Linear Algebra / Dot Product / matrices

[Wildcat]

Homework Statement

Let V=Rⁿ and let AεMnxn(R) Prove that <x,Ay> = <A^T x, y> for all x,yεV

Homework Equations

The Attempt at a Solution

Can someone tell me if I'm on the right track?

<x,Ay> = x^T Ay = (x^TA)y = <A^T x,Y>

 

[micromass]

seems fine!

 

[Wildcat]

Thanks! Can I ask a second part to this or should I open a new thread?

 

[micromass]

No, you can ask in this thread if you want...

 

[Wildcat]

Suppose BεMnxn(R), and <x,Ay> = <Bx,y> for all x,yε V. Prove that B=A^T.

Wouldn't that be the same thing? or do I have to show something else?

 

[micromass]

No, this isn't the same thing (and I can't image that the proof is the same thing either).

What you actually need to show is that

$$<x,Ay>=<Bx,y>~\Leftrightarrow B=A^T$$

One implication is already proven in the OP, now you need to prove the other one.
So to prove this, you'll need to take an arbitrary matrix B such that <x,Ay>=<Bx,y>. And you'll need to show that B is actually the transposed of A. This is something quite different from what you've already done...

 

[Wildcat]

Suppose BεMnxn(R), and <x,Ay> = <Bx,y> for all x,yε V. Prove that B=A^T.

Wouldn't that be the same thing? or do I have to show something else?

Or assume B=A^T then <A^T x,y>= x^T A y = <x,Ay> ??

 

[micromass]

Well, you have proven that B=A^T implies that <x,Ay>=<Bx,y>. But now you have to do the reverse. You don't know that B=A^T, that's what you need to show. So you can't say "assume B=A^T", since there is no reason why you can assume this...

 

[Wildcat]

Well, you have proven that B=A^T implies that <x,Ay>=<Bx,y>. But now you have to do the reverse. You don't know that B=A^T, that's what you need to show. So you can't say "assume B=A^T", since there is no reason why you can assume this...

Thanks, I need to think a little more

 

[Wildcat]

Thanks, I need to think a little more

so I need to show B=A^T

if <x,Ay> = <Bx,y>
x^T Ay = x^TB^Ty
A=B^T then can i say A^T = (B^T)^T = B so B = A^T ??

 

[Dick]

so I need to show B=A^T

if <x,Ay> = <Bx,y>
x^T Ay = x^TB^Ty
A=B^T then can i say A^T = (B^T)^T = B so B = A^T ??

The trouble is that you can't just 'cancel' x^T and y from both sides of the equation to get A=B^T. It's not like x^T and y have inverses or anything like that. I'd suggest you pick a basis and argue from there.

 

[Wildcat]

The trouble is that you can't just 'cancel' x^T and y from both sides of the equation to get A=B^T. It's not like x^T and y have inverses or anything like that. I'd suggest you pick a basis and argue from there.

I was afraid I couldn't cancel them. I'm not sure what you mean by pick a basis. I looked through the chapters on inner product spaces and adjoint of a linear operator and orthonormal basis is in some of the theorems but I'm not sure how to use that.

 

[Dick]

I was afraid I couldn't cancel them. I'm not sure what you mean by pick a basis. I looked through the chapters on inner product spaces and adjoint of a linear operator and orthonormal basis is in some of the theorems but I'm not sure how to use that.

Pick a basis {ei}. The A_ij entry of the matrix in that basis is <ei,Aej>, right? I might have have the indices backwards, but do you get my point?

 

[Wildcat]

Pick a basis {ei}. The A_ij entry of the matrix in that basis is <ei,Aej>, right? I might have have the indices backwards, but do you get my point?

Still not sure I'll study it awhile and see if I can figure it out.

 

[Wildcat]

Pick a basis {ei}. The A_ij entry of the matrix in that basis is <ei,Aej>, right? I might have have the indices backwards, but do you get my point?

then ei^T Aej = (ei^T A) ej = <A^Tei, ej> which would be the ijth entry of A^T
?

 

[Dick]

then ei^T Aej = (ei^T A) ej = <A^Tei, ej> which would be the ijth entry of A^T
?

I'm picking the ei to be the column vector with a 1 in the ith place and zeros elsewhere. So <ei,Aej> would the ijth entry of A. That makes <A^Tei,ej>=<ej,A^Tei> the jith element of A^T. And, sure, they are equal. That's what the transpose is in terms of matrices, right? The point here is that you know (x^T)Ay=(x^T)(B^T)y. If you put x=ei and y=ej, doesn't that show that the ijth entry of A is equal to the ijth entry of B^T? That's how you can conclude A=B^T.

 

[Wildcat]

I'm picking the ei to be the column vector with a 1 in the ith place and zeros elsewhere. So <ei,Aej> would the ijth entry of A. That makes <A^Tei,ej>=<ej,A^Tei> the jith element of A^T. And, sure, they are equal. That's what the transpose is in terms of matrices, right? The point here is that you know (x^T)Ay=(x^T)(B^T)y. If you put x=ei and y=ej, doesn't that show that the ijth entry of A is equal to the ijth entry of B^T? That's how you can conclude A=B^T.

I see. Let me ask you something about my first statement


if <x,Ay> = <Bx,y>
x^T Ay = x^TB^Ty
A=B^T then can i say A^T = (B^T)^T = B so B = A^T

I understand I can't cancel the x^T and y but could I multiply both sides by x then wouldn't x*x^T become a number that I could cancel then do the same with right hand multiply by y^T??

 

[Dick]

I see. Let me ask you something about my first statement


if <x,Ay> = <Bx,y>
x^T Ay = x^TB^Ty
A=B^T then can i say A^T = (B^T)^T = B so B = A^T

I understand I can't cancel the x^T and y but could I multiply both sides by x then wouldn't x*x^T become a number that I could cancel then do the same with right hand multiply by y^T??

x*x^T isn't a number. It's a nxn noninvertible matrix. x^T*x is a number.

 

[Wildcat]

x*x^T isn't a number. It's a nxn noninvertible matrix. x^T*x is a number.

how about y * y^T that would be a number that could cancel?
then could you transpose both sides to get
(x^T Ay)^T = (x^TB^T)^T
then A^Tx = Bx which means A^T = B ?

 

[Dick]

how about y * y^T that would be a number that could cancel?
then could you transpose both sides to get
(x^T Ay)^T = (x^TB^T)^T
then A^Tx = Bx which means A^T = B ?

Why are you still looking for a number to cancel? It's possible that x^TAy=x^TBy for some particular values of x and y even if A is not equal to B. You have to use the fact it's true for ALL x and y. Use an orthonormal basis.

 

[Wildcat]

Why are you still looking for a number to cancel? It's possible that x^TAy=x^TBy for some particular values of x and y even if A is not equal to B. You have to use the fact it's true for ALL x and y. Use an orthonormal basis.

It's not that I'm looking for a number to cancel, necessarily, I guess I'm trying to see if what I originally started with can be done that way. I do understand using the orthonormal basis just wondering if it could be done the other way.