# When does the Cauchy-Schwarz inequality for spectral radii of tensor products become equality?

Let $$V$$ be a complex finite dimensional inner product space. If $$A_{1},\dots,A_{n}:V\rightarrow V$$ are linear operators, then let $$\Phi(A_{1},\dots,A_{n}):L(V)\rightarrow L(V)$$ be the superoperator defined by letting $$\Phi(A_{1},\dots,A_{n})(X)=A_1XA_1^*+\dots+A_nXA_n^*$$. The completely positive superoperators from $$L(V)$$ to $$L(V)$$ are precisely the mappings of the form $$\Phi(A_{1},\dots,A_{n})$$.

If $$A$$ is an operator, then let $$\rho(A)$$ denote the spectral radius of $$A$$. Then

$$\rho(A_{1}\otimes B_{1}+\dots+A_{n}\otimes B_{n})\leq\rho(\Phi(A_{1},\dots,A_{n}))^{1/2}\rho(\Phi(B_{1},\dots,B_{n}))^{1/2}.$$

Is there a characterization of the systems of linear operators $$((A_{1},\dots,A_{n}),(B_{1},\dots,B_{n}))$$ for which $$\rho(A_{1}\otimes B_{1}+\dots+A_{n}\otimes B_{n})=\rho(\Phi(A_{1},\dots,A_{n}))^{1/2}\rho(\Phi(B_{1},\dots,B_{n}))^{1/2}?$$

• If $A_i=\alpha_iU_i,B_i=\beta_iV_i$ where $\alpha_i\geq 0,\beta_i\geq 0$ for all $i$ where $\Phi(A_1,\dots,A_n)\neq 0,\Phi(B_1,\dots,B_n)\neq 0$, then the equality holds precisely when there is some non-zero common $\mathbf{x}\in V\otimes V$ and common eigenvalue $\lambda$ with $(U_{i}\otimes V_{i})\mathbf{x}=\lambda\mathbf{x}$. I have not yet found a nice way to simplify this characterization. Commented May 2, 2022 at 15:25

Yes. We have a necessary and sufficient characterization for when the Cauchy-Schwarz inequality becomes equality.

For this post, $$U,V,W$$ shall denote finite dimensional complex Hilbert spaces. Suppose that $$A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$$ are linear operators. Then define an operator $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n):L(V,U)\rightarrow L(V,U)$$ by letting $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)(X)=A_1XB_1^*+\dots+A_nXB_n^*.$$

Observe that $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$$ is similar to $$A_1\otimes \overline{B_1}+\dots+A_n\otimes\overline{B_n}$$ (Here, $$\overline{C}=(C^{T})^{*}=(C^*)^T$$, so $$\overline{C}$$ is the matrix obtained by replacing every entry in $$C$$ with its complex conjugate), so the spectral radius Cauchy-Schwarz inequality $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)) \leq\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ hold in all cases. It is not too hard to prove the spectral radius Cauchy-Schwarz inequality by using the conventional Cauchy-Schwarz inequality and the characterization of $$\rho(\Phi(A_1,\dots,A_n))^{1/2}$$ given in the 1998 paper The $$p$$-norm joint spectral radius for even integers by Ding-Xuan Zhou.

Observe that if there is a $$\lambda$$ and an invertible $$B$$ with $$B_j=\lambda BA_jB^{-1}$$ for $$1\leq j\leq n$$, then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)) =\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}.$$

I claim that when $$(A_1,\dots,A_n)$$ and $$(B_1,\dots,B_n)$$ have no invariant subspaces, this is the only way in which the spectral radius Cauchy-Schwarz inequality becomes equality. By decomposing $$(A_1,\dots,A_n)$$ and $$(B_1,\dots,B_n)$$ according to their invariant subspaces, we obtain necessary and sufficient conditions for when the Cauchy-Schwarz inequality is actually an equality.

Recall that a linear operator $$\mathcal{E}:L(U)\rightarrow L(V)$$ is called positive if $$\mathcal{E}(P)$$ is positive semidefinite whenever $$P$$ is positive semidefinite. A linear operator $$\mathcal{E}:L(U)\rightarrow L(V)$$ is called completely positive if $$\mathcal{E}\otimes 1_W:L(U\otimes W)\rightarrow L(V\otimes W)$$ is positive whenever $$W$$ is a finite dimensional Hilbert space. We say that an operator $$\mathcal{E}:L(U)\rightarrow L(V)$$ is trace preserving if $$\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$$ for all $$X\in L(U)$$. An operator $$\mathcal{E}:L(U)\rightarrow L(V)$$ is said to be a quantum channel if $$\mathcal{E}$$ is both trace preserving and completely positive.

The completely positive mappings from $$L(V)$$ to $$L(V)$$ are precisely the mappings of the form $$\Phi(A_1,\dots,A_n)$$. Observe that any linear mapping from $$L(V,U)$$ to $$L(V,U)$$ is of the form $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$$. It is not too hard to show using the Frobenius inner product that the mapping $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$$ is trace preserving if and only if $$A_1^*B_1+\dots+A_n^*B_n=1_V$$. In particular, the quantum channels are precisely the mappings of the form $$\Phi(A_1,\dots,A_n)$$ where $$A_1^*A_1+\dots+A_n^*A_n=1_V$$.

Observe that if $$\mathcal{E}$$ is a quantum channel, then $$\rho(\mathcal{E})=1$$.

Let $$E_{U,n}$$ be the collection of tuples $$(A_1,\dots,A_n)\in L(U)^n$$ such that there is some complex number $$\lambda$$ and invertible $$B$$ such that if $$B_j=\lambda BA_jB^{-1}$$ for $$1\leq j\leq n$$, then $$\Phi(B_1,\dots,B_n)$$ is a quantum channel. By this answer, $$L(U)^n\setminus E_{U,n}$$ is a quite small set whenever $$n>1$$. In particular, if $$(A_1,\dots,A_n)$$ has no-invariant subspace, then $$(A_1,\dots,A_n)\in E_{U,n}$$.

Theorem: Suppose that $$(A_1,\dots,A_n)\in L(U)^n,(B_1,\dots,B_n)\in L(V)^n$$ have no invariant subspace. Then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ if and only if there is some invertible matrix $$C$$ along with some complex number $$\eta\neq 0$$ where $$A_j=\eta CB_jC^{-1}$$ for $$1\leq j\leq n$$.

Proof: Since $$(A_1,\dots,A_n),(B_1,\dots,B_n)$$ have no invariant subspace, there are non-zero complex numbers $$\mu,\nu$$ along with invertible matrices $$A,B$$ where if we set $$R_j=\mu AA_jA^{-1},S_j=\nu BB_jB^{-1}$$ for $$1\leq j\leq n$$, then $$\mathcal{E}=\Phi(R_1,\dots,R_n),\mathcal{F}=\Phi(S_1,\dots,S_n)$$ are quantum channels. Let $$W$$ be a complex inner product space with orthonormal basis $$(e_1,\dots,e_n).$$ Now, let $$R,S\in L(V,V\otimes W)$$ be the operators defined by letting $$R=\sum_{j=1}^{n}R_j\otimes e_j,S=\sum_{j=1}^{n}S_j\otimes e_j$$. Then $$\mathcal{E}(X)=\text{Tr}_{W}(RXR^*),\mathcal{F}(X)=\text{Tr}_{W}(SXS^*).$$

Define a mapping $$\mathcal{G}$$ by setting $$\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\sum_{k=1}^nR_kXS_k^*.$$

Since $$R_1^*R_1+\dots+R_n^*R_n=S_1^*S_1+\dots+S_n^*S_n=1_V$$, the mappings $$R,S$$ are isometries.

Now, let $$\lambda$$ be an eigenvalue of $$\mathcal{G}$$. Then there is some eigenvector $$X$$ with $$\mathcal{G}(X)=\lambda X$$. Now perform a polar decomposition of $$X$$ to write $$X=HP$$ where $$H$$ is an isometry and $$P$$ is a positive semidefinite matrix. Therefore, we have $$\lambda HP=\lambda X=\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\text{Tr}_W(RHPS^*).$$ Now, set $$T=(H^*\otimes 1_W)RH$$. Then $$\lambda P=\lambda H^*HP=H^*\text{Tr}_W(RHPS^*)=\text{Tr}_W((H^*\otimes 1_W)RHPS^*)=\text{Tr}_W(TPS^*).$$ Now let $$P=\sum_{k=1}^{n}\sigma_ke_ke_k^*$$. Then $$\text{Tr}(\lambda P)=\lambda\cdot\sum_{k=1}^n\sigma_k$$. Therefore, we have $$\text{Tr}(\lambda P)=\text{Tr}(\text{Tr}_W(TPS^*))=\text{Tr}(TPS^*)=\sum_{k=1}^{n}\sigma_k\text{Tr}(Te_ke_k^*S^*)=\sum_{k=1}^n\sigma_k\text{Tr}(Te_k(Se_k)^*)=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle.$$

Therefore, since $$\lambda\cdot\sum_{k=1}^n\sigma_k=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle,$$ we know that $$|\lambda|\leq 1$$. Furthermore, if $$|\lambda|=1$$, then we know that $$Te_k=\lambda Se_k$$ whenever $$\sigma_k>0$$. Therefore, we have $$T|_{\text{Im}(P)}=\lambda\cdot S|_{\text{Im}(P)}$$. In this case, we have $$\lambda P=\text{Tr}_W(TPS^*)=\text{Tr}_W(\lambda SPS^*)$$, so $$P=\text{Tr}_W(SPS^*)$$. Since $$P=\text{Tr}_W(SPS^*)$$ and since $$(S_1,\dots,S_n)$$ has no invariant subspace, we know that $$P$$ is positive semidefinite. Therefore, $$(H^*\otimes 1_W)RH=T=\lambda S$$.

Now, if $$v\in V$$, then $$\|RH v|=\|v\|=\|\lambda Sv\|=\|(H^*\otimes 1_W)RHv\|$$. Therefore, if $$v\in V$$, then $$RHv\in\text{Im}(H)\otimes 1_W$$. Therefore, since $$\text{Im}(H)$$ is a non-trivial invariance subspace of $$U$$, we know that $$\text{Im}(H)=U$$. Therefore, $$H$$ is a unitary operator.

Thus, $$H^*R_jH=\lambda S_j$$ for $$1\leq j\leq n$$. Thus, $$H^*\mu AA_jA^{-1}H=\lambda \nu BB_jB^{-1}$$. We conclude that $$A_j=\mu^{-1}\lambda\nu A^{-1}HBB_jB^{-1}H^{-1}A=\mu^{-1}\lambda\nu A^{-1}HBB_j(A^{-1}HB)^{-1}.$$

Q.E.D.

Theorem: Suppose that $$A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$$ be linear operators. Then assign $$U,V$$ bases so that $$A_j=\begin{bmatrix} A_{j,1,1}&\dots&A_{j,1,u}\\ \vdots&\ddots&\vdots\\ A_{j,u,1}&\dots&A_{j,u,u} \end{bmatrix}$$ and $$B_j=\begin{bmatrix} B_{j,1,1}&\dots&B_{j,1,v}\\ \vdots&\ddots&\vdots\\ B_{j,v,1}&\dots&B_{j,v,v} \end{bmatrix}$$ and where for each $$\alpha,\beta$$ each of the $$n$$ matrices $$A_{1,\alpha,\beta},\dots,A_{n,\alpha,\beta}$$ have the same dimensions, for each $$\alpha,\beta$$, each of the $$n$$ matrices $$B_{1,\alpha,\beta},\dots,B_{n,\alpha,\beta}$$ have the same dimensions, and where $$A_{j,\alpha,\beta}=0$$ whenever $$\alpha>\beta$$, and where $$B_{j,\alpha,\beta}=0$$ whenever $$\alpha>\beta$$, and where for $$1\leq\alpha\leq u$$, the matrices $$(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha})$$ have no non-trivial invariant subspace, and where if $$1\leq\beta\leq v$$, the matrices $$(B_{1,\beta,\beta},\dots,B_{n,\beta,\beta})$$ have non-trivial no invariant subspace either. Then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ if and only if there are $$\alpha,\beta$$ with $$1\leq\alpha\leq u,1\leq\beta\leq v$$ and where

1. $$\rho(\Phi(A_1,\dots,A_n))=\rho(\Phi(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha}))$$

2. $$\rho(\Phi(B_1,\dots,B_n))=\rho(\Phi(A_{1,\beta,\beta},\dots,A_{n,\beta,\beta}))$$, and

3. There is an invertible matrix $$C$$ and some complex number $$\lambda\neq 0$$ such that $$A_{j,\alpha,\beta}=\lambda CB_{j,\alpha,\beta}C^{-1}$$ for $$1\leq j\leq n$$.

A few observations:

In this answer, we actually have a couple of different proofs of the spectral radius Cauchy-Schwarz inequality. To prove the Cauchy-Schwarz inequality, we need to prove that $$\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$$ whenever $$\Phi(R_1,\dots,R_n),\Phi(S_1,\dots,S_n)$$ are quantum channels, and the general case will follow from the fact that $$E_{V,n}$$ is dense in $$L(V)^n$$ whenever $$n>1$$. We have already shown that $$\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$$, but there is another way of showing this using the induced trace norm.

If $$\mathcal{H}:L(V)\rightarrow L(V)$$, then define the induced trace norm of $$\mathcal{H}$$ to be $$\|\mathcal{H}\|_1=\max\{\|\mathcal{H}(X)\|_1:\|X\|_1\leq 1\}.$$ Recall that if $$R,S$$ are isometries and $$A$$ is a complex matrix, then $$\|RAS^*\|_1=\|A\|_1$$ whenever the matrix multiplication exists. Also, recall that $$\|\text{Tr}_{V}(X)\|_1\leq\|X\|_1$$ whenever this inequality makes sense.

We have $$\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)(X)\|_1=\|\text{Tr}_{W}(RXS^*)\|_1\leq\|RXS^*\|_1=\|X\|_1.$$ Therefore, $$\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)\|_1\leq 1$$, so $$\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$$ since the induced trace norm is submultiplicative.

Empirical verification

I have empirically verified using computer calculations that the conclusions that we have made are reasonable. Define a fitness function $$F:M_d(\mathbb{C})^n\times M_d(\mathbb{C})^n\rightarrow\mathbb{R}$$ by letting $$F(A_1,\dots,A_n;B_1,\dots,B_n)=\frac{\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))}{\rho(\Phi(A_1,\dots,A_n))^{1/2}\rho(\Phi(B_1,\dots,B_n))^{1/2}}.$$

One can maximize the value of $$F(A_1,\dots,A_n;B_1,\dots,B_n)$$ using gradient ascent to obtain examples of tuples $$(A_1,\dots,A_n;B_1,\dots,B_n)$$ with $$F(A_1,\dots,A_n;B_1,\dots,B_n)\approx 1$$, but in each of these examples, we always have a complex number $$\lambda$$ along with some invertible $$B$$ where $$B_j\approx \lambda BA_jB^{-1}$$ for $$1\leq j\leq n$$.