d. Then do I put these together and get a = b - 3c and then use this to work out my basis?

I don't really know what you're asking. But when you have two (linearly independent) equations like this, it will reduce the dimension of the larger space by two. So in this case your larger space has dimension 4, and so the subspace will have dimension 2, which means you'll be able to express each of the coefficients in terms of two variables. You've already shown that c=d, which means you can change all instances of 'd' to 'c' cutting it down to three variables, and the equation a=b-3c means that you can now express 'a' in terms of 'b' and 'c'. Doing this means that everything is now expressed in terms of 'b' and 'c': namely,

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

. So your subspace is

\{ (b-3c)x^3+bx^2+cx+c\, :\, b,c \in \mathbb{R} \}

. You can work out a basis from here. [Note you didn't need to express it in terms of 'b' and 'c'; you could have rearranged differently and expressed it in terms of 'a' and 'd', or 'b' and 'd', or whatever, or even introduced auxiliary variables which are themselves linear combinations of a,b,c,d and expressed it in terms of those.]

Original post by claret_n_blue

How lol? I can't seem to do this as the vectors r in different forms each time. And this messes me up because the question after is to find the dimension of the sum of the spaces and I can't do that without knowing the intersection. Unless there's a way to work that out?

Why don't you post the actual question? I can't make sense of what you're asking. In general you can find the dimension of the sum of two spaces by finding their dimensions and the dimension of their intersection, or you could do it by constructing a basis, or ... there are lots of methods.

Reply 6

claret_n_blue

OP

17

Original post by nuodai

Why don't you post the actual question? I can't make sense of what you're asking. In general you can find the dimension of the sum of two spaces by finding their dimensions and the dimension of their intersection, or you could do it by constructing a basis, or ... there are lots of methods.

I have two subspaces (U and W) of a veector space V where:

U = {ax³ + bx² + cx +d: a = 2(b + d) = 2a + c}

W = {ax³ + bx² + cx + d: c + 2d = a - b = 2c + d}

What are the dimension of U, W,

U \cap W

, and U + W.

To solve it, I need to work out the basis and check the amount of LI vectors in each basis to get a dimension, but I am also struggling as to how they got the basis of each subspace. For example, with U, they said 2x³ + x² - 2x and x² - 1 are basis but I can't seem to see how they worked it. If I had help with that, I could work out the basis (and then dimension) of W (but I still don't know how to calculate the dimension of U intersect W).

Reply 7

nuodai

17

Original post by claret_n_blue

I have two subspaces (U and W) of a veector space V where:

U = {ax³ + bx² + cx +d: a = 2(b + d) = 2a + c}

W = {ax³ + bx² + cx + d: c + 2d = a - b = 2c + d}

What are the dimension of U, W,

U \cap W

, and U + W.

To solve it, I need to work out the basis and check the amount of LI vectors in each basis to get a dimension, but I am also struggling as to how they got the basis of each subspace. For example, with U, they said 2x³ + x² - 2x and x² - 1 are basis but I can't seem to see how they worked it. If I had help with that, I could work out the basis (and then dimension) of W (but I still don't know how to calculate the dimension of U intersect W).

To find a basis it's useful to spot what the dimension of your subspace is and then find that number of linearly independent vectors. Let's take U for example. We have two linear equations:

a=2(b+d)

and

a=2a+c

, and the larger space has dimension 4, so the subspace has dimension 2, so a basis will consist of two vectors. This means we'll have freedom of choice over two of the coefficients of the polynomials in the space.

So consider

ax^3+bx^2+cx+d

. Say we choose

d=0

. Then our equations tell us that

a=2b

and

a=-c

, so putting

b=1

gives

a=2,c=-2

so that

2x^3+x^2-2x

lies in the subspace; this can be our first basis vector. Then we want another basis vector which has to be linearly independent from this one... clearly any vector (polynomial) with nonzero constant term is linearly independent from this, so try putting

a=0

. Then we see that

c=0

and

b=-d

, so putting

b=1

gives

x^2-1

, which is definitely linearly independent since its constant term is nonzero. So a possible basis is

\{ 2x^3+x^2-2x, x^2-1 \}

.

By construction these two vectors are linearly independent; you can verify (if you so wish) that they're a spanning set by checking that if

ax^3+bx^2+cx+d

has

a=2(b+d)=2a+c

then there exist

\lambda, \mu

such that

ax^3+bx^2+cx+d = \lambda(2x^3+x^2-2x) + \mu(x^2-1)

. This is tedious but not hard, but also not necessary since we know it has to be a spanning set since the dimensions match up.

A similar technique shows the dimension of W is 2.

Now for

U \cap W

, suppose

p(x) = ax^3+bx^2+cx+d \in U \cap W

. Then its coefficients must satisfy the equations for both spaces, so

a=2(b+d)=2a+c

and

c+2d=a-b=2c+d

. Simplifying a bit, we have that

a=2a+c

so

a=-c

. Substituting this all round we get

a = 2(b+d)

and

2d-a = a-b = d-2a

. Now we have that

2d-a=d-2a

, and so

d=a

, so substituting all round again we have

a=2(b+a)

and

a=a-b

.

I'm at risk of completing the question for you here... basically, keep playing with these equations and you'll find relations linking a,b,c,d. The number of degrees of freedom you're left with at the end is the dimension of

U \cap W

.

If there's anything I've said or done that you don't understand then you'll need to consult your notes or a textbook -- this stuff is standard and I'm in no position to teach you it.

(edited 12 years ago)

How to calculate the basis of the subspace 'U'

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you