Let $A\in M_n(\Bbb R)$ prove that, $\|A^n\|\le \frac{n}{\ln 2}\|A\|^{n-1}$ when $\lambda_i<1.$ 
Let $n\in \Bbb N$ be fixed  and $A$  be a $n\times n$  complex matrix whose eigenvalues $(\lambda_i)$ satisfy $|\lambda_i|\lt 1$. Prove that:
  $$
\|A^n\|\le \frac{n}{\ln 2}\|A\|^{n-1}
$$

where for any matrix $B$
 $$
\|B\|:= \sup_{\|x\|= 1}\|Bx\| ~~~\text{with}~~\|x\|^2 = \sum^{n}_{i=1} |x_i|^2
$$


Edit: Here $n$ is fixed and represent the dimension of our space and matrix as well. I also thought it could be done by induction but this was  a wrong way.


l don't know how to proceed.
 A: Let $r=\|A\|$.We have to prove $\|A^n\|\le{n \over \ln2}r^{n-1}$.
Since we know $\|XY\|\le\|X\|.\|Y\|$ for any matrices $X,Y$ thus it's easy to see that $\|A^k\|\le\|A\|^k=r^k,k\in Z^+$.
Let $f(t)=(t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n)=t^n+c_1t^{n-1}+\cdots+c_n $ be the characteristic polynomial of $A$. Using Vieta's formula we ob tain:
$$|c_k|=|\sum_{1\le i_1 \cdots\le i_k \le n} \lambda_1\cdots\lambda_k|\le\sum_{1\le i_1 \cdots\le i_k \le n} |\lambda_1\cdots\lambda_k| \le \binom {n} {k},1\le k \le n$$
Cayley-Hamilton theorem indicates that $f(A)=0$ , so:
$$\|A^n\|=\|c_1A^{n-1}+\cdots+c_n\|\le \sum_{k=1}^{n}\binom {n} {k}\|A^k\|\le \sum_{k=1}^{n}\binom {n} {k}r^k=(1+r)^n-r^n$$
Combining this with the trivial estimate $\|A^n\|\le r^n$, so we get:
$$\|A^n\|\le \min(r^n,(1+r)^n-r^n).$$
Let $r_0={1 \over {2^{1 \over n}-1}}$ it is easy to check that two bounds are equal if $r_0=r$,moreever:
$$r_0={1 \over {e^{\ln2 \over n}}-1}< {n\over \ln2}$$
For $r\le r_0$ apply the trivial bound:
$$\|A^n\|\le r^n\le r_0.r^{n-1} \le {n\over \ln2} r^{n-1}$$
For $r>r_0$ we have :
$$\|A^n\|\le (1+r)^n-r^n=r^{n-1}.{{(1+r)^n-r^n}\over {r^{n-1}}}$$
Now the function $g(r)={{(1+r)^n-r^n}\over {r^{n-1}}}$ is decreasing because the numerator has degree $n-1$ and all coeffcients are positive, so:
$${{(1+r)^n-r^n}\over {r^{n-1}}} < {{(1+r_0)^n-{r_0}^n}\over {{r_0}^{n-1}}}=r_0((1+{1\over r_0})^n-1)=r_0 < {n \over \ln2}$$
so $\|A^n\|\le{n \over \ln2}r^{n-1}$.
