Math Is Fun Forum

Anakin

Member

Registered: 2009-10-04

Posts: 145

Span of Vectors

I was having some trouble understanding the whole "span" of vectors concept. I understand that it's the set of all linear combinations for a number of vectors (the definition) but I'm having trouble interpreting it.

Here is one of the examples we were given at the start of the lecture:

For what value(s) of "a" will y be in sp(v1, v2, v3) if v1 = [1, -1, -2], v2 = [5, -4, -7], v3 = [-3, 1, 0] and y = [-4, 3, a]?

I've tried to do some work:

The span of three first vectors is as follows -

r * v1 + s * v2 + t * v3 = [-r + 5s - 3t, -r - 4s + t, -2r - 7s] = sp(v1, v2, v3)

Now to see if y is in the span. I'm not sure how to find that, am I supposed to use the dot product? I don't think so because we covered that later on.

Thank you.

Edit: On second thought, I think my span is incorrect too. :\

Last edited by Anakin (2011-01-17 22:20:07)

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Span of Vectors

I'm gonna try again with the span:

[-r + 5s - 3t, -r - 4s + t, -2r - 7s] = {[x1, x2, x3] where x1, x2, x3 are real numbers} E R^3

x1 = -r + 5s -3t
x2 = x1 - 2r - 9s + 4t
x3 = x1 - 3r - 12s

Let x1 = 2.

We want x2 to be equal to zero (I saw something like this in class but I'm sure I'm doing it wrong lol).
0 = x2 = x1 - 2r - 9s + 4t (make r = 0, s = 0, t = - 1/2 to satisfy x2 = 0)
0 = x2 = 2 - 2

x3 = x1 - 3r - 12s (r = 0 and s = 0 from above)
x3 = 2 - 0 = 2

So the span of the vectors is [2, 0, 2]. That's so off, this is embarrassing but I had to give it a shot.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: Span of Vectors

hi Anakin

I think you want (-4  3  a) = r(1  -1  -2)  + s(5  -4  -7)  + t(-3 1 0)

That is to say, make y a linear combination of the other three vectors.

This leads to three simultaneous equations.  ( with four unknowns; normally a problem; but .....)

Use the first and second to eliminate 't'.

Compare the result with the third equation and the value of 'a' will be obvious.

Post again if this doesn't resolve the question for you.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Span of Vectors

Thanks for the reply Bob. I ended up getting a = 5. Is that the one and only solution to the problem?

And also, why did we want to make y a linear combination of the other three vectors?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: Span of Vectors

hi Anakin,

Yes, I got 5 too.  And no other solution.

The way I see it is this.

Given some vectors, you can make new vectors by making linear combinations from them.  Not necessarily every random vector can be made as a linear combination, so those that can are 'special'.  They are the ones that are 'in the span' of the start vectors.

So if you are given a new vector how do you know whether it is in the span?

Well just try to form it as a linear combination and see if you can.  It may not be possible in which case it isn't in the span.  For instance, just pick a = 100 in the example and you cannot find r, s, and t at all.

But the question says it is in the span so plough (plow) through the equations and see what drops out.  Only if a = 5 does a combination become possible so that's the value of 'a' and it's unique.

I was going to add that linear combinations are always unique but then I realised that's not true.

eg. v1 = ( 1, 2, 3 )  v2 = ( 2, 4, 6 )  then y = ( 3, 6, 9 ) = 3 x v1 = 1.5 x v2, so that's two different combinations for a start. 

You may also be able to see that y2 = ( 3 , 6, 1001 ) cannot be made as a linear combination .... only vectors parallel to v1 or v2 can be, because of the components I chose for these two base vectors.  (I made them parallel).

I expect you are about to learn about vectors that can span the whole set of vectors and are therefore called a basis for the
space.  In the example, v1, v2 and v3 are not a basis.  You should  be learning soon why that is and what to look for to guarantee a basis.

Bob

Last edited by Bob (2011-01-18 02:10:31)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Span of Vectors

Whush, that was a large informative read. I think I understand it now. As a span is all of the linear combinations of the vectors, one vector (y in our case) must form a linear combination with the other vectors in order to be a part of the span, right?

And I think we'll probably be jumping to that soon in class but until then, I'm gonna keep on doing some practice questions until I get a hang of this stuff.

Thanks for your help, Bob.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: Span of Vectors

hi Anakin,

As a span is all of the linear combinations of the vectors, one vector (y in our case) must form a linear combination with the other vectors in order to be a part of the span, right?

That's exactly right.

Good luck with the studies.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob