# Proving matrix equality $(K_{n,n}\otimes I_n)a^{\otimes3}=a^{\otimes3}$

How to prove the matrix equality $$(K_{n,n}\otimes I_n)a^{\otimes3}=a^{\otimes3}$$?

Here $$K_{n,n}$$ is a $$n^2\times n^2$$ commutation matrix, $$I_n$$ is a $$n\times n$$ identity matrix and $$a$$ is a $$n\times1$$ vector, $$\otimes$$ denotes the kronecker product and $$a^{\otimes3}$$ is defined as $$a\otimes(a\otimes a)$$.

## 1 Answer

The commutation matrix (based on what I found) satisfies $$K_{n,n}(u \otimes v) = v \otimes u$$ for any $$n \times 1$$ vectors $$u,v$$.

Noting that the Kronecker product is associative, we may compute $$(K_{n,n} \otimes I) a^{\otimes 3} = (K_{n,n} \otimes I)((a \otimes a) \otimes a) = (K_{n,n} (a \otimes a)) \otimes (I a) = (a \otimes a) \otimes a$$

In response to the comment: note that $$K_{n,n}(u \otimes v) = K_{n,n}(\operatorname{vec}(vu^T)) = \operatorname{vec}([vu^T]^T) = \operatorname{vec}(uv^T) = v \otimes u$$ where vec denotes the vectorization operator.

• Why does K_{n,n}(u⊗v)=v⊗u hold? – user570271 Apr 8 at 15:59
• @user570271 see my latest edit – Omnomnomnom Apr 8 at 17:24