Proof Since $A\perp B$ , we can apply Lemma 2.5 to conclude that there exist $\hat {A}\in M_m $ , $\hat {B}\in M_{n-m}$ and unitary matrices $U,V\in M_n$ such that

$$ \begin{align*}UAV=\hat{A}\oplus 0\quad \text{and}\quad UBV=0\oplus \hat{B}.\end{align*} $$

Let $UCV$ be partitioned as

$$ \begin{align*}UCV=\begin{bmatrix} C_{11}&C_{12}\\ C_{21}&C_{22} \end{bmatrix} \end{align*} $$

with $C_{11}\in M_m$ and $C_{22}\in M_{n-m}.$ It follows from $(A+B)\perp C$ that

$$ \begin{align*} (U(A+B)V)^{*}(UCV)=0 \quad \text{and}\quad (U(A+B)V)(UCV)^*=0, \end{align*} $$

that is,

$$ \begin{align*}\begin{bmatrix} \hat{A}^*C_{11}&\hat{A}^*C_{12}\\ \hat{B}^{*}C_{21}&\hat{B}^{*}C_{22} \end{bmatrix}=0\quad \text{and}\quad \begin{bmatrix} \hat{A}C_{11}^*&\hat{A}C_{21}^*\\ \hat{B}C_{12}^*&\hat{B}C_{22}^* \end{bmatrix}=0. \end{align*} $$

Then we have

$$ \begin{align*}V^*A^*CV=(UAV)^*(UCV)= \begin{bmatrix} \hat{A}^*C_{11}&\hat{A}^*C_{12}\\ 0&0 \end{bmatrix}=0 \end{align*} $$

and

$$ \begin{align*}UAC^*U^*=(UAV)\ (UCV)^*=\begin{bmatrix} \hat{A}C_{11}^*&\hat{A}C_{21}^*\\ 0&0 \end{bmatrix}=0 .\end{align*} $$

Thus, $A^{*}C=0$ and $AC^{*}=0$ , i.e., $A\perp C$ . Similarly, we can also conclude that $B\perp C$ .

Proof Let $U\in M_n$ be a unitary matrix such that

$$ \begin{align*}U^*CU=\mathrm{diag}(\lambda_1(C),\lambda_2(C),\ldots,\lambda_n({C})).\end{align*} $$

Denote by $u_i$ the ith column of U for $i=1,\ldots ,n$ . Let $\hat {U}=[u_1,u_2,\ldots ,u_k].$ Then, applying Lemma 2.4, we have

$$ \begin{align*} \sum_{i=1}^k\lambda_{i}^{\gamma}(C+D)=\sum_{i=1}^k\lambda_{i}\big((C+D)^{\gamma}\big)\geq \mathrm{tr}(\hat{U}^*(C+D)^{\gamma}\hat{U}) \end{align*} $$

and

$$ \begin{align*}\quad \sum_{i=1}^k\lambda_{i}^{\gamma}(C-D)=\sum_{i=1}^k\lambda_{i}\big((C-D)^{\gamma}\big)\geq \mathrm{tr}(\hat{U}^*(C-D)^{\gamma}\hat{U}).\end{align*} $$

Since $-C\leq D\leq C$ , we have

$$ \begin{align*}-x^*Cx\leq x^*Dx\leq x^*Cx \quad\text{ for all}\quad x\in \mathbb{C}^{n}.\end{align*} $$

By Lemma 2.3, we have

$$ \begin{align*} u_i^{*}(C+D)^{\gamma}u_{i}\geq (u_i^*(C+D)u_i)^{\gamma} \quad \text{and} \quad u_i^{*}(C-D)^{\gamma}u_{i}\geq (u_i^*(C-D)u_i)^{\gamma} \end{align*} $$

for $i=1,\ldots ,n$ . Applying Lemma 2.2 with $a=u_i^{*}Cu_i$ and $b=u_i^{*}Du_i$ , we get

$$ \begin{align*} (u_i^{*}(C+D)u_i)^{\gamma}+(u_i^{*}(C-D)u_i)^{\gamma}\geq 2(u_i^{*}Cu_i)^{\gamma}\quad \text{for}\quad i=1,\ldots, n. \end{align*} $$

It follows from the above inequalities that

$$ \begin{align*} \sum_{i=1}^k\lambda_{i}^{\gamma}(C+D)+\sum_{i=1}^k\lambda_{i}^{\gamma}(C-D)\geq& \mathrm{tr}(\hat{U}^*(C+D)^{\gamma}\hat{U})+\mathrm{tr}(\hat{U}^*(C-D)^{\gamma}\hat{U})\\ =&\sum_{i=1}^{k}u_i^{*}(C+D)^{\gamma}u_i+\sum_{i=1}^{k}u_i^{*}(C-D)^{\gamma}u_i \\ \geq&\sum_{i=1}^{k}(u_i^{*}(C+D)u_i)^{\gamma}+\sum_{i=1}^{k}(u_i^{*}(C-D)u_i)^{\gamma} \\ \geq& 2 \sum_{i=1}^{k}(u_i^{*}Cu_i)^{\gamma}=2\sum_{i=1}^k\lambda_i^{\gamma}(C). \\[-42pt] \end{align*} $$

Proof Notice that

$$ \begin{align*}\|A+B\|_{(p,k)}^p=\sum_{i=1}^ks_i^p(A+B)=\sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big((A^*A+B^*B)+(A^*B+B^*A)\big)\end{align*} $$

and

$$ \begin{align*}\|A-B\|_{(p,k)}^p=\sum_{i=1}^ks_i^p(A-B)=\sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big((A^*A+B^*B)-(A^*B+B^*A)\big).\end{align*} $$

Let $C=A^*A+B^*B$ and $D= A^*B+B^*A$ . Then $C+D=(A+B)^*(A+B)$ and $C-D=(A-B)^*(A-B)$ are both positive semidefinite, that is, $-C\leq D\leq C$ . Applying Lemma 2.7, we get (2.5).

Proof With the assumption that $A\perp B$ , we can assume that the largest k singular values of $A+B$ are $s_1(A),\ldots ,s_{\ell }(A),s_1(B),\ldots ,s_{k-\ell }(B)$ for some $0\leq \ell \leq k$ . Then

(2.6) $$ \begin{align} \|A+B\|_{(p,k)}^p=\sum_{i=1}^{\ell}s_i^p(A)+\sum_{i=1}^{k-\ell}s_i^p(B)\leq \sum_{i=1}^{k}s_i^p(A)+\sum_{i=1}^{k}s_i^p(B). \end{align} $$

On the other hand, we have

$$ \begin{align*}\|A+B\|_{(p,k)}^p=\|A\|_{(p,k)}^p+\|B\|_{(p,k)}^p=\displaystyle\sum_{i=1}^{k}s_i^p(A)+\displaystyle\sum_{i=1}^{k}s_i^p(B).\end{align*} $$

Thus, the equality in (2.6) holds, which implies

(2.7) $$ \begin{align} \sum_{i=1}^{\ell}s_i^p(A)=\sum_{i=1}^{k}s_i^p(A)\quad \text{and}\quad \sum_{i=1}^{k-\ell}s_i^p(B)=\sum_{i=1}^{k}s_i^p(B). \end{align} $$

Since A and B are both nonzero, we have

$$ \begin{align*}\displaystyle\sum_{i=1}^{k}s_i^p(A)>0\quad \mathrm{and}\quad \displaystyle\sum_{i=1}^{k}s_i^p(B)>0,\end{align*} $$

which implies $\ell \geq 1$ and $k-\ell \geq 1$ , i.e., $1\leq \ell \leq k-1.$ With (2.7), it follows that

$$ \begin{align*} \sum_{i=\ell+1}^{k}s_i^p(A)=0\quad \text{and} \quad \displaystyle \sum_{i=k-\ell+1}^{k}s_k^p(B)=0, \end{align*} $$

which implies $s_{\ell +1}(A)=0$ and $s_{k-\ell +1}(B)=0.$ Therefore,

$$ \begin{align*}\mathrm{rank}(A)\leq \ell \quad\mathrm{and } \quad\mathrm{rank}(B)\leq k-\ell.\end{align*} $$

Since $A\perp B$ , we have

$$ \begin{align*}\mathrm{rank}(A+B)=\mathrm{rank}(A)+\mathrm{rank}(B)\leq \ell+k-\ell=k.\\[-35pt]\end{align*} $$

Proof Denote the ith diagonal entry of $U^*BU$ by $b_i.$ Then $\lambda _i(A)+\alpha b_{i}$ is the ith diagonal entry of $U^*(A+\alpha B)U$ . It follows that

$$ \begin{align*}\big(\lambda_1(A+\alpha B),\ldots,\lambda_k(A+\alpha B) \big )\succ_w \big (\lambda_{1}(A)+\alpha b_{1},\ldots,\lambda_{k}(A)+\alpha b_{k}\big). \end{align*} $$

Notice that $g(x)=x^{\gamma } (x>0)$ is an increasing convex function when $\gamma>1$ . We can apply Theorem 3.26 in [Reference Zhan31] to obtain

$$ \begin{align*}\big(\lambda_1^{\gamma}(A+\alpha B),\ldots,\lambda_k^{\gamma}(A+\alpha B) \big)\succ_w \big ((\lambda_{1}(A)+\alpha b_{1})^{\gamma},\ldots,(\lambda_{k}(A)+\alpha b_{k})^{\gamma}\big).\end{align*} $$

Thus, $\displaystyle \sum _{i=1}^{k}\lambda _{i}^{\gamma }(A+\alpha B)\geq \displaystyle \sum _{i=1}^{k}(\lambda _i(A)+\alpha b_i)^{\gamma }.$ With the assumption in (2.8), we can conclude that

(2.9) $$ \begin{align} \sum_{i=1}^{k}(\lambda_i(A)+\alpha b_{i})^{\gamma} \leq \sum_{i=1}^{k}\lambda_i^{\gamma}(A)+\sum_{i=1}^{k}\lambda_i^{\gamma}(\alpha B)\quad \text{for all}\quad 0<\alpha<1. \end{align} $$

Let $f(\alpha )=\displaystyle \sum _{i=1}^{k}(\lambda _i(A)+\alpha b_{i})^{\gamma } -\displaystyle \sum _{i=1}^{k}\lambda _i^{\gamma }(A)-\displaystyle \sum _{i=1}^{k}\lambda _i^{\gamma }(\alpha B)$ be a function on $\alpha $ . Then we have

(2.10) $$ \begin{align} f(\alpha)=f(0)+f^{'}(0)\alpha+o(\alpha)= \left[\sum_{i=1}^{k}\lambda_i^{\gamma-1}(A)b_{i}\gamma\right]\alpha+o(\alpha), \end{align} $$

where a function $g(\alpha )=o(\alpha )$ means $ \lim \limits _{\alpha \to 0}\frac {g(\alpha )}{\alpha }=0$ . Since A and B are both positive semidefinite, we have $\lambda _i(A)\geq 0$ and $ b_{i}\geq 0$ for all $i=1,\ldots ,n.$ It follows that $\displaystyle \sum _{i=1}^{k}\lambda _i^{\gamma -1}(A)b_{i}\gamma \geq 0.$ We claim that $\displaystyle \sum _{i=1}^{k}\lambda _i^{\gamma -1}(A)b_{i}\gamma =0$ . Otherwise, $\displaystyle \sum _{i=1}^{k}\lambda _i^{\gamma -1}(A)b_{i}\gamma>0$ leads to $f(\alpha )>0$ when $\alpha>0$ is sufficiently small, which contradicts (2.9). It follows that

$$ \begin{align*}\lambda_i(A)b_{i}=0 \quad \text{for}\quad i=1,\ldots,k.\end{align*} $$

For the case $\lambda _k(A)=0$ , we may assume that $ t$ is the largest integer such that ${\lambda _t(A)>0}$ . Then $U^*AU=\mathrm {diag}(\lambda _1(A),\ldots ,\lambda _t(A))\oplus 0_{n-t}$ and $b_i=0$ for $i=1,\ldots ,t.$ Recall that B is positive semidefinite. Thus, $U^*BU=0_{t}\oplus \hat {B}$ with $\hat {B}\in M_{n-t}.$ It follows that $A\perp B.$

For the case $\lambda _k(A)>0,$ we first have $b_i=0$ for all $i=1,\ldots ,k$ . Since B is positive semidefinite, it follows that $U^*BU=0_k\oplus C$ with $C\in M_{n-k}.$ Recall that $\ell $ is the largest integer such that $\lambda _{k+\ell }(A)=\lambda _k(A)$ . If $\ell =0$ , then the proof is completed. If $\ell>0$ , then for any $i=k+1,\ldots ,k+\ell $ , replacing the role of $\lambda _k(A)+\alpha b_k$ with $\lambda _i(A)+\alpha b_i$ in the above argument, we can conclude $b_i=0$ . Thus, we have $b_i=0$ for $i=1,\ldots ,k+\ell .$ It follows that $U^*BU=0_{k+\ell }\oplus \hat {B}$ with $\hat {B}\in M_{n-k-\ell }$ .

Proof If $s_k(T)=0$ , then $\lambda _k(T^*T)=\lambda _k(TT^*)=0$ . We can use Lemma 2.11 twice to conclude that $T^*T\perp S^*S$ and $TT^*\perp SS^*$ , and hence $T\perp S.$

If $s_k(T)>0$ , then we have

$$ \begin{align*}V^*T^*TV=\mathrm{diag}(s_1^2(T),\ldots,s_{n}^2(T))\quad \text{and}\quad UTT^*U^*=\mathrm{diag}(s_1^2(T),\ldots,s_{n}^2(T)).\end{align*} $$

Notice that $\lambda _{i}(TT^*)=\lambda _{i}(T^*T)=s^2_i(T)$ for $i=1,\ldots ,n$ . Thus,

$$ \begin{align*}\lambda_k(T^*T)=\lambda_k(TT^*)=s_k^2(T)>0\end{align*} $$

and $\ell $ is the largest integer such that

$$ \begin{align*}\lambda_{k+\ell}(TT^*)=\lambda_{k}(TT^*)\quad \text{and} \quad \lambda_{k+\ell}(T^*T)=\lambda_{k}(T^*T).\end{align*} $$

Then we use Lemma 2.11 twice to conclude that

$$ \begin{align*}VS^*SV=0_{k+\ell}\oplus C\quad \text{and}\quad USS^*U^*=0_{k+\ell}\oplus D.\end{align*} $$

It follows that $USV=0_{k+\ell }\oplus \hat {S}.$

Proof For any real symmetric matrix $S\in M_n$ and any unitary matrix $X\in M_m$ ,

$$ \begin{align*}{\phi}(I_m\otimes S)=\sum_{i=1}^m{\phi}(XE_{ii}X^*\otimes S)={W}_X(I_m\otimes S){W}_X^*.\end{align*} $$

Since $W_I=I_{mn}$ , it follows that

$$ \begin{align*}{W}_X(I_m\otimes S){W}_X^*=W_I(I_m\otimes S){W}_I^*=I_m\otimes S.\end{align*} $$

Thus, ${W}_X$ commutes with $I_m\otimes S$ for all real symmetric $S\in M_n.$ This yields that ${{W}_X=Z_X\otimes I_n}$ for some unitary matrix $Z_X\in M_m$ , and hence

(2.11) $$ \begin{align} {\phi}(XE_{ii}X^*\otimes B)=(Z_XE_{ii}Z_X^*)\otimes \varphi_{i,X}(B)\quad \text{for all}\quad i=1,\ldots,m\text{ and }B\in M_n. \end{align} $$

Define linear maps $\text {tr}_1 : M_{mn}\to M_{n}$ and $ \text {Tr}_1: M_{mn}\to M_{n}$ as

$$ \begin{align*}\text{tr}_1(A\otimes B)=\text{tr} (A)B\quad \text{and} \quad \text{Tr}_1(A\otimes B)=\text{tr}_1({\phi}(A\otimes B))\end{align*} $$

for all $A\in M_m$ and $B\in M_n.$ The map $\text {tr}_1$ is also called the partial trace function in quantum science. Then

$$ \begin{align*}\text{Tr}_1(XE_{ii}X^*\otimes B)=\varphi_{i,X}(B).\end{align*} $$

Note that $\text {Tr}_1$ is linear and therefore continuous and the set

$$ \begin{align*}\{XE_{ii}X^*\mid 1\leq i\leq m, X\in M_m \text{ is unitary}\}=\{xx^*\in M_m\mid x^*x=1\}\end{align*} $$

is connected. So, all the maps $\varphi _{i, X}$ are the same, and hence we can rewrite (2.11) as

$$ \begin{align*}{\phi}(XE_{ii}X^*\otimes B)=(Z_XE_{ii}Z_X^*)\otimes \varphi_{2}(B)\quad \text{for all}\quad i=1,\ldots,m\text{ and }B\in M_n,\end{align*} $$

where $\varphi _2$ is the identity map or the transposition map. By the linearity of ${\phi }$ , it follows that

$$ \begin{align*}{\phi}(A\otimes B)=\varphi_1(A)\otimes \varphi_{2}(B)\quad \text{for all}\quad A\in M_m\text{ and }B\in M_n\end{align*} $$

for some linear map $\varphi _1.$

Proof of Theorem 2.1.

Notice that the $(p,k)$ -norm reduces to the spectral norm when $k=1$ . It was shown in [Reference Fošner, Huang, Li and Sze10] that a linear map $\phi $ preserves the spectral norm of tensor products $A\otimes B$ for all $A\in M_m$ and $B\in M_n$ if and only if $\phi $ has form $A\otimes B\mapsto U(\varphi _1(A)\otimes \varphi _2(B))V$ for some unitary matrices $U,V\in M_{mn}$ , where $\varphi _s$ is the identity map or the transposition map for $s=1,2$ . So we need only consider the case when $k\geq 2$ in the following discussion. Since the sufficiency part is clear, we consider only the necessity part.

Suppose a linear map $\phi : M_{mn}\to M_{mn}$ satisfies (2.1) and $k \geq 2$ . We need the following three claims.

Claim 1 For any unitary matrices $X\in M_m\text { and } Y\in M_n$ , we have

$$ \begin{align*}\phi (XE_{ii}X^*\otimes YE_{jj}Y^*)\perp \phi(XE_{ii}X^*\otimes YE_{ss}Y^*)\end{align*} $$

and

$$ \begin{align*}\phi (XE_{jj}X^*\otimes YE_{ii}Y^*)\perp \phi(XE_{ss}X^*\otimes YE_{ii}Y^*)\end{align*} $$

for any possible $i,j,s$ with $j\neq s$ . Moreover,

$$ \begin{align*}\textrm{rank}(\phi(XE_{ii}X^*\otimes Y(E_{jj}+E_{ss})Y^*))\leq k\end{align*} $$

and

$$ \begin{align*}\textrm{rank}(\phi(X(E_{jj}+E_{ss})X^*\otimes YE_{ii}Y^*))\leq k,\end{align*} $$

for any possible $i,j,s$ with $j\neq s $ .

Proof of Claim 1. For simplicity, we denote

$$ \begin{align*}T=\phi (XE_{ii}X^*\otimes YE_{jj}Y^*)\quad \text{and}\quad S=\phi(XE_{ii}X^*\otimes YE_{ss}Y^*).\end{align*} $$

We need to show

$$ \begin{align*}T\perp S \quad\mathrm{and }\quad \mathrm{rank}(T+S)\leq k.\end{align*} $$

With the assumption in (2.1), we have

$$ \begin{align*}\|\phi(XFX^{*}\otimes YGY^{*})\|_{(p,k)}=\|XFX^{*}\otimes YGY^{*}\|_{(p,k)}=\|F\otimes G\|_{(p,k)}\end{align*} $$

for all $F\in M_m$ and $G\in M_n$ . It follows that

$$ \begin{align*}\|T+xS\|_{(p,k)}^p=\|E_{ii}\otimes (E_{jj}+xE_{ss})\|^{p}_{(p,k)}=1+x^p,\end{align*} $$

$$ \begin{align*}\|T-xS\|_{(p,k)}^p=\|E_{ii}\otimes (E_{jj}-xE_{ss})\|^{p}_{(p,k)}=1+x^p,\end{align*} $$

$$ \begin{align*}\|T\|^p_{(p,k)}=1 \quad \text{and}\quad \|xS\|^p_{(p,k)}=x^p\end{align*} $$

for all $0<x<1$ . We can conclude from the above equalities that

(2.12) $$ \begin{align} \|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p=2\|T\|_{(p,k)}^p+2\|xS\|_{(p,k)}^p\quad \text{for all}\quad 0<x<1. \end{align} $$

Applying Corollary 2.9 with $A=T$ and $B=xS$ , we get

(2.13) $$ \begin{align} \|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p\geq 2\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T+x^2S^*S) \end{align} $$

for all $0<x<1.$ Since $\|T\|_{(p,k)}^p=\displaystyle \sum _{i=1}^k\lambda _i^{\frac {p}{2}}(T^*T)$ and $\|xS\|_{(p,k)}^p=\displaystyle \sum _{i=1}^k\lambda _i^{\frac {p}{2}}(x^2S^*S),$ it follows from (2.12) and (2.13) that

(2.14) $$ \begin{align} \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T+x^2S^*S)\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(x^2S^*S) \end{align} $$

for all $0<x<1.$

Replacing the role of $(T,S)$ with $(T^*,S^*)$ in the above argument, we have

(2.15) $$ \begin{align} \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(TT^*+x^2SS^*)\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(TT^*)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(x^2SS^*) \end{align} $$

for all $0<x<1.$ We claim that $s_k(T)=0$ . Otherwise, suppose $s_k(T)>0$ . Then, by (2.14) and (2.15), we can apply Corollary 2.12 to conclude that there exist unitary matrices $U,V\in M_{n}$ such that

$$ \begin{align*}UTV=\mathrm{diag}(s_1(T),\ldots,s_{mn}(T))\quad \text{and}\quad USV=0_{k+\ell}\oplus \hat{S},\end{align*} $$

where $\ell $ is the largest integer such that $s_{k+\ell }(T)=s_k(T).$ Thus, there exists a sufficiently small number $t>0$ such that the largest k singular values of $T+tS$ are $s_1(T),\ldots ,s_k(T)$ . Since $\|T\|_{(p,k)}^p=\|E_{ii}\otimes E_{jj}\|_{(p,k)}^p=1,$ we have

$$ \begin{align*}\|T+tS\|_{(p,k)}^p=\sum_{i=1}^ks_i^p(T+tS)=\sum_{i=1}^ks_i^p(T)=\|T\|_{(p,k)}^p=1,\end{align*} $$

which contradicts the fact that

$$ \begin{align*}\|T+xS\|_{(p,k)}^p=\|E_{ii}\otimes (E_{jj}+xE_{ss})\|_{(p,k)}^p=1+x^p\quad \text{for all}\quad 0<x<1.\end{align*} $$

So, our claim is correct, that is, $s_k(T)=0.$

Now, applying Corollary 2.12 again, we have $T\perp S.$ Notice that

$$ \begin{align*}\|T+S\|_{(p,k)}^p=\|T\|^p_{(p,k)}+\|S\|_{(p,k)}^p.\end{align*} $$

Applying Lemma 2.10 on S and T, we have

$$ \begin{align*}\mathrm{rank}(T+S)\leq k.\end{align*} $$

Similarly, we can also show that

$$ \begin{align*}\phi (XE_{jj}X^*\otimes YE_{ii}Y^*)\perp \phi(XE_{ss}X^*\otimes YE_{ii}Y^*)\end{align*} $$

and

$$ \begin{align*}\textrm{rank}(\phi(X(E_{jj}+E_{ss})X^*\otimes YE_{ii}Y^*))\leq k\end{align*} $$

for any possible $i,j,s$ with $j\neq s$ .

Claim 2 For any unitary matrices $X\in M_m\text { and } Y\in M_n$ , we have

$$ \begin{align*}\phi (XE_{ii}X^*\otimes Y(E_{jj}+E_{ss})Y^*)\perp \phi(XE_{tt}X^*\otimes Y(E_{jj}+ E_{ss})Y^*)\end{align*} $$

whenever $i\neq t.$

Proof of Claim 2. For simplicity, we denote

$$ \begin{align*}T=\phi (XE_{ii}X^*\otimes Y(E_{jj}+E_{ss})Y^*) \quad \mathrm{and }\quad S=\phi(XE_{tt}X^*\otimes Y(E_{jj} +E_{ss})Y^*).\end{align*} $$

We need to show $S\perp T.$ Applying Corollary 2.9 on T and $xS$ , we get

(2.16) $$ \begin{align} \|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p\geq 2\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T+x^2S^*S) \end{align} $$

for all $0<x<1.$ With the assumption in (2.1), we have

1. (i) $\|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p=2\|T\|_{(p,k)}^p+2\|xS\|_{(p,k)}^p$ for the case $k\geq 4$ ;

2. (ii) $\|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p=2\|T\|_{(p,k)}^p+\|xS\|_{(p,k)}^p$ for the case $k=3$ ;

3. (iii) $\|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p=2\|T\|_{(p,k)}^p$ for the case $k=2.$

So we can conclude that for any integer $k\geq 2$ ,

(2.17) $$ \begin{align} \nonumber \|T+xS\|_{(p,k)}^p+\|T-xS\|_{(p,k)}^p&\leq 2\|T\|_{(p,k)}^p+2\|xS\|_{(p,k)}^p \\ &=2\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T)+2\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(x^2S^*S). \end{align} $$

It follows that

(2.18) $$ \begin{align} \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T+x^2S^*S)\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(T^*T)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(x^2S^*S) \end{align} $$

for all $0<x<1.$ The above observations also hold if $(T,S)$ is replaced by $(T^*,S^*)$ , that is,

(2.19) $$ \begin{align} \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(TT^*+x^2SS^*)\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}(TT^*)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}(x^2SS^*) \end{align} $$

for all $0<x<1.$

If $s_k(T)=0$ , then applying Corollary 2.12, we have $T\perp S$ . Otherwise, $s_k(T)>0$ . Notice that Claim 1 implies $\mathrm {rank}(T)\leq k.$ Thus, by (2.18) and (2.19), we can apply Corollary 2.12 to conclude that there exist unitary matrices $U,V\in M_{mn}$ such that

$$ \begin{align*}UTV=\mathrm{diag}(s_1(T),\ldots,s_{k}(T))\oplus 0_{mn-k}\quad \text{and}\quad USV=0_{k}\oplus \hat{S}.\end{align*} $$

It follows that $T\perp S.$ This completes the proof.

Claim 3 For any unitary matrices $X\in M_m$ and $Y\in M_n$ ,

$$ \begin{align*}\phi(XE_{ii}X^*\otimes YE_{jj}Y^*)\perp \phi(XE_{rr}X^*\otimes YE_{ss}Y^*)\quad \text{for any }(i,j)\neq (r,s).\end{align*} $$

Proof of Claim 3. If $i=r$ or $j=s$ , then applying Claim 1 directly, we have

$$ \begin{align*}\phi(XE_{ii}X^*\otimes Y E_{jj}Y^*)\perp \phi (XE_{rr}X^*\otimes YE_{ss}Y^*).\end{align*} $$

Next, we suppose that $i\neq r$ and $j\neq s$ . With Claim 1, we have

(2.20) $$ \begin{align} \phi(XE_{ii}X^*\otimes Y E_{jj}Y^*)\perp \phi (XE_{ii}X^*\otimes YE_{ss}Y^*) \end{align} $$

and

(2.21) $$ \begin{align} \phi(XE_{rr}X^*\otimes Y E_{jj}Y^*)\perp \phi (XE_{rr}X^*\otimes YE_{ss}Y^*). \end{align} $$

With Claim 2, we have

(2.22) $$ \begin{align} \phi(XE_{ii}X^*\otimes Y (E_{jj}+E_{ss})Y^*)\perp \phi (XE_{rr}X^*\otimes Y(E_{jj}+E_{ss})Y^*). \end{align} $$

Applying Lemma 2.6, we conclude from (2.20) and (2.22) that

(2.23) $$ \begin{align} \phi(XE_{ii}X^*\otimes Y E_{jj}Y^*)\perp \phi (XE_{rr}X^*\otimes Y(E_{jj}+E_{ss})Y^*). \end{align} $$

Then, applying Lemma 2.6 again, we can conclude from (2.21) and (2.23) that

$$ \begin{align*}\phi(XE_{ii}X^*\otimes Y E_{jj}Y^*)\perp \phi (XE_{rr}X^*\otimes YE_{ss}Y^*).\end{align*} $$

Now we prove that $\phi $ has the desired form (2.2). For any unitary matrix $Y\in M_n$ , applying Claims 1 and 3, we know

$$ \begin{align*}\mathscr{F}=\{\phi(E_{ii}\otimes YE_{jj}Y^*):i=1,\ldots,m\text{ and }j=1\ldots,n\}\end{align*} $$

is a set of $mn$ orthogonal matrices in $M_{mn}$ . By Claim 1, each matrix in $\mathscr {F}$ has exactly one nonzero singular value, which equals 1. Thus, there exist unitary matrices $U_Y,V_Y\in M_{mn}$ such that

(2.24) $$ \begin{align} \phi(E_{ii}\otimes YE_{jj}Y^*)=U_Y(E_{ii}\otimes E_{jj})V_Y^* \end{align} $$

for all $ i=1,\ldots , m \text { and } j=1,\ldots , n.$ Without loss of generality, we may assume that $U_I=V_I=I_{mn}$ , i.e.,

(2.25) $$ \begin{align} \phi(E_{ii}\otimes E_{jj})=E_{ii}\otimes E_{jj} \end{align} $$

for all $i=1,\ldots , m \text { and } j=1,\ldots , n.$ By (2.24) and (2.25), we have:

1. (i) $I_{mn}=\phi (I_m\otimes I_n)=U_Y(I_m\otimes I_n)V_Y^*$ ;

2. (ii) $E_{ii}\otimes I_n=\phi (E_{ii}\otimes I_n)=U_Y(E_{ii}\otimes I_n)V_Y^* \text { for all } i=1,\ldots ,m.$

It follows that $U_Y=V_Y$ and $U_Y$ commutes with $E_{ii}\otimes I_n$ for all $i=1,\ldots , m$ . Therefore, $U_Y$ commutes with $E_{11}\otimes I_n+2E_{22}\otimes I_n+\cdots +mE_{mm}\otimes I_n,$ which implies that $U_Y=\bigoplus \limits _{i=1}^{m}U_{i,Y}$ with unitary matrices $U_{i,Y}\in M_n$ . It follows that

$$ \begin{align*}\phi(E_{ii}\otimes YE_{jj}Y^*)=E_{ii}\otimes U_{i,Y}E_{jj}U_{i,Y}^*.\end{align*} $$

So far, we have showed that for any unitary matrix $Y\in M_n$ , there exists a unitary matrix $U_{i,Y}\in M_n$ depending on i and Y such that

$$ \begin{align*}\phi(E_{ii}\otimes YE_{jj}Y^*)=E_{ii}\otimes U_{i,Y}E_{jj}U_{i,Y}^*\quad \text{for}\quad j=1,\ldots,n.\end{align*} $$

By the linearity of $\phi $ , we conclude from the above equation that for any $i=1,\ldots , m$ , there exists a linear map $\psi _i$ such that

$$ \begin{align*}\phi(E_{ii}\otimes B)=E_{ii}\otimes \psi_{i}( B) \quad \text{for all}\quad B \in M_n.\end{align*} $$

Let $\hat {k}=\min \{k,n\}$ . Then it is easy to check that

$$ \begin{align*}\|\psi_i(B)\|_{(p,\hat{k})}=\|E_{ii}\otimes \psi_i(B)\|_{(p,k)}=\|E_{ii}\otimes B\|_{(p,k)}=\|B\|_{(p,\hat{k})}\end{align*} $$

for all $B\in M_n$ . That is, $\psi _i$ is a linear map on $M_n$ preserving the $(p,\hat {k})$ -norm. Thus, by Theorem 1 in [Reference Li and Tsing23], $\psi _i$ has form $B\mapsto W_i B\widetilde {W}_i$ or $B\mapsto W_i B^T\widetilde {W}_i$ for some unitary matrices $W_i, \widetilde {W}_i\in M_n.$ Let $W=\bigoplus \limits _{i=1}^m W_i$ and $\widetilde {W}=\bigoplus \limits _{i=1}^m\widetilde {W}_i$ . It follows that for any ${i=1,\ldots , m,}$

$$ \begin{align*}\phi(E_{ii}\otimes B)=W(E_{ii}\otimes \varphi_i (B))\widetilde{W}\quad \text{for all}\quad B\in M_n,\end{align*} $$

where $\varphi _i$ is the identity map or the transposition map. Recall that $I_{mn}=\phi (I_m\otimes I_n). $ Thus, we have $\widetilde {W} = W^*$ .

Applying Claim 3 again, we can repeat the same argument above to show that for any unitary matrix $X \in M_m$ and any integer $1\leq i\leq m$ , there exists a unitary matrix $W_X$ such that

$$ \begin{align*} \phi(XE_{ii}X^*\otimes B)=W_X(E_{ii}\otimes \varphi_{i,X} (B)){W}_X^*\quad \text{for all}\quad B\in M_n, \end{align*} $$

where $\varphi _{i,X}$ is the identity map or the transposition map. We may further assume that $W_I=I_{mn}$ . Then, applying Lemma 2.13, we have

$$ \begin{align*}\phi(A\otimes B)=\varphi_1(A)\otimes \varphi_2(B) \quad \text{for all}\quad A\in M_m\text{ and }B\in M_n,\end{align*} $$

where $\varphi _2$ is the identity map or the transposition map and $\varphi _1$ is a linear map on $M_m$ . Let $\widetilde {k}=\min \{k,m\}.$ It is easy to verify that $\varphi _1$ is a linear map on $M_m$ preserving the $(p,\widetilde {k})$ -norm. Hence, $\varphi _1$ also has the form $A\mapsto UAV$ or $A\mapsto UA^TV$ for some unitary matrices $U, V\in M_m$ . This completes the proof.

Proof Considering all symmetric real matrices as in the proof of Lemma 2.13, one can conclude that there exists some unitary matrix $Z_{X}$ such that

(3.2) $$ \begin{align} \phi\left (\bigotimes\limits_{i=1}^{m-1}X_iE_{j_ij_i}X_i^*\otimes B\right )= \left(Z_{X}\left (\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\right )Z_{X}^*\right) \otimes \varphi_{j_1,\ldots,j_{m-1},X}(B) \end{align} $$

for all $B\in M_{n_m}$ and integers $1\leq j_i\leq n_i $ with $ 1\leq i\leq m-1$ . Define linear maps $\mathrm {tr_1}:M_N\to M_{n_m}$ and $\mathrm {Tr_1}:M_N\to M_{n_m}$ by

$$ \begin{align*}\mathrm{tr_1}(A\otimes B)=\mathrm{tr}(A)B\quad \text{and} \quad \mathrm{Tr_1}(A\otimes B)=\mathrm{tr_1}(\phi(A\otimes B ))\end{align*} $$

for all $A\in M_{n_1\cdots n_{m-1}}$ and $B\in M_{n_m}$ . Then

$$ \begin{align*}\mathrm{Tr_1}\left(\bigotimes\limits_{i=1}^{m-1}X_iE_{j_ij_i}X_i^*\otimes B\right )=\varphi_{j_1,\ldots,j_{m-1},X}(B).\end{align*} $$

Notice that $\text {Tr}_1$ is linear and therefore continuous. Besides, the set

(3.3) $$ \begin{align} \nonumber \left\{\bigotimes\limits_{i=1}^{m-1}X_iE_{j_ij_i}X_i^*\mid 1\leq j_i\leq n_i \text{ and } X_i\in M_{n_i} \text{ is unitary for }i=1,\ldots,m-1\right\} \\ =\left\{\bigotimes\limits_{i=1}^{m-1}x_ix_i^*\mid x_i\in \mathbb{C}^{n_i} \text{ with } x_i^*x_i=1 \text{ for }i=1,\ldots,m-1\right\} \end{align} $$

is connected. So, all the maps $\varphi _{j_1,\ldots ,j_{m-1},X}$ are the same. Then (3.2) can be rewritten as

$$ \begin{align*}\phi\left (\bigotimes\limits_{i=1}^{m-1}X_iE_{j_ij_i}X_i^*\otimes B\right )= \left(Z_{X}\left (\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\right )Z_{X}^*\right) \otimes \varphi_{2}(B),\end{align*} $$

where $\varphi _2$ is the identity map or the transposition map. With the linearity of $\phi $ , it follows that

$$ \begin{align*}\phi(A_1\otimes \cdots \otimes A_{m-1}\otimes B)=\varphi_1(A_1\otimes \cdots \otimes A_{m-1})\otimes \varphi_2(B)\end{align*} $$

for some linear map $\varphi _1.$

Proof By Theorem 3.2 of [Reference Fošner, Huang, Li and Sze10], we know the result holds for $k=1$ . So we may assume $k\ge 2$ .

We use induction on m. By Theorem 2.1, the result holds for $m=2$ . Now suppose that $m\geq 3$ and the result holds for any $(m-1)$ -partite system. We need to show that the result holds for any m-partite system.

We first show that for any unitary matrices $X_i\in M_{n_i}, i=1,\ldots ,m,$

(3.6) $$ \begin{align} \phi(X_1E_{i_1i_1}X_1^*\otimes\cdots \otimes X_mE_{i_mi_m}X_m^*)\perp\phi(X_1E_{j_1j_1}X_1^*\otimes \cdots\otimes X_mE_{j_mj_m}X_m^*) \end{align} $$

for any $(i_1,\ldots ,i_m)\neq (j_1,\ldots ,j_m).$ Without loss of generality, we need only to prove that (3.6) holds when $X_i=I_{n_i}$ for $i=1,\ldots ,m$ . By Lemma 2.6, it suffices to show that for any integer $1\leq s\leq m,$ we have

(3.7) $$ \begin{align} \nonumber &\phi\left (\bigotimes\limits_{u=1}^{s-1}(E_{i_ui_u}+E_{j_uj_u})\otimes E_{i_si_s}\otimes \bigotimes\limits_{u=s+1}^mE_{i_ui_u}\right ) \\ &\qquad\qquad\qquad\perp \phi\left (\bigotimes\limits_{u=1}^{s-1}(E_{i_ui_u}+E_{j_uj_u})\otimes E_{j_sj_s}\otimes \bigotimes\limits_{u=s+1}^mE_{i_ui_u}\right ) \end{align} $$

for all $\mathrm {i}=(i_1,\ldots ,i_m)$ and $\mathrm {j}=(j_1,\ldots ,j_m)$ with $i_u\neq j_u$ for $ u=1,\ldots , s.$ Denote by $A_s(\mathrm {i,j})$ and $B_s(\mathrm {i,j})$ the two matrices in (3.7) accordingly. It is easy to check that for any integer $1\leq s\leq m$ and real number $0<x<1$ ,

$$ \begin{align*} &\|A_s(\mathrm{i,j})+xB_s(\mathrm{i,j})\|_{(p,k)}^p+\|A_s(\mathrm{i,j}) -xB_s(\mathrm{i,j})\|_{(p,k)}^p\qquad\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\leq 2\|A_s(\mathrm{i,j})\|^p_{(p,k)}+2\|xB_s(\mathrm{i,j})\|^p_{(p,k)}.\notag \end{align*} $$

Then, applying the same argument as in the proof of Theorem 2.1, we conclude that for any integer $1\leq s \leq m$ and real number $0<x<1$ ,

(3.8) $$ \begin{align} \nonumber &\sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(A_s^*(\mathrm{i,j})A_s(\mathrm{i,j}\big)+x^2B_s^*(\mathrm{i,j})B_s(\mathrm{i,j})\big) \\ &\qquad\qquad\qquad\qquad\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(A^*_s(\mathrm{i,j})A_s(\mathrm{i,j})\big)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(x^2B_s^*(\mathrm{i,j})B_s(\mathrm{i,j})\big) \end{align} $$

and

(3.9) $$ \begin{align} \nonumber& \sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(A_s(\mathrm{i,j})A^*_s(\mathrm{i,j})+x^2B_s(\mathrm{i,j})B_s^*(\mathrm{i,j})\big) \\&\qquad\qquad\qquad\qquad\leq \sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(A_s(\mathrm{i,j})A^*_s(\mathrm{i,j})\big)+\sum_{i=1}^k\lambda_i^{\frac{p}{2}}\big(x^2B_s(\mathrm{i,j})B^*_s(\mathrm{i,j})\big) \end{align} $$

for all $\mathrm {i}=(i_1,\ldots ,i_m)$ and $\mathrm {j}=(j_1,\ldots ,j_m)$ with $i_u\neq j_u$ for $ u=1,\ldots , s.$ Now we distinguish two cases.

Case 1. Suppose $k>2^{m-1}$ . Then

(3.10) $$ \begin{align} \|A_s(\mathrm{i,j})+xB_s(\mathrm{i,j})\|_{(p,k)}^p=2^{s-1}+ a_s x^p\quad \text{ for}\quad 0<x<1, \end{align} $$

where $a_s=2^{s-1}$ for $s=1,\ldots ,m-1$ and $a_m=\min \{k-2^{m-1},2^{m-1}\}$ . We claim that

(3.11) $$ \begin{align} s_k(A_{s}(\mathrm{i,j}))=0 \mathrm{\quad for}\quad s=1,\ldots,m. \end{align} $$

Otherwise, $s_k(A_s(\mathrm {i,j}))>0$ for some $1\leq s\leq m$ . Then, by (3.8) and (3.9), we can use the same argument as in Claim 1 in the proof of Theorem 2.1 to conclude that there exists a sufficiently small $x>0$ such that

$$ \begin{align*}\|A_s(\mathrm{i,j})+xB_s(\mathrm{i,j})\|_{(p,k)}^p=\|A_s(\mathrm{i,j})\|_{(p,k)}^p=2^{s-1},\end{align*} $$

which contradicts (3.10). Thus, (3.11) holds. Applying Corollary 2.12, we have

$$ \begin{align*}A_s(\mathrm{i,j})\perp B_s(\mathrm{i,j}) \mathrm{\quad for}\quad s=1,\ldots,m.\end{align*} $$

Case 2. Suppose $k\leq 2^{m-1}$ . Let $s_0$ be the integer such that $2^{s_0-1}< k\leq 2^{s_0}.$ We can use the same argument as in Case 1 to show that for any integer $1\leq s\leq s_0$ ,

(3.12) $$ \begin{align} A_{s}(\mathrm{i,j})\perp B_s(\mathrm{i,j})\quad\text{ and}\quad s_k(A_s(\mathrm{i,j}))=0 \end{align} $$

for all $\mathrm {i}=(i_1,\ldots ,i_m)$ and $\mathrm {j}=(j_1,\ldots ,j_m)$ with $i_u\neq j_u$ for $u=1,\ldots ,s.$

Next, we use induction on s to prove that for any $s_0+1\leq s \leq m, \mathrm {i}=(i_1,\ldots ,i_m )$ and $\mathrm {j}=(j_1,\ldots ,j_m)$ with $i_u\neq j_u$ for $u=1,\ldots , s,$ there exist unitary matrices $U,V\in M_N$ depending on s and $(\mathrm {i,j})$ such that

(3.13) $$ \begin{align} UA_{s}(\mathrm{i,j})V=I_{2^{s-1}}\oplus 0_{N-2^{s-1}}\quad \text{and}\quad A_{s}(\mathrm{i,j})\perp B_s(\mathrm{i,j}). \end{align} $$

First, by (3.12), we have $A_{s_0}(\mathrm {i,j})\perp B_{s_0}(\mathrm {i,j})$ and there exist unitary matrices $U,V\in M_N$ and an integer $0\leq r< k$ such that

$$ \begin{align*}UA_{s_0}(\mathrm{i,j})V=\mathrm{diag}(a_1,\ldots,a_r)\oplus 0\end{align*} $$

and

$$ \begin{align*}UB_{s_0}(\mathrm{i,j})V=0_r\oplus \mathrm{diag}(b_{r+1},\ldots,b_N),\end{align*} $$

with $a_1\geq \cdots \geq a_r>0$ and $b_{r+1}\geq \cdots \geq b_N\geq 0$ . If $a_1>1$ , then by (3.12), applying Lemma 2.6, we have $s_1\Big (\phi \Big (\bigotimes \limits _{u=1}^mE_{i_ui_u}\Big )\Big )>1$ for some $(i_1,\ldots ,i_m)$ . It follows that $\Big \|\phi \Big (\bigotimes \limits _{u=1}^mE_{i_ui_u}\Big )\Big \|_{(p,k)}>1$ , which contradicts (3.4). Thus, $a_1\leq 1$ , and similarly $b_{r+1}\leq 1.$ Then we have

(3.14) $$ \begin{align} \sum_{j=1}^ra_j^p+\sum_{j=r+1}^kb_j^p\leq k. \end{align} $$

Clearly, $a_1\geq \cdots \geq a_r\geq xb_{r+1}\geq \cdots \geq xb_k$ are the largest k singular values of $A_{s_0}(\mathrm {i,j})+xB_{s_0}(\mathrm {i,j})$ for all $0<x\leq \frac {a_r}{b_{r+1}}$ . Hence,

$$ \begin{align*}\|A_{s_0}(\mathrm{i,j})+xB_{s_0}(\mathrm{i,j})\|^p_{(p,k)}=\sum_{j=1}^ra_j^p+x^p\sum_{j=r+1}^kb_j^p\quad \text{for all}\quad 0< x\leq \frac{a_r}{b_{r+1}}.\end{align*} $$

On the other hand, with (3.4), we have

$$ \begin{align*}\|A_{s_0}(\mathrm{i,j})+xB_{s_0}(\mathrm{i,j})\|^p_{(p,k)}=2^{s_0-1}+x^p(k-2^{s_0-1})\quad \text{for all}\quad 0<x\leq 1.\end{align*} $$

It follows from the above two equations that

$$ \begin{align*}\displaystyle\sum_{j=1}^ra_j^p=2^{s_0-1} \quad\mathrm{and}\quad \displaystyle\sum_{j=r+1}^kb_j^p=k-2^{s_0-1}.\end{align*} $$

Therefore, $ \sum _{j=1}^ra_j^p+ \sum _{j=r+1}^kb_j^p=k,$ i.e., the equality in (3.14) holds, which implies that $a_j=1$ for $j=1,\ldots ,r$ . Notice that $\|A_{s_0}(\mathrm {i,j})\|_{(p,k)}^p=2^{s_0-1}$ . Thus, $r=2^{s_0-1},$ that is, $UA_{s_0}(\mathrm {i,j})V=I_{2^{s_0-1}}\oplus 0_{N-2^{s_0-1}}$ . Hence, (3.13) holds for $s_0$ .

Suppose that (3.13) holds for $s-1$ with $s_0< s\leq m.$ We will show that (3.13) also holds for s. Notice that $A_{s}(\mathrm {i,j})=A_{s-1}(\mathrm {\hat {i},\hat {j}})+B_{s-1}(\mathrm {\hat {i},\hat {j}})$ for some $\mathrm {\hat {i}}=(\hat {i}_1,\ldots ,\hat {i}_m)$ and $\mathrm {\hat {j}}=(\hat {j}_1,\ldots ,\hat {j}_m)$ with $\hat {i}_u\neq \hat {j}_u$ for $u=1,\ldots , s-1.$ By our assumption, we have

$$ \begin{align*}UA_{s-1}(\mathrm{\hat{i},\hat{j}})V=I_{2^{s-2}}\oplus 0_{N-2^{s-2}}\quad\text{and}\quad UB_{s-1}(\mathrm{\hat{i},\hat{j}})V=0_{2^{s-2}}\oplus I_{2^{s-2}}\oplus 0_{N-2^{s-1}}\end{align*} $$

for some unitary matrices $U,V\in M_N$ . It follows that

$$ \begin{align*}UA_{s}(\mathrm{i,j})V=I_{2^{s-1}}\oplus 0_{N-2^{s-1}}.\end{align*} $$

Then, with (3.8) and (3.9), we apply Corollary 2.12 to conclude that

$$ \begin{align*}UB_s(\mathrm{i,j})V=0_{2^{s-1}}\oplus \hat{B}\end{align*} $$

for some $\hat {B}\in M_{N-2^{s-1}}$ . It follows that $A_s(\mathrm {i,j})\perp B_s(\mathrm {i,j}).$ Now we have proved that (3.13) holds for s. Then we can conclude from the above discussion that for any $s=1,\ldots ,m$ ,

$$ \begin{align*}A_s(\mathrm{i,j})\perp B_s(\mathrm{i,j})\end{align*} $$

for all $\mathrm {i}=(i_1,\ldots ,i_m)$ and $\mathrm {j}=(j_1,\ldots ,j_m)$ with $i_u\neq j_u$ for $ u=1,\ldots , s,$ that is, (3.6) holds.

It follows that for any unitary matrix $X_m\in M_{n_m}$ , there exist unitary matrices $U_{X_m}$ and $V_{X_m}$ such that

(3.15) $$ \begin{align} \phi\left(\bigotimes\limits_{i=1}^{m-1} E_{j_ij_i}\otimes X_mE_{j_{m}j_{m}}X_m^*\right)= U_{X_m}(E_{j_1j_1}\otimes \cdots \otimes E_{j_mj_m})V_{X_m}^* \end{align} $$

for all $ j_i=1,\ldots , n_i$ with $1 \leq i\leq m.$ For simplicity, we may assume that $U_{I}=V_{I}=I$ , i.e.,

(3.16) $$ \begin{align} \phi(E_{j_1j_1}\otimes\cdots\otimes E_{j_{m}j_{m}})= E_{j_1j_1}\otimes \cdots \otimes E_{j_mj_m}. \end{align} $$

Then we have $\phi (I_N)=I_N$ . Applying a similar argument as in the last two paragraphs of the proof of Theorem 2.1, one can conclude from (3.15) and (3.16) that there are unitary matrices $W,\widetilde {W}\in M_N$ such that for all $ j_i=1,\ldots ,n_i$ with $1\leq i\leq m-1$ ,

$$ \begin{align*}\phi\left(\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\otimes B\right)= W\left(\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\otimes \varphi_{j_1,\ldots,j_{m-1}}(B)\right)\widetilde{W},\end{align*} $$

where $\varphi _{j_1,\ldots ,j_{m-1}}$ is the identity map or the transposition map. It follows that ${\phi (I_N)=W\widetilde {W}}$ . Recall that $\phi (I_N)=I_N$ and W and $\widetilde {W}$ are both unitary matrices. We have $\widetilde {W}=W^*.$

Following a similar argument as above, one can show that for any ${X=(X_1,\ldots ,X_{m-1})}$ and any integer $ j_i=1,\ldots , n_i$ with $1\leq i\leq m-1$ , there exists a unitary matrix $W_X\in M_N$ such that

(3.17) $$ \begin{align} \phi\left (\bigotimes\limits_{i=1}^{m-1}X_iE_{j_ij_i}X_i^*\otimes B\right )=W_X\left (\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\otimes\varphi_{j_1,\ldots,j_{m-1},X}(B)\right )W_X^* \end{align} $$

for all $B\in M_{n_m},$ where $\varphi _{j_1,\ldots ,j_{m-1},X}$ is the identity map or transposition map. Denote $\mathrm {I}=(I_{n_1},\ldots ,I_{n_{m-1}}).$ For simplicity, we may further assume that $W_{\mathrm {I}}=I_N$ , i.e., for any integer $j_i=1,\ldots , n_i$ with $1\leq i\leq m-1$ ,

$$ \begin{align*} \phi\left (\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\otimes B\right )=\bigotimes\limits_{i=1}^{m-1}E_{j_ij_i}\otimes\varphi_{j_1,\ldots,j_{m-1},\mathrm{I}}(B)\quad \text{for all}\quad B\in M_{n_m}. \end{align*} $$

Then we apply Lemma 3.1 to conclude that

$$ \begin{align*}\phi(A_1\otimes \cdots\otimes A_{m-1}\otimes B)=\psi(A_1\otimes \cdots\otimes A_{m-1})\otimes \varphi_m(B)\end{align*} $$

for all $B\in M_{n_m}$ and $A_i\in M_{n_i}$ , $1\leq i\leq m-1$ , where $\varphi _m$ is the identity map or the transposition map and $\psi $ is a linear map. Let $\hat {k}=\min \Big \{k,\prod \limits _{i=1}^{m-1}n_i\Big \}.$ It is easy to check that

$$ \begin{align*}\|\psi(A_1\otimes\cdots \otimes A_{m-1})\|_{(p,\hat{k})}=\|A_1\otimes\cdots \otimes A_{m-1}\|_{(p,\hat{k})}\end{align*} $$

for all $A_i \in M_{n_i}, i=1,\ldots ,m-1.$ Then, by the induction hypothesis, we conclude that there exist unitary matrices $\widetilde {U}$ and $\widetilde {V} $ such that

$$ \begin{align*}\psi(A_1\otimes\cdots \otimes A_{m-1})=\widetilde{U}(\varphi_1(A_1)\otimes \cdots\otimes \varphi_{m-1}(A_{m-1}))\widetilde{V},\end{align*} $$

where $\varphi _i$ is the identity map or the transposition map for $i=1,\ldots ,m-1.$ Therefore, $\phi $ has the desired form and the proof is completed.