Let U be the subspace of R4 given by:

U = nullspace of the matrix

[0 0 2 3 ]

[0 -3 -2 -2 ]

Find a spanning set for U.

Any and all help will be awesome.

Not sure how to start at all (Worried)

Cheers

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- October 2nd 2009, 01:38 AMMathmaticiousFinding a spanning set of a null space
Let U be the subspace of R4 given by:

U = nullspace of the matrix

[0 0 2 3 ]

[0 -3 -2 -2 ]

Find a spanning set for U.

Any and all help will be awesome.

Not sure how to start at all (Worried)

Cheers - October 2nd 2009, 04:41 AMSwlabr
The first thing to do is to find out what the nullspace of your matrix actually is - can you find a general form that these vectors take?

Can you think of a spanning set for this vector space? For instance, if they were of the form then your space would be spanned by the vectors , , and .

I hope that that helps... - October 2nd 2009, 06:47 AMaman_cc
- October 2nd 2009, 07:17 AMSwlabr
Yes. Yes it is. No excuses, other than to point that I was still correct...I mean, 4+2=80712/13452...

Quote:

2. Nevertheless, is there a good/structured approach to find dimension and basis of a space given by something like

and span the space you gave. - October 2nd 2009, 07:24 AMaman_cc
- October 2nd 2009, 07:32 AMSwlabr
In the vectors you gave you had an and a , so two variables. Say you had a subspace of vectors where you can write them like you did but with variables. Then you can easily split your general form into vectors each with precisely one variable. In each vector you can take out this variable as a common factor, and the set of all of these vectors (the ones with the variables removed) forms a spanning set. You may be able to reduce this spanning set, if not then it is a basis.

Does that make sense?

Note also that a vector space cannot have a basis of order greater than the length of the vector... - October 2nd 2009, 07:35 AMaman_cc
Thanks. I need to think this carefully though. I was also wondering if it has to do something with the fact that each component is a linear combination of the variables. What happens if we relax that - for e.g. {a^2,b^3,a+b^-1,b} (I have not checked if it is a sub-space.)

- October 2nd 2009, 02:48 PMHallsofIvy
I presume you mean the matrix

The kernel is, by definition, the set of vectors such that

which means we must have 2z+ 3t= 0 and -3y- 2z- 2t= 0. Those two equations drop the dimension from 4 to 2. We can write z= (-3/2)t and then y= (-1/3)(2z+ 2t)= (-1/3)(-3t+ 2t)= (1/3)t. That is, y and z depend on t while x, since it does not appear in the equations can be anything. Use x and t as "free variables". Any vector in the kernel are of the form