- #1

- 3

- 0

A*Q*B=C

Here A is 2X4 known matrix

Q is a 4X4 unknown matrix

B is a 4X2 known matrix

C is a 2X2 known matrix

How can I solve for unknown matrix Q;

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- Thread starter tapashettisr
- Start date

- #1

- 3

- 0

A*Q*B=C

Here A is 2X4 known matrix

Q is a 4X4 unknown matrix

B is a 4X2 known matrix

C is a 2X2 known matrix

How can I solve for unknown matrix Q;

- #2

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- #3

HallsofIvy

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A

Q= A

Of course, if either A or B does not have an inverse, there may be no solution or there may be an infinite number of solutions.

- #4

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Er, B and A are not square matrices, so they do not have inverses, no?

- #5

HallsofIvy

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- #6

radou

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Just curious, how would these look like?

- #7

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- #8

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thanx all for the suggestions

- #9

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Inv(A)*A*Q*B*inv(B)=iinv(A)*C*inv(B)=Q

- #10

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and i have an equation like AQA(dagger)=block diagnol matrix

all the matrices are 4*4

how can i evaluate 16 unknowns from these 16 equations using mathematica,as terms involved are much complicated.

- #11

hotvette

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A*Q*B=C

Here A is 2X4 known matrix

Q is a 4X4 unknown matrix

B is a 4X2 known matrix

C is a 2X2 known matrix

How can I solve for unknown matrix Q;

This problem can be treated as a succession of two under determined problems that can be solved for least norm solutions. If you let Z = QB, the result is AZ=C, which can be solved as a mininum norm problem using QR factorization. If A

Solve R

You then have QB = Z. Taking transpose of both sides, you get B

- #12

hotvette

Homework Helper

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I think my last reply was too convoluted. Simply put, if Ax=b is an underdetermined system (where A has more columns than rows), it is easy to show that the minimum norm solution (assuming AA^{T} isn't singular) is:

x_{min norm} = A^{T}(AA^{T})^{-1}b

x

Last edited:

- #13

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That works. That is a special case of the Moore-Penrose psuedoinverse cited in post #7.

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