# Inequality related to spectral radius

Let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

For $A = (A_1,\cdots,A_d)\in\mathcal{L}(E)^d$, the algebraic spectral radius of $A$ was given by $$r_a(A)=\inf_{n\in \mathbb{N}^*}\left\|\sum_{f\in F(n,d)} A_f^* A_f\right\|^{\frac{1}{2n}} =\lim_{n\to+\infty}\left\|\sum_{f\in F(n,d)} A_f^* A_f\right\|^{\frac{1}{2n}} ,$$ where $F(n,d):=\{f:\,\{1,\cdots,n\}\longrightarrow \{1,\cdots,d\}\}$ and $A_f:=A_{f(1)}\cdots A_{f(n)}$, for $f\in F(n,d)$.

I don't understand why if $n=1$, we have $$r_a(A)\leq \|A\|:=\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{k=1}^d\|A_kx\|^2\bigg)^{\frac{1}{2}}?$$

• $n=1$ or $d=1$? Commented Apr 23, 2018 at 12:21
• @s.harp $n=1$. I see this property in a paper. Commented Apr 23, 2018 at 12:24
• jstor.org/stable/2159479?seq=1#page_scan_tab_contents Commented Apr 23, 2018 at 12:25
• @s.harp Do you understand me? Commented Apr 23, 2018 at 15:38
• It might help to note that when $n=1$, $$\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{k=1}^d\|A_kx\|^2\bigg)^{\frac{1}{2}}=\bigg(\displaystyle\sum_{k=1}^d|A_k|^2\bigg)^{\frac{1}{2}}$$
– Surb
Commented Apr 23, 2018 at 19:00

For some reason, you switched the indices from the paper. The "$n=1$" from the paper is your "$d=1$".
So what the paper is saying is that, when $d=1$, $$r_a(A)=\lim_n\|A_1^{n*}A_1^n\|^{1/2n}=\lim_n\|A_1^n\|^{1/n}=r(A_1).$$ Also $$\|A_1^n\|^{1/n}\leq\|A_1\|,$$ so $$r_a(A)\leq r(A_1)\leq \|A_1\|=\sup_{\|x\|=1}\|A_1x\|.$$
One has $$r_a(A)\leq \left(\left\|\sum_{f\in \mathbf{F}(n,d)} \mathbf{A}_f^*\mathbf{A}_{f}\right\|^{\frac{1}{2n}} \right),$$ for all $n\in \mathbb{N}^*$.
If $n=1$, then $$r_a(A)\leq \left\|\displaystyle\sum_{k=1}^dA_k^* A_k \right\|^{1/2}=||A||.$$