(1)

行列\(A,B,C\)をそれぞれ\(k\times l,l\times m,m\times n\)行列とする。
vec作用素とクロネッカー積は
\[ \left(\mathrm{vec}\left(B\right)\right)_{\left(q-1\right)l+p}=\left(B\right)_{p,q} \] \[ \left(C^{T}\otimes A\right)_{\left(p-1\right)k+r,\left(q-1\right)l+s}=\left(C^{T}\right)_{p,q}\left(A\right)_{r,s} \] なので、
\begin{align*} \left(\mathrm{vec}\left(ABC\right)\right)_{i} & =\left(ABC\right)_{\mod\left(i-1,k\right)+1,\left\lfloor \frac{i-1}{k}\right\rfloor +1}\\ & =\sum_{t=1}^{m}\sum_{s=1}^{l}\left(A\right)_{\mod\left(i-1,k\right)+1,t}\left(B\right)_{t,s}\left(C\right)_{s,\left\lfloor \frac{i-1}{k}\right\rfloor +1}{}_{,}\\ & =\sum_{t=1}^{m}\sum_{s=1}^{l}\left(C\right)_{s,\left\lfloor \frac{i-1}{k}\right\rfloor +1}\left(A\right)_{\mod\left(i-1,k\right)+1,t}\left(B\right)_{t,s}{}_{,}\\ & =\sum_{t=1}^{m}\sum_{s=1}^{l}\left(C^{T}\right)_{\left\lfloor \frac{i-1}{k}\right\rfloor +1,s}\left(A\right)_{\mod\left(i-1,k\right)+1,t}\left(\mathrm{vec}\left(B\right)\right)_{\left(s-1\right)l+t}\\ & =\sum_{t=1}^{m}\sum_{s=1}^{l}\left(C^{T}\otimes A\right)_{\left(\left\lfloor \frac{i-1}{k}\right\rfloor +1-1\right)k+\mod\left(i-1,k\right)+1,\left(s-1\right)l+t}\left(\mathrm{vec}\left(B\right)\right)_{\left(s-1\right)l+t}\\ & =\sum_{t=1}^{m}\sum_{s=1}^{l}\left(C^{T}\otimes A\right)_{i,\left(s-1\right)l+t}\left(\mathrm{vec}\left(B\right)\right)_{\left(s-1\right)l+t}\\ & =\sum_{j=1}^{lm}\left(C^{T}\otimes A\right)_{i,j}\left(\mathrm{vec}\left(B\right)\right)_{j}\\ & =\left(\left(C^{T}\otimes A\right)\mathrm{vec}\left(B\right)\right)_{i} \end{align*}

(1)-2

行列\(A,B,C\)をそれぞれ\(k\times l,l\times m,m\times n\)行列とする。
\(i\in\left\{ 1,2,\cdots,k\right\} ,j\in\left\{ 1,2,\cdots,n\right\} \)とすると、1から\(nk\)までの自然数は\(\left(j-1\right)k+i\)で表され、vec作用素とクロネッカー積は
\[ \left(\mathrm{vec}\left(B\right)\right)_{\left(q-1\right)l+p}=\left(B\right)_{p,q} \] \[ \left(C^{T}\otimes A\right)_{\left(p-1\right)k+r,\left(q-1\right)l+s}=\left(C^{T}\right)_{p,q}\left(A\right)_{r,s} \] なので、
\begin{align*} \left(\left(C^{T}\otimes A\right)\mathrm{vec}\left(B\right)\right)_{\left(j-1\right)k+i} & =\sum_{t=1}^{mn}\left(C^{T}\otimes A\right)_{\left(j-1\right)k+i,t}\left(\mathrm{vec}\left(B\right)\right)_{t}\\ & =\sum_{v=1}^{m}\sum_{u=1}^{l}\left(C^{T}\otimes A\right)_{\left(j-1\right)k+i,\left(v-1\right)l+u}\left(\mathrm{vec}\left(B\right)\right)_{\left(v-1\right)l+u}\\ & =\sum_{v=1}^{m}\sum_{u=1}^{l}\left(C^{T}\otimes A\right)_{\left(j-1\right)k+i,\left(v-1\right)l+u}\left(B\right)_{u,v}\\ & =\sum_{v=1}^{m}\sum_{u=1}^{l}\left(C^{T}\right)_{j,v}\left(A\right)_{i,u}\left(B\right)_{u,v}\\ & =\sum_{v=1}^{m}\sum_{u=1}^{l}\left(A\right)_{i,u}\left(B\right)_{u,v}\left(C\right)_{v,j}\\ & =\left(ABC\right)_{i,j}\\ & =\left(\mathrm{vec}\left(ABC\right)\right)_{\left(j-1\right)k+i} \end{align*} となる。
従って、
\[ \left(C^{T}\otimes A\right)\mathrm{vec}\left(B\right)=\mathrm{vec}\left(ABC\right) \] となり与式は成り立つ。

(2)

(1)より、
\[ \mathrm{vec}\left(ABC\right)=\left(C^{T}\otimes A\right)\mathrm{vec}\left(B\right) \] なので、\(C=I\)とおくと、
\[ \mathrm{vec}\left(AB\right)=\left(I\otimes A\right)\mathrm{vec}\left(B\right) \] \(A\rightarrow I,B\rightarrow A,C\rightarrow B^{T}\)とおくと、
\[ \mathrm{vec}\left(ABC\right)=\left(B^{T}\otimes I\right)\mathrm{vec}\left(A\right) \] \(B\rightarrow I,C\rightarrow B\)とおくと、
\[ \mathrm{vec}\left(ABC\right)=\left(B^{T}\otimes A\right)\mathrm{vec}\left(I\right) \] となるので与式は成り立つ。

(3)

行列\(A,B\)をそれぞれ\(m\times n,n\times m\)行列とする。
\begin{align*} \tr\left(AB\right) & =\sum_{i=1}^{m}\sum_{j=1}^{n}\left(A\right)_{i,j}\left(B\right)_{j,i}\\ & =\sum_{k=1}^{mn}\left(\mathrm{vec}\left(A^{T}\right)\right)_{k}\left(\mathrm{vec}\left(B\right)\right)_{k}\\ & =\sum_{k=1}^{mn}\left(\mathrm{vec}\left(A^{T}\right)\right)_{k}^{T}\left(\mathrm{vec}\left(B\right)\right)_{k}\\ & =\left(\mathrm{vec}\left(A^{T}\right)\right)^{T}\left(\mathrm{vec}\left(B\right)\right) \end{align*} となるので与式は成り立つ。

(4)

\begin{align*} \tr\left(ABC\right) & =\left(\mathrm{vec}\left(A^{T}\right)\right)^{T}\mathrm{vec}\left(BC\right)\cmt{\because\tr\left(AB\right)=\left(\mathrm{vec}\left(A^{T}\right)\right)^{T}\mathrm{vec}\left(B\right)}\\ & =\left(\mathrm{vec}\left(A^{T}\right)\right)^{T}\left(I\otimes B\right)\mathrm{vec}\left(C\right)\cmt{\because\mathrm{vec}\left(AB\right)=\left(I\otimes A\right)\mathrm{vec}\left(B\right)} \end{align*} となるので与式は成り立つ。

(5)

\begin{align*} \tr\left(AD^{T}BDC\right) & =\tr\left(D^{T}BDCA\right)\\ & =\left(\mathrm{vec}\left(D\right)\right)^{T}\mathrm{vec}\left(BDCA\right)\cmt{\because\tr\left(AB\right)=\left(\mathrm{vec}\left(A^{T}\right)\right)^{T}\mathrm{vec}\left(B\right)}\\ & =\left(\mathrm{vec}\left(D\right)\right)^{T}\left(\left(CA\right)^{T}\otimes B\right)\mathrm{vec}\left(D\right)\cmt{\because\mathrm{vec}\left(ABC\right)=\left(C^{T}\otimes A\right)\mathrm{vec}\left(B\right)}\\ & =\left(\mathrm{vec}\left(D\right)\right)^{T}\left(A^{T}C^{T}\otimes B\right)\mathrm{vec}\left(D\right) \end{align*} 更に計算を進めると、
\begin{align*} \tr\left(AD^{T}BDC\right) & =\left(\mathrm{vec}\left(D\right)\right)^{T}\left(A^{T}C^{T}\otimes B\right)\mathrm{vec}\left(D\right)\\ & =\left(\left(\mathrm{vec}\left(D\right)\right)^{T}\left(A^{T}C^{T}\otimes B\right)\mathrm{vec}\left(D\right)\right)^{T}\\ & =\left(\mathrm{vec}\left(D\right)\right)^{T}\left(\left(A^{T}C^{T}\right)^{T}\otimes B^{T}\right)\mathrm{vec}\left(D\right)\\ & =\left(\mathrm{vec}\left(D\right)\right)^{T}\left(CA\otimes B^{T}\right)\mathrm{vec}\left(D\right) \end{align*} となるので与式は成り立つ。