# Calculation with Khatri rao product and moore penrose pseudoinverse

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

• 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? Commented Jun 24, 2019 at 8:36
• @Florian As I understood, I can rewrite Khatri-Rao product using a Kronecker. So...why I can not use properties of Kronecker product?
– user682102
Commented Jun 24, 2019 at 9:23
• 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. Commented Jun 24, 2019 at 10:47

• 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 $$M \cdot C \geq N^2$$.
• Can I write the followin$g: (A \diamond B)= (A I \diamond B) I$ , where I is identity matrix