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ttyme411gl 2022-07-09 Answered
Let { X 1 , , X K } is a set of random matrices, where X k R M × N , k = 1 , , K, and U R M × r and V R N × r are two matrices containing orthogonal columns (i.e., U U = I , V V = I). I was wondering, if the following question has a analytical solution:
max U , V k = 1 K U X k V F 2
If not, how should I solve it? Alternating optimization?

(At first, I thought it may be related to the SVD of the sum of the matrices { X k }, but so far I have no hint to prove it.)
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Answers (2)

Nicolas Calhoun
Answered 2022-07-10 Author has 15 answers
maximize k = 1 K U X k V F 2 subject to U U = I V V = I
Let
f k ( U , V ) := U X k V F 2 = tr ( U X k V V X k U ) = tr ( V X k U U X k V )
Hence,
U f k ( U , V ) = 2 X k V V X k U
V f k ( U , V ) = 2 X k U U X k V
Let the Lagrangian be
L ( U , V , Λ 1 , Λ 2 ) := k = 1 K f k ( U , V ) Λ 1 , U U I Λ 2 , V V I
where the Lagrange multipliers Λ 1 and Λ 2 are symmetric matrices. Taking the partial derivatives with respect to U and V ,
U L ( U , V , Λ 1 , Λ 2 ) = 2 k = 1 K X k V V X k U 2 U Λ 1
V L ( U , V , Λ 1 , Λ 2 ) = 2 k = 1 K X k U U X k V 2 V Λ 2
Finding where the partial derivatives vanish, we obtain two cubic matrix equations in U , V , Λ 1 , Λ 2 and two quadratic matrix equations in U and V
k = 1 K X k V V X k U = U Λ 1 k = 1 K X k U U X k V = V Λ 2 U U = I V V = I
How can one solve these matrix equations? I do not know.

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Ximena Skinner
Answered 2022-07-11 Author has 7 answers
Define the variables
Y = U U T y = v e c ( Y ) Z = V V T z = v e c ( Z ) S = k X k X k
Note that ( Y , Z ) are not orthogonal but they are ortho-projectors
Y 2 = Y = Y T Z 2 = Z = Z T
Similarly, the vectors ( y , z ) are not unit vectors but they satisfy simple normalization conditions
y T y = M z T z = N
Write the objective function as
f = k U T X k V F 2 = k ( U T X k V ) : ( U T X k V ) = k ( Y X k ) : ( X k Z ) = k v e c ( Y X k ) T v e c ( X k Z ) = k y T ( X k I M ) ( I N X k ) z = y T S z
where colon denotes the inner/Frobenius product.

The main result of these manipulations was to recast the problem as a single matrix-vector equation. The vector gradients are then
f y = S z , f z = S T y
This still isn't a solution to the problem, just another way of thinking about it.

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