# Spanning set

1. Oct 8, 2008

### d_b

Hi! I'm a bit confuse about spanning set.

could anyone give me an example of how to prove that a matrix with 1 column be a spanning set of R4? There isn't a clear example in the text book.

ex: if A={[c1, c2, c3....,cn]} prove that it is span R4.

Last edited: Oct 8, 2008
2. Oct 8, 2008

### JG89

If you mean a set with one matrice that spans all of R4, there is no such thing. If it doesn't matter how many matrices are in it, as long as it's one column, then the simplest spanning set I an think of is {[1 0 0 0], [0 1 0 0], [0 0 1 0], [0 0 0 1]}

3. Oct 8, 2008

### d_b

No its not it. Let say I have four vectors A,B,C,D (i.e.ABCD could be anything of the sort [1,2,3,4], any real numbers). Now how do i prove that it would span R4?

The book said to used block multiplication theorem but i'm not sure how to use it.

4. Oct 8, 2008

### JG89

If let's say A,B,C,D span R4 then c1A + c2B + c3C + c4D = [x,y,z,w] where the ci's are real numbers. You just have to prove (I would do it through Gaussian elimination) that it has solutions.

5. Oct 8, 2008

### Defennder

What is "block multiplication theorem" ? Certainly I haven't heard of that. One way you could do this is to invoke the dimension theorem for say matrix A whose columns are vectors a,b,c,d which you have to prove they span R4.

So dim(Colspace(A)) + dim( nullspace(A) ) = no. of columns of A.

So we need to show that nullity(A) = 0, since if rank(A) = 4 = dim(R4) then the column vectors span R4. So this reduces to showing that only the trivial solution satisfies Ax=0, since otherwise nullity(A) $$\neq 0$$.

EDIT: Given the above, note that you can easily check if A is of full rank. It's computationally quicker than verifying it by Gaussian elimination.

Last edited: Oct 9, 2008
6. Oct 9, 2008

### Tac-Tics

To show that a set of vectors spans a space, you must show that any vector in the space can be written as a linear combination of vectors in that set.

If the set is something easy, such as S = {[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]}, you would prove it like this.

Let v = [x, y, z, w] be any vector in R^4. Since v = [x, y, z, w] = x * [1, 0, 0, 0] + y * [0, 1, 0, 0] + z * [0, 0, 1, 0] + w * [0, 0, 0, 1], v can be written as a linear combination of vectors in S. Thus, S spans R^4.

For less obvious sets, the proof might be a little more difficult, but ultimately, it involves showing that given [x, y, z, w], you can find proper coefficients for the linear combination to make it true.

7. Oct 24, 2008

### d_b

that's awesome...thank you!!! that is excately what i was looking for, it make perfect sense because it is excately the samething as in R*3 ....

8. Oct 25, 2009

### bassmish

First ever Post!...

hi, just wondering what it would mean for a set of matrices to for a spanning set for C^4 as opposed to R^4

Also would it be right to say that a set of matrices form a spanning set when their determinant is non-zero???

thanks

9. Oct 25, 2009

### bassmish

just noticed this thread is a year old!...