Finding a Solution to a Complex Homework Problem

[jumbogala]

Homework Statement

I'm having a lot of trouble with this question. I don't even know where to begin.
span{X₁, X₂, ... ,X_q}. a matrix A is m x n, and it is a zero matrix. Prove that AX_p is NOT zero for some p.

Homework Equations

The Attempt at a Solution

I'm guessing I need to show that X_i 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?

 

[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.

 

[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.

 

[Dick]

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.

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?

 

[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?

 

[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?

 

[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?

 

[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?

 

[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?

 

[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?

 

[jumbogala]

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

 

[Dick]

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

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?

 

[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!

 

[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?

 

[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?

 

[Dick]

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?

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??

 

[jumbogala]

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

 

[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?

 

[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?

 

[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?

 

[jumbogala]

I thought the definition of span{X1 ... Xq} was a1x1 + ... q1Xq though =/ My prof didn't really explain what span was, so maybe I'm unclear on that.

x is in R^n... so therefore x is in span{X1 ... Xq}.

 

[Dick]

Try to get clear on that. span{X1...Xq} is the set of ALL vectors that can be written as a1*X1+...aq*Xq where the a's range over ALL their possible values. So yes, if x is in R^n and R^n is the span{X1...Xq} then x can be written in the form x=a1*X1+...+aq*Xn for SOME choice of a1...aq. Is this making sense?

 

[jumbogala]

oh okay, so it doesn't mean x is equal to the span of x, but rather equal to one particular linear combination of the vectors that span R^n, right?

 

[Dick]

oh okay, so it doesn't mean x is equal to the span of x, but rather equal to one particular linear combination of the vectors that span R^n, right?

Right, right, right! I think you are getting it. And X1...Xq span R^n. I wouldn't say 'one particular', but you can say 'at least one'. I'd be ok with that.

 

[jumbogala]

okay, that makes sense to me now! so going back to the proof, that vector x is in the span, so Ax = Aa1X1 + ... + AqXq?

 

[Dick]

okay, that makes sense to me now! so going back to the proof, do I need to somehow show that x is non zero?

Didn't we pick x to be nonzero? Like it was a column vector with a 1 in a certain place and zeros every else? So that Ax was nonzero. We have been digressing a bit.

 

[jumbogala]

yes, I edited my post after that. instead I think I might be able to use Ax = Aa1X1 + ... + AqXq to prove this, although I'm not quite sure how... A is not zero so multiplying it by a1 or q shouldn't change things, so then Ax = 0 = AXi? I somehow doubt that's right though haha.

 

[Dick]

Isn't A(a1*X1)=a1(AX1) for example? Matrix times real number times column vector=real number times matrix times column vector. Visualize matrix multiplication again.

 

[jumbogala]

so instead 0 = a1(A*X1) + ... + q(A*Xq). And since A is non zero A*Xi shouldn't be zero either.

 

[Dick]

so instead 0 = a1(A*X1) + ... + q(A*Xq). And since A is non zero A*Xi shouldn't be zero either.

0=a1*(A*X1)+...+aq*(A*Xq)? NOO. Ax=a1*(A*X1)+...+aq*(A*Xq)! Ax is nonzero. Now what are you trying to prove again?

 

[jumbogala]

Aah Ax is non zero! for some reason I was thinking Ax was equal to zero.

So instead a1*(A*X1)+...+aq*(A*Xq) is NOT equal to zero, meaning the A*Xis can't be equal to zero?

 

[Dick]

Right. SOME of the A*X's can be zero. But certainly not ALL of them. We PICKED x so that Ax was nonzero, right? Whew! Thanks for playing along.

 

[jumbogala]

so when the question says to prove that A*Xi is not equal to zero, that doesn't hold for EVERY Xi in the span? the way it is phrased makes it sound like I'm supposed to show that it's always true that A*Xi is not 0, but if Xi = 0 that would be untrue.

is that the end of the proof then?

 

[Dick]

The question says "Prove that AXi is NOT zero for some i.". I just looked it up and quoted it exactly. Wouldn't you agree that's the same thing as proving that "Not all of AXi are equal to zero"? The proof's not over until you say "I understand" and can repeat it. You don't have to repeat it here. Just say you can.

 

[jumbogala]

Yes I suppose that does make sense. I was thinking 'some i' was synonymous with 'all i'. I am pretty new to proofs and math that doesn't use numbers, as you can probably (certainly) tell, hah.

And yes, I could repeat it now. I'm not sure I could do other similar proofs on my own yet though... I suppose I'll have to practice a whole bunch. and get a tutor.

Anyway, 2 and a half hours later - thank you so so much! you have been a great help.