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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 $a\otimes(a\otimes a)$.

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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.

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  • $\begingroup$ Why does K_{n,n}(u⊗v)=v⊗u hold? $\endgroup$ – user570271 Apr 8 '19 at 15:59
  • $\begingroup$ @user570271 see my latest edit $\endgroup$ – Omnomnomnom Apr 8 '19 at 17:24

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