# Separability of the outer product of a vector with itself, where the vector is a tensor product of two other vectors

Is it true that $$(|{\psi}\rangle \otimes |{\phi}\rangle)(|{\psi}\rangle \otimes |{\phi}\rangle)^{\dagger} = (|\psi\rangle\langle\psi|)\otimes (|\phi\rangle\langle\phi|)$$

where

• $^{\dagger}$ is the conjugate transpose symbol

• $\otimes$ is the tensor product symbol

• $|{\psi}\rangle \otimes |{\phi}\rangle$ is the tensor product of two vectors

• and the left hand side is the outer product of $(|{\psi}\rangle \otimes |{\phi}\rangle)$ with itself

I am trying to determine if the left hand side is separable. If it is, and my equation is wrong, what is the correct equation? If my equation is correct, please tell me how to prove it; I cannot find that identity anywhere.

Suppose that $\:|\psi\rangle \in \mathrm F\:$, where $\:\mathrm F\:$ a $\:m$-dimensional Hilbert space with basic state vectors $\:|f_{i}\rangle, \quad i=1,2\cdots,m\:$ and $\:|\phi\rangle \in \mathrm H\:$, where $\:\mathrm H\:$ a $\:n$-dimensional Hilbert space with basic state vectors $\:|h_{j}\rangle, \quad j=1,2\cdots,n$.

To prove the identity $$\left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} = \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right) \tag{01}$$ it's sufficient to prove the identity for the basic state vectors $$\left(|f_{i}\rangle \otimes |h_{j}\right)\left(|f_{k}\rangle \otimes |h_{\ell}\right)^{\dagger} = \left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) \tag{02}$$ since if (using Einstein's summation convention) \begin{align} |\psi\rangle & = a_{i}|f_{i}\rangle \qquad i=1,2\cdots,m \tag{03a}\\ |\phi\rangle & = b_{j}|h_{j}\rangle \qquad j=1,2\cdots,n \tag{03b} \end{align} then \begin{align} \left(|\psi\rangle \otimes |\phi\rangle\right) & = a_{i}b_{j}\left(|f_{i}\rangle \otimes |h_{j}\rangle\right) \tag{04a}\\ \left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} & = a^*_{k}b^*_{\ell}\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\dagger} \tag{04b} \end{align} and \begin{align} \left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} & = a_{i}a^*_{k}b_{j}b^*_{\ell}\left(|f_{i}\rangle \otimes |h_{j}\right)\left(|f_{k}\rangle \otimes |h_{\ell}\right)^{\dagger} \tag{05a}\\ \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right) & = a_{i}a^*_{k}b_{j}b^*_{\ell}\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) \tag{05b} \end{align} Note that \begin{align} & \left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} \tag{06a}\\ & \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right) \tag{06b}\\ & \left(|f_{i}\rangle \otimes |h_{j}\rangle\right)\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\dagger} \tag{06c}\\ & \left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) \tag{06d} \end{align} are all linear transformations in the product $\:(m\!\cdot\! n)$-dimensional space $\: \mathrm F \otimes \mathrm H$.

To prove identity (02) we may take without loss of generality as basic state vectors those having a component equal to 1 and all other components equal to 0 \begin{align} \left(|f_{i}\rangle \right)_{\mu} & = \delta_{i\mu} , \quad \left(|f_{k}\rangle \right)_{\rho} = \delta_{k\rho} \qquad i,k,\mu,\rho =1,2,\cdots, m \tag{07a}\\ \left(|h_{j}\rangle \right)_{\nu} & = \delta_{j\nu} , \quad \left(|h_{\ell}\rangle \right)_{\sigma} = \delta_{\ell\sigma} \qquad j,\ell,\nu,\sigma =1,2,\cdots, n \tag{07b} \end{align} Then the transformations (06c) and (06d) are equal and are represented by a $\:(m\!\cdot\! n)\times(m\!\cdot\! n)$ matrix with one element equal to 1 and all other elements equal to 0.

By example : Let $\:m=3,n=2\:$ and $\:i=3,k=2, j=2, \ell=1\:$. Then $$\left(|f_{i}\rangle \otimes |h_{j}\rangle\right) =\left(|f_{3}\rangle \otimes |h_{2}\rangle\right) = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \otimes \begin{bmatrix} 0\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 1 \end{bmatrix} \tag{08}$$ $$\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right) =\left(|f_{2}\rangle \otimes |h_{1}\rangle\right) = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1\\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ 0\\ 0 \end{bmatrix} \tag{09}$$ so $$\left(|f_{3}\rangle \otimes |h_{2}\rangle\right)\left(|f_{2}\rangle \otimes |h_{1}\rangle\right)^{\dagger} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 1 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ 0\\ 0 \end{bmatrix}^{\boldsymbol{\dagger}} = \begin{bmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&0\\ \end{bmatrix} \tag{10}$$ Now $$\left(|f_{i}\rangle\langle f_{k}|\right) = \left(|f_{3}\rangle\langle f_{2}|\right)= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \begin{bmatrix} 0&1&0 \end{bmatrix} = \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&1&0 \end{bmatrix} \tag{11}$$ and $$\left(|h_{j}\rangle\langle h_{\ell}|\right) = \left(|h_{1}\rangle\langle h_{2}|\right)= \begin{bmatrix} 0\\ 1 \end{bmatrix} \begin{bmatrix} 1&0 \end{bmatrix} = \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix} \tag{12}$$ so $$\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) = \left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right) = \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&1&0 \end{bmatrix} \otimes \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix} \tag{13}$$ that is $$\left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right) = \begin{bmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&0\\ \end{bmatrix} \tag{14}$$ From ((10) and (14) $$\left(|f_{3}\rangle \otimes |h_{2}\rangle\right)\left(|f_{2}\rangle \otimes |h_{1}\rangle\right)^{\dagger} = \begin{bmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&0\\ \end{bmatrix} = \left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right) \tag{15}$$

Note :

I would prefer the following symbols that give an elegant expression of identity (01). More exactly, since $\:\left(|\psi\rangle \otimes |\phi\rangle\right)\:$ is a state in the product space $\: \mathrm F \otimes \mathrm H\:$ we define it as a ket $$|\psi\otimes \phi\rangle \stackrel{\text{def}}{\equiv} \left(|\psi\rangle \otimes |\phi\rangle\right) \tag{N-01}$$ Then $$\left(|\psi\rangle \otimes |\phi\rangle\right)^{\boldsymbol{\dagger}}= \langle \psi\otimes \phi| \tag{N-02}$$ and identity (01) is expressed as $$\color{blue}{|\psi\otimes \phi\rangle\langle \psi\otimes \phi| = \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right)} \color{black}{} \tag{N-03}$$ Now, if each of $\:|\psi\rangle \in \mathrm F,|\phi\rangle \in \mathrm H\:$ is a unit ket in its own Hilbert space $$\langle\psi|\psi\rangle = 1 = \langle\phi|\phi\rangle \tag{N-04}$$ then in $\: \mathrm F \otimes \mathrm H\:$ $$\langle \psi\otimes \phi|\psi\otimes \phi\rangle = 1 \tag{N-05}$$ In this case we have the following interpretation of (N-03) :

The projection on the unit ket $\:|\psi\otimes \phi\rangle\:$ in the product space $\: \mathrm F \otimes \mathrm H\:$ is the product of the projection on the unit ket $\:|\psi\rangle\:$ in $\:\mathrm F\:$ by the projection on the unit ket $\:|\phi\rangle\:$ in $\:\mathrm H$.

From this example above, we could give a proof of identity (02) in general under the choice (07) for the basic state vectors. So

1. $\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)$ is a $(m\!\cdot\! n)$-column vector with its $(i\!-\!1)n\!+\!j$ component equal to 1 and all the other components equal to 0 $$\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)= \begin{bmatrix} 0\\ 0\\ \vdots\\ 0\\ \vdots\\ \color{blue}{\bf 1}\\ \vdots\\ 0 \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:m\cdot n\hphantom{(i\!-\!1)n} \end{matrix} \tag{16}$$ 2. $\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}}\:$ is a $(m\!\cdot\! n)$-row vector with its $(k\!-\!1)n\!+\!\ell$ component equal to 1 and all the other components equal to 0 \begin{align} \left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}} & = \begin{bmatrix} 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\ \end{bmatrix} \tag{17}\\ & \hphantom{=\!=} \begin{matrix} \:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow \end{matrix} \nonumber\\ & \hphantom{=\!=} \begin{matrix} \:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n \end{matrix} \nonumber \end{align} 3. $\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}}$, product of equations (16) and(17), is a $(m\!\cdot\! n)\times(m\!\cdot\! n)$-square matrix with its element in the $(i\!-\!1)n\!+\!j$ row and $(k\!-\!1)n\!+\!\ell$ column equal to 1 and all the other elements equal to 0 \begin{align} \left(|f_{i}\rangle \otimes |h_{j}\rangle\right)&\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}} = \tag{18}\\ & \begin{bmatrix} 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0 \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:m\cdot n\hphantom{(i\!-\!1)n} \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n \end{matrix} \nonumber \end{align} 4. $\left(|f_{i}\rangle\langle f_{k}|\right)$ is a $m\!\times\! m$-square matrix with its element in the $i$-row and $k$-column equal to 1 and all the other elements equal to 0 \begin{align} \left(|f_{i}\rangle\langle f_{k}|\right) & = \tag{19}\\ & \begin{bmatrix} 0&0&\cdots&0&\cdots&0&\cdots&0\\ 0&0&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\color{blue}{\bf 1}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&0&\cdots&0 \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{i}\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:m\hphantom{(i\!-\!1)n\!+\!j} \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\uparrow&\cdots&\uparrow \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{k}&\,\cdots&\!\!m \end{matrix} \nonumber \end{align} 5. $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $n\!\times\!n$-square matrix with its element in the $j$-row and $\ell$-column equal to 1 and all the other elements equal to 0 \begin{align} \left(|h_{j}\rangle\langle h_{\ell}|\right) & = \tag{20}\\ & \begin{bmatrix} 0&0&\cdots&0&\cdots&0&\cdots&0\\ 0&0&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\color{blue}{\bf 1}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&0&\cdots&0 \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{j}\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:n\hphantom{(i\!-\!1)n\!+\!j} \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\uparrow&\cdots&\uparrow \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{\ell}&\,\cdots&\!\!n \end{matrix} \nonumber \end{align} 6. $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $m\!\times\!m$-square matrix in block form. All its block elements are $n\!\times\!n$-square null matrices (symbol $\mathcal{O}_{n}$) except the block in the $i$-row and $k$-column which is equal to the $n\!\times\!n$-square matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$. Of course $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $(m\!\cdot\! n)\times(m\!\cdot\! n)$-square matrix. In block form \begin{align} \left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) & = \tag{21}\\ & \begin{bmatrix} \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\ \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots& \color{blue}{\left(|h_{j}\rangle\langle h_{\ell}|\right)}&\cdots&\mathcal{O}_{n}\\ \vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\ \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\ \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{i}\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:m\hphantom{(i\!-\!1)n\!+\!j} \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\:\,\uparrow & \:\:\:\,\uparrow &\,\,\cdots &\:\,\uparrow &\,\,\cdots &\hphantom{==}\uparrow &\hphantom{==}\cdots &\:\:\uparrow \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\:\,1 & \:\:\:\,2 & \,\,\cdots &\:\, \lambda &\,\,\cdots &\hphantom{==}\,\color{red}{k} &\hphantom{==}\cdots &\:\: m \end{matrix} \nonumber \end{align} Now, in this block form we note that above the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ there are $(i\!-\!1)$ rows of $n\!\times\!n$-square null matrices and to the left of the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ there are $(k\!-\!1)$ columns of $n\!\times\!n$-square null matrices. This means that the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ is located after the element in the $(i-1)n$ row and the $(k-1)n$ column. But inside this matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ the unique nonzero element (=1) is located in the $j$-row and $\ell$-column and so in the $(i\!-\!1)n\!+\!j$ row and $(k\!-\!1)n\!+\!\ell$ column of the matrix $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$. So \begin{align} \left(|f_{i}\rangle\langle f_{k}|\right)\otimes & \left(|h_{j}\rangle\langle h_{\ell}|\right) = \tag{22}\\ & \begin{bmatrix} 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\ 0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0 \end{bmatrix} \begin{matrix} \longleftarrow \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \\ \vdots \\ \longleftarrow \end{matrix} \begin{matrix} 1\hphantom{(i\!-\!1)n\!+\!j}\\ 2\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j} \\ \kappa\hphantom{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \color{red}{(i\!-\!1)n\!+\!j}\\ \vdots\hphantom{(i\!-\!1)n\!+\!j}\\ \:m\cdot n\hphantom{(i\!-\!1)n} \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow \end{matrix} \nonumber\\ & \hphantom{.\,} \begin{matrix} \:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n \end{matrix} \nonumber \end{align}

Equations (18) and (22) give a proof of the identity (02) and so a proof of the identity (01).