Exercise 2.28

Let \(A = \sum_{i,j}a_{ij}|i\rangle \langle j|\) be a \(p\times q\) matrix and \(B = \sum_{m,n}b_{mn}|m\rangle\langle n|\) be a \(r\times s\) matrix, we can write the tensor product \(A\otimes B\) as

(1)\[ A\otimes B = \sum_{i,m;j,n} a_{ij}b_{mn}(|i\rangle \otimes |m\rangle)(\langle j|\otimes \langle n|) \]

The matrix representation of \(A\otimes B\) is given by

(2)\[\begin{split} A\otimes B = \begin{pmatrix} a_{11}B & a_{12}B & \cdots & a_{1q}B \\ a_{21}B & a_{22}B & \cdots & a_{2q}B \\ \vdots & \vdots & \ddots & \vdots \\ a_{p1}B & a_{p2}B & \cdots & a_{pq}B \\ \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} & a_{11}b_{12} & \cdots & a_{1q}b_{1s} \\ a_{11}b_{21} & a_{11}b_{22} & \cdots & a_{1q}b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ a_{p1}b_{r1} & a_{p2}b_{r2} & \cdots & a_{pq}b_{rs} \\ \end{pmatrix} \end{split}\]

I will prove the following relations about tensor product,

(3)\[ (A\otimes B)^{t} = A^{t}\otimes B^{t}, (A\otimes B)^{*} = A^{*}\otimes B^{*}, (A\otimes B)^{\dagger} = A^{\dagger}\otimes B^{\dagger} \]

For \((A\otimes B)^{t}\), eq. (2) tell us that \((A\otimes B)^{t}\) is given by

(4)\[\begin{split} (A\otimes B)^{t} = \begin{pmatrix} a_{11}b_{11} & a_{11}b_{21} & \cdots & a_{p1}b_{r1} \\ a_{11}b_{12} & a_{11}b_{22} & \cdots & a_{p2}b_{r2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1q}b_{1s} & a_{1q}b_{2s} & \cdots & a_{pq}b_{rs} \\ \end{pmatrix} \end{split}\]

If we write eq. (4) into a form like eq. (1),

(5)\[ (A\otimes B)^{t} = \sum_{i,m;j,n} a_{ji}b_{nm }(|i\rangle \otimes |m\rangle)(\langle j\otimes \langle n|) \]

For \(A^{t}\otimes B^{t}\), we have \(A^{t} = \sum_{i,j}a_{ji}|i\rangle \langle j|\) and \(B^{t} = \sum_{m,n}b_{nm}|m\rangle\langle n|\), then according to eq. (1)$​, we have

(6)\[ A^{t}\otimes B^{t} = \sum_{i,m;j,n} a_{ji}b_{nm}(|i\rangle \otimes |m\rangle)(\langle j\otimes \langle n|) \]

From eq. (5) and (6), we have \((A\otimes B)^{t} = A^{t}\otimes B^{t}\).

For \((A\otimes B)^{*}\), according to eq. (1), we have

(7)\[ (A\otimes B)^{*} = \sum_{i,m;j,n} a^{*}_{ij}b^{*}_{mn}(|i\rangle \otimes |m\rangle)(\langle j\otimes \langle n|) \]

For \(A^{*}\otimes B^{*}\), I notice that \(A^{*} = \sum_{i,j}a^{*}_{ij}|i\rangle \langle j|\) and \(B^{*} = \sum_{m,n}b^{*}_{mn}|m\rangle\langle n|\), then according to eq. (1), we have

(8)\[ A^{*}\otimes B^{*} = \sum_{i,m;j,n} a^{*}_{ij}b^{*}_{mn}(|i\rangle \otimes |m\rangle)(\langle j\otimes \langle n|) \]

From eq. (7) and (8), we have \((A\otimes B)^{*} = A^{*}\otimes B^{*}\).

For \((A\otimes B)^{\dagger}\), since the hermitian conjugate of a matrix \(A^{\dagger}\) is equivalent with the transpose of complex conjugate of matrix \(A\), that is, \(A^{\dagger} = (A^{*})^{T}\). So we can make use of the above two relations and get

(9)\[ (A\otimes B)^{\dagger} = ((A\otimes B)^*)^{T} = ((A^*\otimes B^*))^{T} = (A^*)^T\otimes (B^*)^T = A^\dagger\otimes B^\dagger \]