Strange form of diagonalization

[Ojisan]

I'm trying to solve the following problem (not homework ) which is a strange form of diagonalization problem. Standard references and papers didn't turn up anything for me. Does anyone see possible approach for this?

- Given n x n full rank random matrices A1, A2, ... A9
Find length n unitary vectors x1, x2, x3, y1, y2, and y3 such that [y1^H 0 0; 0 y2^H 0; 0 0 y3^H] [A1 A2 A3; A4 A5 A6; A7 A8 A9] [x1 0 0; 0 x2 0; 0 0 x3] reduces to a 3 x 3 diagonal matrix.

^H is the Hermitian transpose and 0's indicate appropriate zero vectors.

It's like a constrained form of SVD but I can't seem to get a handle of it. Thanks in advance for any thoughts!

 

[CompuChip]

If you are careful not to commute anything (like you would when they were ordinary numbers), I think you can simply do the multiplication and get
$$\left( \begin{array}{lll} {y_1^H} & 0 & 0 \\ 0 & {y_2^H} & 0 \\ 0 & 0 & {y_3^H} \end{array} \right) \left( \begin{array}{lll} {A_1} & {A_2} & {A_3} \\ {A_4} & {A_5} & {A_6} \\ {A_7} & {A_8} & {A_9} \end{array} \right) \left( \begin{array}{lll} {x_1} & 0 & 0 \\ 0 & {x_2} & 0 \\ 0 & 0 & {x_3} \end{array} \right) = \left( \begin{array}{lll} {y_1^H} {A_1} {x_1} & 0 & 0 \\ 0 & {y_2^H} {A_5} {x_2} & 0 \\ 0 & 0 & {y_3^H} {A_9} {x_3} \end{array} \right)$$

 

[Ojisan]

Thanks, but I'm not sure if I follow you. You should get the following with straight computation.

$$\left( \begin{array}{lll} {y_1^H} & 0 & 0 \\ 0 & {y_2^H} & 0 \\ 0 & 0 & {y_3^H} \end{array} \right) \left( \begin{array}{lll} {A_1} & {A_2} & {A_3} \\ {A_4} & {A_5} & {A_6} \\ {A_7} & {A_8} & {A_9} \end{array} \right) \left( \begin{array}{lll} {x_1} & 0 & 0 \\ 0 & {x_2} & 0 \\ 0 & 0 & {x_3} \end{array} \right) = \left( \begin{array}{lll} {y_1^H} {A_1} {x_1} & {y_1^H} {A_2} {x_2} & {y_1^H} {A_3} {x_3} \\ {y_2^H} {A_4} {x_1} & {y_2^H} {A_5} {x_2} & {y_2^H} {A_6} {x_3} \\ {y_3^H} {A_7} {x_1} & {y_3^H} {A_8} {x_2} & {y_3^H} {A_9} {x_3} \end{array} \right)$$

and the objective is to null out the off-diagonals.

 

[Hurkyl]

Are you sure it has a nontrivial solution? Consider the case where n=1...


I suppose you're interested in a particular case where you know it really does have a solution? Well, I suppose the thing to do is to start small.

Can you figure out how to zero out y₂^HA₄x₁? Can you figure out all possible ways to do so?

Now, what about zeroing out both y₂^HA₄x₁ and y₃^HA₇x₁ at the same time?

And keep going until you manage to do it all.



Alternatively, is the matrix of A's diagonalizable? Maybe looking at that might help.

 

[Ojisan]

Thanks for your suggestion Hurkyl.

Solutions do exist, although probably not unique. I can search with a brute force conditioning to find a solution, by imposing partial conditions as you suggest.

The interpretation amounts to, for example, making A₂x₂ and A₃x₃ orthogonal to y₁ while keeping A₁x₁ nonorthogonal to y₁. Unfortunately, fixing x₂ and x₃ in this manner affects the other rows, and vice versa.

A is diagonalizable, so I have tried looking at SVD of individual A as well as combined SVD which didn't lead to much insight. I wish there was block unitary diagonalization procedure, but I had no luck in that direction.

I'd appreciate any thoughts.

 

[CompuChip]

The interpretation amounts to, for example, making A₂x₂ and A₃x₃ orthogonal to y₁ while keeping A₁x₁ nonorthogonal to y₁.

Is the latter necessary? A null block is diagonal as well.

 

[Ojisan]

True, but necessary for the problem that I am interested in. I should have stated my problem to include non-zero diagonal. In fact, I would like to maximize the Frobenius norm of the resultant diagonal matrix, but I was thinking that might be harder.