Inner Products and Matrices

m.ooo

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Dec 1, 2021
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Hi everyone! I have an assignment that really confused me.



Define an appropriate matrix A of the corresponding inner product xTAyx^T A y that sets the angle between the vectors x=[1,1]Tx=[1,1]T and y=[2,3]Ty=[2,3]T to 60 degrees.



can someone help me out with this?
.
 
Is there anything in particular in the assignment which confuses you the most?
First hint: do you know how to compute the angle between two vectors?
 
Is there anything in particular in the assignment which confuses you the most?
First hint: do you know how to compute the angle between two vectors?
Yes, I know how to compute the angle.
I also know that the cosθ has to be 1/2 since the angle has to be 60 degrees. However, I don’t know how to define the matrix A.
 
Yes, I know how to compute the angle.
I also know that the cosθ has to be 1/2 since the angle has to be 60 degrees. However, I don’t know how to define the matrix A.
Good. What should be the value of the inner product of 'x' and 'y' if the cosine is 1/2 ?
 
Good. What should be the value of the inner product of 'x' and 'y' if the cosine is 1/2 ?
I found that the inner product of x,y is 5. So far I’ve came in conclusion that A=sqrt(26)/10. I’m just not sure if the matrix A has to be 1 x 1.
 
I found that the inner product of x,y is 5. So far I’ve came in conclusion that A=sqrt(26)/10. I’m just not sure if the matrix A has to be 1 x 1.
You are right, it is 5 in the "default" inner product, i.e. when A is identity (1001)\left(\begin{array}{rr}1&0\\0&1\end{array}\right). BTW, what angle do you get in this case?
You are right not to be sure that A is 1x1 because it is not. What dimensions does A have to have so that you could compute xTAyx^T A y ?
 
You are right, it is 5 in the "default" inner product, i.e. when A is identity (1001)\left(\begin{array}{rr}1&0\\0&1\end{array}\right). BTW, what angle do you get in this case?
You are right not to be sure that A is 1x1 because it is not. What dimensions does A have to have so that you could compute xTAyx^T A y ?
First of all, thank you for helping me out :)
The angle I found for the default inner product is approx 11.309 degrees.
A must be 2 x 2, I’m quite sure about that. But I’m really stuck for the next steps.

*edit : I think that its form has to be (a, 0, 0, b). Excuse me if I’m wrong
 
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Great! I got 11.3099 degrees -- close enough. And you are right, A must be 2 x 2!
Can you show the steps/formula how you got the angle?
 
Good, thank you. As you can see it has 3 inner products in there. Now you need to replace the implied A=IA=I in each of them with an unknown AA and solve the equation for AA's coefficients.
When I did this I got a pretty hairy equation. But I believe the solution is not unique since there are 3 unknown coefficients in AA. BTW, do you know why 3 and not 4? So I assigned fixed values to two of them and computed the third.
Hope this helps, and let us know how it goes.
 
Good, thank you. As you can see it has 3 inner products in there. Now you need to replace the implied A=IA=I in each of them with an unknown AA and solve the equation for AA's coefficients.
When I did this I got a pretty hairy equation. But I believe the solution is not unique since there are 3 unknown coefficients in AA. BTW, do you know why 3 and not 4? So I assigned fixed values to two of them and computed the third.
Hope this helps, and let us know how it goes.
Actually that’s my problem for 5 days now, this doesn’t seem to work for me. I cant compute any of the coefficients
 
What have you gotten so far?
I tried to modify the identity matrix A=(1001)A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} with so many ways, replacing the 1s with unknown variables, while at the same time I was multiplying it with each inner product. But I can't really come to a conclusion with any of the results I get.
 
Yeah, I've also learned that diagonal matrices don't do the job. I believe I've managed to solve it by looking at matrices with one unknown non-diagonal element, i.e.A=(1cc1)A =\left(\begin{array}{rr}1 & c \\ c & 1\end{array}\right)

A side note: I wonder how far, i.e., how large an angle, can one get with diagonal A's? I could not get 60 degrees, but I did not try to find the maximum -- did you get some insight there?
 
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