How to take a matrix outside the diagonal operator?

In summary, the conversation is about deriving the equation trace(A*Diag(B*B^T)*A^T) = norm(W,2), where W = vec(sqrt(diag(A^T*A))*B) and sqrt(diag(A^T*A)) is the square root of diag(A^T*A). The conversation also discusses the conversion of equation 70 into equation 71 using the formula ||vec(A)||_2^2 = tr{AA^H}. The question is about the rearrangement of terms in equations 70 and 71 and how to prove the equality of the yellow highlighted terms in both equations.
  • #1
Umesh
3
0
How to derive (proof) the following

trace(A*Diag(B*B^T)*A^T) = norm(W,2),

where W = vec(sqrt(diag(A^T*A))*B)
&
sqrt(diag(A^T*A)) is the square root of diag(A^T*A),

B & A are matrix.

Please see the equation 70 and 71 on page 2068 of the supporting matrial.
 

Attachments

  • MSE-Based_Transceiver_Designs_for_Full-Duplex_MIMO_Cognitive_Radios.pdf
    1.3 MB · Views: 90
Physics news on Phys.org
  • #2
diag(A) can be expressed in suffix notation as [itex]\delta_{ijk}A_{jk}[/itex] where [tex]
\delta_{ijk} = \begin{cases} 1 & i = j = k, \\ 0 & \mbox{otherwise}. \end{cases}[/tex]
 
  • #3
Umesh said:
where W = vec(sqrt(diag(A^T*A))*B)
&
sqrt(diag(A^T*A)) is the square root of diag(A^T*A),
diag(##A^TA##) is a square matrix. If we let D = diag(##A^T A##), then what you're asking about is ##\sqrt D##. I don't see anywhere in equations 70 and 71 that they are taking the square root of a matrix. The only square roots I see are the square roots of what appear to be some constants.

Although it does make sense to take the square root of a square matrix (##A = \sqrt B \Rightarrow A^2 = B##) it's not clear to me whether this is what you mean or that you're taking the square root of each entry on the diagonal matrix.

Also, it's not clear to me why you are taking the square root of what I'm calling D.
 
  • #4
Please check only yellow highlighted terms on page number 2068.
Eq. 70 is converted into eq. 71 using formula ||vec(A)||_2^2 = tr{AA^H} (3rd line of appendix).
My question is
In eq 70, the term H*V*V^H*H^H is inside, and R is outside the diagonal operator as
tr{R* diag(H*V*V^H*H^H)*R^H} and
In eq 71, R is inside, and the HV term is outside the diagonal operator and written as
|| vec( ( diag(R^H*R) )^0.5*H*V ||_2^2.
My question is how or what properties is used to take H*V outside and R^H*R inside the diagonal operator. ?
Or How can we proof yellow highlighted terms in eq 70 is equal to yellow highlighted terms in eq 71?
Please see the attached pdf file.
 

Attachments

  • MSE-Based_Transceiver_Designs_for_Full-Duplex_MIMO_Cognitive_Radios.pdf
    1.3 MB · Views: 91
  • #5
Umesh said:
Please check only yellow highlighted terms on page number 2068.
Eq. 70 is converted into eq. 71 using formula ||vec(A)||_2^2 = tr{AA^H} (3rd line of appendix).
My question is
In eq 70, the term H*V*V^H*H^H is inside, and R is outside the diagonal operator as
tr{R* diag(H*V*V^H*H^H)*R^H} and
In eq 71, R is inside, and the HV term is outside the diagonal operator and written as
|| vec( ( diag(R^H*R) )^0.5*H*V ||_2^2.

Start from this expression. You are given [itex]\|\operatorname{vec}(A)\|_2^2 = \operatorname{tr}(AA^H)[/itex], so first apply that.
 
  • #6
pasmith said:
Start from this expression. You are given [itex]\|\operatorname{vec}(A)\|_2^2 = \operatorname{tr}(AA^H)[/itex], so first apply that.
Please try to understand the question and then reply.
 

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
1K
Replies
27
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
969
  • Linear and Abstract Algebra
Replies
6
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
18
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
5
Views
909
  • Set Theory, Logic, Probability, Statistics
Replies
11
Views
2K
Back
Top