2
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.