Using eigenvectors and eigenvalues to a find power of a transformation Let $T:R^2 \to R^2$ be a linear operator with eigenvectors $(1,-1)$ and $(1,1)$, and eigenvalues $1/2$ and $2$, respectively. How do I evaluate $T^{10}(v)$, given $v = (5,1)$?
I tried to write the transformation matrix, but I think I am missing some detail related to the basis...
 A: We can write the eigenvalue decomposition $T = VDV^{-1}$. Now if we calculate $TT$ we get: $$T^2 = TT = VDV^{-1}VDV^{-1} = VD\underset{=I}{\underbrace{\left(V^{-1}V\right)}}DV = VDDV^{-1} = VD^2V^{-1}$$ If we keep going in the same way you will see that $T^k = VD^kV^{-1}$. So you will get away with calculating the diagonal matrix $D$ raised to 10, which is a much faster thing to do than to do 10 full matrix multiplications.
If this seems confusing at start, try and think of it as multiplying with $V^{-1}$ does the "translate to eigenbasis" part which the answer by Paul explains well and multiplying with $V$ does the "translate back to original basis" part.
A: Note that $$\tag{1}v=(5,1)=2(1,-1)+3(1,1).$$ 
Since $(1,−1)$ and $(1,1)$ are eigenvectors of $T$ with eigenvalues $1/2$ and $2$ respectively, we have 
$$\tag{2}T(1,-1)=\frac{1}{2}(1,-1)\mbox{ and }T(1,1)=2(1,1).$$
By $(1)$ and $(2)$, we have 
$$T(v)=2T(1,-1)+3T(1,1)=2(\frac{1}{2})(1,-1)+3(2)(1,1)=\cdots$$
where we have used the fact that $T$ is linear.
Similarly, we have 
$$T^{10}(v)=2T^{10}(1,-1)+3T^{10}(1,1)=2(\frac{1}{2})^{10}(1,-1)+3(2)^{10}(1,1)=\cdots.$$
I think you can finish from this point. 
