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

I hope that that helps... - Oct 2nd 2009, 06:47 AMaman_cc
@Swlabr: Few questions please

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

2. Nevertheless, is there a good/structured approach to find dimension and basis of a space given by something like $\displaystyle \{a,2a,a+b,b): a, b \in \mathbb{F} \}$ - Oct 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 $\displaystyle \{a,2a,a+b,b): a, b \in \mathbb{F} \}$

$\displaystyle (1, 2, 1, 0)$ and $\displaystyle (0, 0, 1, 1)$ span the space you gave. - Oct 2nd 2009, 07:24 AMaman_cc
- Oct 2nd 2009, 07:32 AMSwlabr
In the vectors you gave you had an $\displaystyle a$ and a $\displaystyle b$, so two variables. Say you had a subspace of vectors where you can write them like you did but with $\displaystyle n>0$ variables. Then you can easily split your general form into $\displaystyle n$ 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, 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.)

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

$\displaystyle \begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}$

The kernel is, by definition, the set of vectors $\displaystyle \begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}$ such that

$\displaystyle \begin{bmatrix}0 & 0 & 2 & 3 \\ 0 & -3 & -2 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ t\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

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