TITLE: Commutation matrix proof
QUESTION [3 upvotes]: Prove that each commutation matrix $K$ is invertible and that $K^{-1} = K^{T}$
We found that $K$ is a square matrix and because we assume that $K$ only has distinct elements it has the maximal rank and is therefore an invertible square matrix.
We don't know how to prove the last part. Could someone please help?

REPLY [1 votes]: Let $E_{ij}$ denotes the $(i,j)$-th elementary matrix of size $m\times n$ and $e_k$ denotes the vector of length $mn$ that has a $1$ in the $k$-th entry and zeros elsewhere. Then the commutation matrix $K=K^{(m,n)}$ is an $mn\times mn$ matrix that satisfies $K\operatorname{vec}(E_{ij}) = \operatorname{vec}(E_{ij}^T)$. That is, $Ke_{(j-1)n+i}=e_{(i-1)m+j}$ for every $1\le i\le m$ and every $1\le j\le n$. Therefore $K$ is a permutation matrix and in turn $K^{-1}=K^T$.
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