Calculation with Khatri rao product and moore penrose pseudoinverse

Question

It is given:

$ \left [(A\diamond B\diamond C)^T \right ]^\dagger$

$\diamond$ is Kharti -rao product

$\dagger$ -moore penrose pseudoinverse

I have started:

$ \left [(A\diamond B\diamond C)^T \right ]^\dagger =\left [(A\diamond (B\diamond C))^T \right ]^\dagger = \left [(A^T \diamond (B\diamond C))^T \right ]^\dagger= ((A^T)^\dagger \diamond ((B\diamond C)(C^TC*B^TB) ^\dagger)$

What Can i do with $(A^T)^\dagger $?

the second question is if i have done correctly?

Edit 1:

A: MxN

B: NxN

C: CxN

Answer 1

Some hints:

• Transpose and pseudo inverse are interchangeable. Therefore, you can ignore the transpose, compute the pseudo inverse first, and then transpose the result.
• If a matrix has full column rank, you can expand the pseuso-inverse via $X^\dagger = (X^T X)^{-1} X^T $.
• The Khatri-Rao product satisfies $(A \diamond B)^T(A \diamond B) = (A^TA * B^TB)$. Applying this rule twice gives $(A \diamond B \diamond C)^T(A \diamond B\diamond C) = (A^TA * B^TB * C^TC)$.
• Overall, this could give something like $(A \diamond B \diamond C) \cdot (A^TA * B^TB * C^TC)^{-1}$ but whether or not this works depends on the dimensions of the matrices and on the rank of the Khatri-Rao product...
• *edit: For the dimensions you posted, a necessary condition would be $M \cdot C \geq N^2$.