The problem involves proving a relationship between the dot product and matrix transformations in the context of linear algebra, specifically focusing on the properties of matrices and their transposes. The original poster presents a statement regarding the equality of dot products involving matrices and vectors.
The discussion is ongoing, with participants providing feedback on the original poster's reasoning and suggesting further exploration of the implications of the established relationships. Some participants have offered guidance on using specific mathematical concepts, such as orthonormal bases, to aid in the proof.
Participants express uncertainty about the assumptions made in the problem and the implications of the relationships being explored. There is a focus on ensuring that the proofs hold for all vectors in the vector space, not just specific instances.
Wildcat said: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 said: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 said:Thanks, I need to think a little more
Wildcat said: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 said: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 said: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 said: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?
Dick said: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 said:then ei^T Aej = (ei^T A) ej = <A^Tei, ej> which would be the ijth entry of A^T
?
Dick said: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 said: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 said:x*x^T isn't a number. It's a nxn noninvertible matrix. x^T*x is a number.
Wildcat said: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 said: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.