A diagonalizable operator with positive eigenvalues and existence of a projection as a limit Let $A: \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ be a linear map and assume it is diagonalizable with positive real eigenvalues. Show that there is a projection $\pi: \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ and a number $ 0 < c \leq 1$ such that $\lim_{n \to \infty} \frac{A^{n}(v)}{||A^{n}||} = c \pi(v)$. 
This is an old qual problem of which the solution I am aware of for defining the projection is a little messy. It uses the spectral theorem and a certain estimate about the spectral radius and the matrix norm. 
 A: Let $A=RDR^{-1}$, where $D=\text{diagonal}(d_1,\ldots,d_m)$,  where $d_1=\ldots=d_i>d_{i+1}\geq\ldots d_m>0$.
Notice that
$$\lim_{n\rightarrow \infty}\dfrac{A^n}{\|A^n\|}=\lim_{n\rightarrow \infty}\left(\dfrac{A}{d_1}\right)^n     \left\|\left(\dfrac{A}{d_1}\right)^n\right\|^{-1}=$$ $$=\lim_{n\rightarrow \infty}R(\text{diag}(1,\ldots,1,(\frac{d_{i+1}}{d_1})^n,\ldots, ,(\frac{d_{m}}{d_1})^n))R^{-1}\|R(\text{diag}(1,\ldots,1,(\frac{d_{i+1}}{d_1})^n,\ldots ,(\frac{d_{m}}{d_1})^n))R^{-1}\|^{-1}$$
$$=R(\text{diag}(1,\ldots,1,0,\ldots,0))R^{-1}\|R(\text{diag}(1,\ldots,1,0,\ldots ,0))R^{-1}\|^{-1}.$$
Notice that $(R(\text{diag}(1,\ldots,1,0,\ldots,0))R^{-1})^2=R(\text{diag}(1,\ldots,1,0,\ldots,0))R^{-1}$, therefore $R(\text{diag}(1,\ldots,1,0,\ldots,0))R^{-1}$ is a projection. 
Let $c=\|R(\text{diag}(1,\ldots,1,0,\ldots ,0))R^{-1}\|^{-1}$. Now, the question is whether the norm of a projection is bigger or equal to 1 or not. 
If your norm satisfy the multiplicative inequality $\|AB\|\leq\|A\|\|B\|$ and $P^2=P$ then $\|P\|=\|P^2\|\leq\|P\|^2$. Thus, $1\leq\|P\|$.
