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basis for a subspace watch

1. Hi,

I'm struggling with question 3 (shown in photos). I've managed to prove that it is a subspace. However my calculations aren't working out to find a basis.

Photos attached of solutions
Attached Images

2. (Original post by Roxanne18)
Hi,

I'm struggling with question 3 (shown in photos). I've managed to prove that it is a subspace. However my calculations aren't working out to find a basis.

Photos attached of solutions
You seem to have gone wrong earlier on, and don't have the correct subspace. Any vector in U must satisfy both criteria.

For a vector in the subspace, you've shown x=(3/4) y

And also have 0x -7y-3z=0

From this you can express both y and z in terms of x, for example.

So, a general vector in U will be....
3. (Original post by ghostwalker)
You seem to have gone wrong earlier on, and don't have the correct subspace. Any vector in U must satisfy both criteria.

For a vector in the subspace, you've shown x=(3/4) y

And also have 0x -7y-3z=0

From this you can express both y and z in terms of x, for example.

So, a general vector in U will be....

However I am a bit confused. The subspace was given to me so unsure how I don't have the correct subspace?
I'm also unsure about the second equation

Should I be soling for a different variable ?
4. (Original post by Roxanne18)

However I am a bit confused. The subspace was given to me so unsure how I don't have the correct subspace?
Well, you were given *conditions* satisfied by elements of the subspace, but it was up to you to interpret those conditions correctly.

In particular, *both* conditions 4x-3y=0 *and* x+y+z=0 must be satisfied by every element of the subspace, whereas you seem to have thought it was sufficient for *either* condition to be satisfied.

This is a fundamental error and nothing done subsequently makes sense.
5. (Original post by DFranklin)
Well, you were given *conditions* satisfied by elements of the subspace, but it was up to you to interpret those conditions correctly.

In particular, *both* conditions 4x-3y=0 *and* x+y+z=0 must be satisfied by every element of the subspace, whereas you seem to have thought it was sufficient for *either* condition to be satisfied.

This is a fundamental error and nothing done subsequently makes sense.

So how would I ensure both conditions were satisfied? I though that's why I subtracted them ?

I'm only in the second week of this year, so haven't had many lectures and struggling to follow.
6. (Original post by Roxanne18)

So how would I ensure both conditions were satisfied? I though that's why I subtracted them ?

I'm only in the second week of this year, so haven't had many lectures and struggling to follow.
We try to give hints rather than solutions, but given that you don't seem to have come across this stuff before, here's a bit of a guide ...

You have potentially 3 independent values x, y and z, but the fact that you are given 2 restrictions (equations) linking them means that you can write any two in terms of the third one. (N.B. this isn't absolutely guaranteed - in a different type of question you might be given equations that are inconsistent or duplicate each other!).

So, pretend that you are given a fixed value for one of them - say x = t. Can you then write y and z in terms of t using your familiar GCSE / A level techniques for solving simultaneous equations?

This process should give you a general element of your subspace.
7. (Original post by davros)
We try to give hints rather than solutions, but given that you don't seem to have come across this stuff before, here's a bit of a guide ...

You have potentially 3 independent values x, y and z, but the fact that you are given 2 restrictions (equations) linking them means that you can write any two in terms of the third one. (N.B. this isn't absolutely guaranteed - in a different type of question you might be given equations that are inconsistent or duplicate each other!).

So, pretend that you are given a fixed value for one of them - say x = t. Can you then write y and z in terms of t using your familiar GCSE / A level techniques for solving simultaneous equations?

This process should give you a general element of your subspace.
Okay, I think that kind of makes sense. I now understand how I completely ignored one of the conditions.

So using one equation I found the value for y, and then substituted that into the second equation, and found an expression as my photo shows.

But I dont understand (if that is even right) how I figure out a basis.
Attached Images

8. The vector you have at the end (1, 3/4, -7/4) is a basis (as is any non-zero multiple of it).

[It's actually fairly easy to find a basis when the basis only has one element. If instead you only had one condition (e.g. x+y+z = 0), then the basis would have had two elements (If your starting space is , and you have m equations, the basis will have at least n-m elements (and usually but not always exactly n-m)). This becomes a bit trickier...]
9. (Original post by DFranklin)
The vector you have at the end (1, 3/4, -7/4) is a basis (as is any non-zero multiple of it).

[It's actually fairly easy to find a basis when the basis only has one element. If instead you only had one condition (e.g. x+y+z = 0), then the basis would have had two elements (If your starting space is , and you have m equations, the basis will have at least n-m elements (and usually but not always exactly n-m)). This becomes a bit trickier...]
Oh is it? I always thought a basis was a collection of vectors, say 2 or more, as they should be linearly independent and span.

So how should I know how many elements are in the basis ?

10. (Original post by Roxanne18)
Oh is it? I always thought a basis was a collection of vectors, say 2 or more, as they should be linearly independent and span.
A collection can be any size (you could even have a collection with 0 elements). And if you only have 1 vector, it's certainly lin. indep.

So how should I know how many elements are in the basis ?
You have to check that all (simultaneous) solutions to the conditions can be written as a linear combination of the elements of the basis. If you find a solution that *can't* be written like this, then you've got a new vector to add to your basis.

It's not actually terribly easy to do this in general (and it's one of those things that is very hard to explain without actually doing an example), but you'll usually only get questions where it's not too bad.

Edit: there's also some stuff you can do based on the rank of the matrix you can make from the coefficients of the linear equations defining the subspace. But I am 100% sure you haven't covered this yet, so don't worry about it yet.

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