# Homework Help: Spanning proof

1. Jan 20, 2010

### jumbogala

1. The problem statement, all variables and given/known data
I'm having a lot of trouble with this question. I don't even know where to begin.
span{X1, X2, ... ,Xq}. a matrix A is m x n, and it is a zero matrix. Prove that AXp is NOT zero for some p.

2. Relevant equations

3. The attempt at a solution
I'm guessing I need to show that Xi is never zero or something, because that way AXi will also not be zero. Since the span there is equal to aX1 + bX2 + ... + qXq, it doesn't make sense that Xi would be zero ever.

I'm sure that's wrong, because there are cases where two non zero matrices multiplied give a zero matrix. Also, I think my proof should be more mathematical. Can anyone steer me in the right direction?

Last edited: Jan 21, 2010
2. Jan 20, 2010

### Dick

What have {X1, X2, ... Xq} got to do with the matrix A?? I think you've left a lot of the question out. If the span of the X's is in the null space of A that's certainly not true.

3. Jan 20, 2010

### jumbogala

Reading the question again, the only thing I left out was that the R should be to the power of n.

The only link I can see between A and span{X1, X2, ... , Xq} is that any of those Xs multiplied by A cannot give zero.

Also the question doesn't mention the null space of A.

4. Jan 20, 2010

### Dick

That helps a lot. So the X's span all of R^n. Define f(x)=Ax. f(x) maps R^n->R^m. If A is nonzero, can the range of f(x) be the zero vector in R^m?

5. Jan 20, 2010

### jumbogala

Okay, my first problem is: what's the difference between R^n and R^m? R^n is all the columns? I'm pretty new to all of this.

I'm guessing the range of f(x) can't be the zero vector?

6. Jan 20, 2010

### Dick

I'm just not using the funky font to write R. R^n is the same space you referred to in your original post. It's the space of all column vectors with n components. Similar for R^m. So you can see how Ax maps R^n to R^m, right?

7. Jan 20, 2010

### jumbogala

oh, I think I might see it now. if A is non zero then there can't be any zero vectors in R^m OR R^n, right? So f(x) = Ax means that x can't be zero. is that right?

if so where are you getting f(x) = Ax though? is that a given fact?

8. Jan 20, 2010

### Dick

No, you aren't getting it. Of course there's is a zero vector in R^n and R^m. It's the column filled with zeros. I'm just writing f(x)=Ax to emphasize Ax is a function of x. If x is a column vector in R^n then Ax is a column vector R^m. Are you following that?

9. Jan 20, 2010

### jumbogala

okay yes, that makes sense. because when you multiply a column vector with n components by A, you get a column vector with m components. correct?

10. Jan 20, 2010

### Dick

Ok, now can you PROVE that if A is nonzero then there is SOME vector in R^n such that Ax is nonzero? That's half the proof. Hint: if A is nonzero then some column must contain a nonzero entry. Suppose it's column number k. Can you think of a vector in R^n such that Ax is nonzero?

11. Jan 20, 2010

### jumbogala

Wouldn't any vector except the zero vector in R^n give an Ax that is nonzero?

12. Jan 20, 2010

### Dick

No! You aren't really visualizing this. Suppose A is the 2x2 matrix [[1,0],[0,0]]. I.e. first row 1 0 second row 0 0. Ok? A is nonzero, right? Take x to be the 2x1 column vector [0,1]. I.e 0 on the top and 1 on the bottom. x is nonzero, also right? What is Ax?

13. Jan 20, 2010

### jumbogala

Ax is [0,0].

but obviously there are some vectors that do not give [0,0], like if x is [2,3]... I don't know how to actually prove that though!

Last edited: Jan 20, 2010
14. Jan 20, 2010

### Dick

That was my hint. If column number k of A contains a nonzero entry and you define a column vector x to be all zeros except for a 1 in row k, can Ax be zero? Are you still visualizing matrix multiplication? What does Ax have to do with the columns in A?

15. Jan 20, 2010

### jumbogala

No, Ax cannot be zero, you will always get a one in row k of Ax.

And... if there are 2 columns in A, then Ax has that many rows?

16. Jan 20, 2010

### Dick

Hmm. I'm going to hope you are seeing this. If x is as I described it before, would you agree Ax IS column number k of A??

17. Jan 20, 2010

### jumbogala

wow, I definitely didn't see that before! but yes, I would agree with that.

18. Jan 20, 2010

### Dick

Ok, we're getting there! So if A is nonzero, then it must have a nonzero column. If A has a nonzero column you can pick a vector in R^n such Ax is that column, so Ax is nonzero. Now your question says span{X1,...Xq} is R^n. So x=a1*X1+...+aq*Xq for some choice of numbers a1...aq. Can you see where this is going?

19. Jan 20, 2010

### jumbogala

Nope, I'm having trouble making the connection now as to why x is equal to span{X1... Xq}.

I'm guessing but that one entry in A which is non zero would be multiplied by x?

20. Jan 20, 2010

### Dick

No, no, no again. x isn't EQUAL to span{X1...Xq}. The span is a SET of vectors. x is a single vector. x is IN the span. Please think hard about this, ok? Remember they told you span{X1...Xq} is EQUAL to R^n. Is x in R^n?