It is a known result that
$$\lvert\lvert A \rvert\rvert \leq \sqrt{\rho(A^TA)} $$
Is there a way to bound $\lvert\lvert A \rvert\rvert $ using spectrum of $A$ only (not $A^TA$).
Ultimately, I am trying to express $$\lim_{t\to\infty} \lvert\lvert A \rvert\rvert^t $$ in terms of spectrum of A.
I also tried with diagonalization of $A$ into $P^{-1}DP$
Then $$\lim_{t\to\infty} \lvert\lvert A \rvert\rvert^t = \lim_{t\to\infty} \lvert\lvert P^{-1}DP \rvert\rvert^t$$
and I got struck .
No, because there are nonzero $A$ with spectrum $\{0\}$. Consider e.g. the matrix $$ \pmatrix{0 & 1\cr 0 & 0\cr}$$
EDIT: On $\ell^2$, consider the operator $A$ corresponding to the matrix with diagonal blocks $$ \pmatrix{0 & 1\cr 0 & 0\cr},\ \pmatrix{0 & 1 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr},\ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr}, \ldots $$ Again the spectrum is just $\{0\}$, but for all $t$ we have $\|A^t\| = 1$. So no condition on the spectrum can guarantee $\|A^t\| \to 0$.