# Finding a spanning set of a null space

• Oct 2nd 2009, 02:38 AM
Mathmaticious
Finding 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
• Oct 2nd 2009, 05:41 AM
Swlabr
Quote:

Originally Posted by Mathmaticious
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

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 $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ then your space would be spanned by the vectors $(2,0,0,0)$, $(0,5,0,0)$, $(0,0,6,0)$ and $(0,0,0,1)$.

I hope that that helps...
• Oct 2nd 2009, 07:47 AM
aman_cc
Quote:

Originally Posted by Swlabr
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 $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ then your space would be spanned by the vectors $(2,0,0,0)$, $(0,5,0,0)$, $(0,0,6,0)$ and $(0,0,0,d)$.

I hope that that helps...

1. In your example $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ what is the relevance of 2,5,6? Isn't it just same as $\{a,b,c,d): a, b, c, d \in \mathbb{F} \}$

2. Nevertheless, is there a good/structured approach to find dimension and basis of a space given by something like $\{a,2a,a+b,b): a, b \in \mathbb{F} \}$
• Oct 2nd 2009, 08:17 AM
Swlabr
Quote:

Originally Posted by aman_cc
1. In your example $\{(2a,5b, 6c,d): a, b, c, d \in \mathbb{F} \}$ what is the relevance of 2,5,6? Isn't it just same as $\{a,b,c,d): a, b, c, d \in \mathbb{F} \}$

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 $\{a,2a,a+b,b): a, b \in \mathbb{F} \}$
Yes. The answer to your problem is either one or two, but is clearly two as the entries in the 1st and 4th positions are independent of one another. It is no more than the number of variables given, but can be less if they cancel out with one another.

$(1, 2, 1, 0)$ and $(0, 0, 1, 1)$ span the space you gave.
• Oct 2nd 2009, 08:24 AM
aman_cc
Quote:

Originally Posted by Swlabr
Yes. Yes it is.

Yes. The answer to your problem is either one or two, but is clearly two as the entries in the 1st and 4th positions are independent of one another. It is no more than the number of variables given, but can be less if they cancel out with one another.

$(a, 2a, a, 0)$ and $(0, 0, b, b)$ span the space you gave.

Thanks Swalbr
If you won't mind can you explain why you say - "It is no more than the number of variables given"?
Also is there is a formal way to solve such a problem - I am looking at something equivalent to row-reduction to find the rank of matrix. Thanks
• Oct 2nd 2009, 08:32 AM
Swlabr
Quote:

Originally Posted by aman_cc
Thanks Swalbr
If you won't mind can you explain why you say - "It is no more than the number of variables given"?
Also is there is a formal way to solve such a problem - I am looking at something equivalent to row-reduction to find the rank of matrix. Thanks

In the vectors you gave you had an $a$ and a $b$, so two variables. Say you had a subspace of vectors where you can write them like you did but with $n>0$ variables. Then you can easily split your general form into $n$ 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...
• Oct 2nd 2009, 08:35 AM
aman_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.)
• Oct 2nd 2009, 03:48 PM
HallsofIvy
I presume you mean the matrix
$\begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}$

The kernel is, by definition, the set of vectors $\begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}$ such that
$\begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

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 $\begin{bmatrix}x \\ (1/3)t \\ (-3/2)t \\ t\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}+ t\begin{bmatrix}0 \\ 1/3 \\ -3/2 \\ 1\end{bmatrix}$