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

  • $\begingroup$ In your second step, you are changing $(A \diamond (B\diamond C))^T$ into $(A^T \diamond (B\diamond C))^T$. This is not correct. $(A \diamond B)^T$ is also not equal to $A^T \diamond B^T$. This only works for Kronecker products. Could you indicate the dimensions of your matrices? Does the Khatri-Rao product have full column rank? $\endgroup$
    – Florian
    Commented Jun 24, 2019 at 8:36
  • $\begingroup$ @Florian As I understood, I can rewrite Khatri-Rao product using a Kronecker. So...why I can not use properties of Kronecker product? $\endgroup$
    – user682102
    Commented Jun 24, 2019 at 9:23
  • $\begingroup$ You can. $A \diamond B = (A \otimes B) \cdot \Gamma$, where $\Gamma$ is a selection matrix (equal to $I \diamond I$ btw). So $(A \diamond B)^T = ((A \otimes B) \cdot \Gamma)^T = \Gamma^T (A^T \otimes B^T)$. But that is not equal to $A^T \diamond B^T$ as you can easily verify. $\endgroup$
    – Florian
    Commented Jun 24, 2019 at 10:47

1 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 $M \cdot C \geq N^2$.
  • $\begingroup$ Can I write the followin$g: (A \diamond B)= (A I \diamond B) I $ , where I is identity matrix $\endgroup$
    – user682102
    Commented Jun 28, 2019 at 7:44
  • $\begingroup$ Sure... you just added two identity matrices, they do not change the result... $\endgroup$
    – Florian
    Commented Jun 28, 2019 at 8:52

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