I am reading about the connection between spectral norm and spectral radius and more particularly about
Lemma: let $A\in M_n$ and $\epsilon>0$ be given. There is a matrix norm $||\cdot||$ such that $\rho(A) \leq ||A|| \leq \rho(A) + \epsilon$
Proof: For unitary $U$ and upper triangular $N$ we have $A=UNU^{H}$. Set $D_t = diag(t, t^2, \dots,t^n)$ and compute $$D_tND_t^{-1}=\left[ \begin{array}{ccccc} \lambda_1& t^{-1}N_{12} & t^{-2}N_{13} & \dots & t^{-n+1}N_{1n} \\ & \lambda_2 & t^{-1}N_{23} & \dots &t^{-n+2}N_{2n} \\ & & \lambda_3 \\ & \text{0} & & \ddots & t^{-1}N_{n-1,n} \\ & & & & \lambda_n \end{array} \right].$$ Thus, for $t>0$ large enough, the sum of all the absolute values of the off-diagonal entries of $D_tND_t^{-1}$ is less than $\epsilon$. In particular, $||D_tND_t^{-1}||_1 \leq \rho(A)+\epsilon$ for all large enough $t$. Thus, if we define the matrix norm $||\cdot||$ by $||B||=||D_tU^HBUD_t^{-1}||_1=||B||=||(D_tU^H)B(D_t^{-1}U^H)^{-1}||_1$ and replace $B$ with $A=UNU^{H}$ we conclude that $$||A|| \leq \rho(A) + \epsilon$$ for sufficiently large $\epsilon$.
My questions are
- Why the author states that the absolute values of the off-diagonal entries of $D_tND_t^{-1}$ is less than $\epsilon$ and thus $||D_tND_t^{-1}||_1 \leq \rho(A)+\epsilon$? Can we show that more analyticaly?
- Can we define $$||A||^t \leq (\rho(A) + \epsilon_t)^t$$ with $\lim_{t \to\infty}\epsilon_t=0$, i.e., with a diminishing sequence $\epsilon_t$? If the aforementioned inequality is defined should we have $\rho(A) + \epsilon_t < 1$?
EDIT: $N_{ij}$ was previously $d_{ij}$ I changed it to make more clear the multiplication. So in comments consider $d_{ij}=N_{ij}$.