# Are these $L_2$-spectral radii approximations strictly increasing?

Suppose that $$V$$ is a finite dimensional complex Hilbert space. Let $$L(V)$$ denote the collection of all linear mappings from $$V$$ to $$V$$. Let $$A_1,\dots,A_r:V\rightarrow V$$ be linear operators. Then define a completely positive superoperator $$\Phi(A_1,\dots,A_r):L(V)\rightarrow L(V)$$ by letting $$\Phi(A_1,\dots,A_r)(X)=A_1XA_1^*+\dots+A_rXA_r^*$$, and define the $$L_2$$-spectral radius of $$A_1,\dots,A_r$$ by letting $$\rho_2(A_1,\dots,A_r)=\rho(\Phi(A_1,\dots,A_r))^{1/2}$$.

The Cauchy-Schwarz inequality holds for the $$L_2$$-spectral radius. If $$V,W$$ are finite dimensional complete Hilbert spaces, and $$A_i:V\rightarrow V,B_i:W\rightarrow W$$ for $$1\leq i\leq r$$, then $$\rho(A_1\otimes B_1+\dots+A_r\otimes B_r)\leq\rho(\Phi(A_1,\dots,A_r))^{1/2}\rho(\Phi(B_1,\dots,B_r))^{1/2}.$$

Now, define the $$L_{2,d}$$-spectral radius of $$A_1,\dots,A_r$$ by letting

$$\rho_{2,d}(A_1,\dots,A_r)$$ $$=\sup\{\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho(\Phi(X_1,\dots,X_r))^{1/2}}\mid \rho(\Phi(X_1,\dots,X_r))\neq 0,X_1,\dots,X_r\in M_d(\mathbb{C})\}.$$

From the Cauchy-Schwarz inequality, we observe that $$\rho_{2,d}(A_1,\dots,A_r)\leq\rho_{2,g}(A_1,\dots,A_r)\leq\rho_{2,\dim(V)}(A_1,\dots,A_r)=\rho_2(A_1,\dots,A_r)$$ whenever $$1\leq d\leq g$$. One should therefore think of $$\rho_{2,d}$$ as an approximation to $$\rho_{2}$$.

Suppose that $$1\leq d<\dim(V)$$. Then is there some $$r$$ along with linear operators $$A_1,\dots,A_r:V\rightarrow V$$ such that $$\rho_{2,d}(A_1,\dots,A_r)<\rho_{2,d+1}(A_1,\dots,A_r)?$$

Example 0: $$\rho_{2,d}(A)=\rho_{d}(A)=\rho(A).$$ Therefore, $$\rho_{2,d},\rho_{2}$$ should be considered to be a sort of spectral radius but for multiple linear operators.

Example 1: $$\rho_{2,1}(A_1,\dots,A_r)=\sup\{\rho(\alpha A_1+\dots+\alpha_r A_r):|\alpha_1|^2+\dots+|\alpha_r|^2=1\}.$$

Example 2: Observe that if $$d\geq\dim(V)$$, then $$\rho_{2,d}(A_1,\dots,A_r)=\rho(\Phi(A_1,\dots,A_r))^{1/2}.$$

Example 3: $$\rho_{2,1}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix})=1/\sqrt{2}<1=\rho_{2,2}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}).$$

Example 4: Let $$C_1,\dots,C_r$$ be the $$r\times r$$-matrices, where $$(C_i)_{i,i+1\mod r}=1$$ and where all the other entries in $$C_i$$ are zero. Then $$\rho(C_1\otimes X_1+\dots+C_r\otimes X_r)=\rho(X_1\dots X_r)^{1/r}$$. Therefore, $$\rho_{2,d}(C_1,\dots,C_r)=\sup\{\frac{\rho(X_1\dots X_r)^{1/r}}{\rho(\Phi(X_1,\dots,X_r))^{1/2}}:\rho(\Phi(X_1,\dots,X_r))\neq 0,X_1,\dots,X_r\in M_d(\mathbb{C})\}.$$

Now, suppose that $$j_1,\dots,j_r$$ are natural numbers with $$1\leq j_1\leq\dots\leq j_r=d$$ and where $$j_{i+1}-j_{i}\in\{0,1\}$$ for $$1\leq i\leq r$$. For $$1\leq i\leq r$$, let $$X_i$$ be the $$d\times d$$ matrix where $$(X_i)_{j_{i},j_{i+1}}=1$$ (here $$i$$ is taken modulo $$r$$) and where all the other entries in $$X_i$$ are zero. Then $$\rho(X_1\dots X_r)=1$$. Let $$n_{1},\dots,n_{d}$$ be the natural numbers defined by letting $$n_{j}=\{i\mid 1\leq i\leq r,j_{i}=j\}$$. Then $$r=n_1+\dots+n_d$$. Suppose now that $$n_j=r/d$$ for all $$j$$. Then $$\rho(\Phi(X_1,\dots,X_r))=\frac{r}{d}.$$ Therefore, $$\rho_{2,d}(C_1,\dots,C_r)\geq\sqrt{\frac{d}{r}}$$ whenever $$d$$ is a factor of $$r$$.

Yes. If $$C_1,\dots,C_r$$ are the matrices in Example 4, then by my answer to my other question, $$\rho_{2,d}(C_1,\dots,C_r)=\sqrt{d/r}$$ whenever $$1\leq d\leq r$$.