Prove that $C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$, for every $n \in N$ EDIT: I forgot to mention $C = 0.5uu^T + 0.33vv^T$ and now if I use it, I solve it easily.

Given: $Cx = \frac{1}{2}au + \frac{1}{3}bv$, $x \in R^2$, $u,v$ are orthonormal vectors in $R^2$, $x = au + bv$ and $a,b \in R$, $C$ is a matrix $2x2$. Note: $a,b$ are scalars.
Prove that $$C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$$, for every $n \in N$

Well, it's trivial that we need to show that with induction, so for $n = 1$ this works because it is given (actually I proved it and it is really true).
Now I assume that it works for $k = n$, and I want to prove for $k = n + 1 $. I use $k$ because it is more comfortable for me that way, and I got stuck with $a^2$ and $u,v$ are gone in the last equation, instead of $a$ and multiplied with $u,v$ as required.
$C^{k+1}x=\frac{1}{2^{k+1}}au + \frac{1}{3^{k+1}}bv$ I need to prove.
Going from the left side of the equation and I try to prove the right side:
$C^{k+1}x = C^k(Cx) = C^k(\frac{1}{2}au + \frac{1}{3}bv) = \frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) = 
\frac{1}{2^{k+1}}a^2 + \frac{1}{3^{k+1}}b^2$.
Edit: if I take $au$ out, then I get was it required. can I do that?
I mean: $\frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =$ long equation $= au\frac{1}{2^{k+1}} + bu\frac{1}{3^{k+1}}$
 A: Think of $C$ as a linear transformation on $\mathbb R^2$.
Then, the matrix of $C$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2 & 0 \\ 0 &1/3}$.
Therefore, the matrix of $C^n$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2^n & 0 \\ 0 &1/3^n}$. 

Concretely, if $Cx=\frac{1}{2}au + \frac{1}{3}bv$ whenever $x = au + bv$, then by taking $a=1, b=0$ we get $Cu=\frac{1}{2}u$, and by taking $a=0, b=1$ we get $Cv=\frac{1}{3}v$, which solves the problem in your induction.
A: Your problem description is formulated rather confusingly. Here is how I would state it

Given: Two orthonormal vectors $\def\R{\Bbb R}u,v\in\R^2$, and $C:\R^2\to\R^2$ a linear map, given by a $2\times2$ a matrix, such that for all $a,b\in\R$ one has $C(au + bv)= \frac12au + \frac13bv$.
Prove that $$C^n(au + bv) = \frac1{2^n}au + \frac1{3^n}bv,$$ for every $n \in\Bbb N$

Now you can see that your step $C^k(\frac12au + \frac13bv) = \frac1{2^{k}}au + \frac1{3^{k}}bv(\frac12au + \frac13bv)$ is quite wrong; I think what you did is textually replace $C^k$ by the value given by the induction hypothesis for) $C^k(x)$, without even adding grouping parentheses, and then get a nonsensical formula (you cannot multiply $\frac13bv$ and another vector together). But $C^k$ is not the same as $C^k(x)$, it is a linear operator  that is here applied to the vector $(\frac12au + \frac13bv)$. Your induction hypothesis is in term of a vector of the form $a'u+b'v$ (I have to use different names because in the induction proof I already used $a,b$), so one here take $a'=\frac12a$ and $b'=\frac13b$. Now your induction hypothesis applies and gives $C^k(a'u + b'v) = \frac1{2^k}a'u + \frac1{3^k}b'v$ which works out to $C^k(\frac12au + \frac13bv) = \frac1{2^k}\frac12au + \frac1{3^k}\frac13bv$, and you are essentially done.
A: $Cx = \frac{1}{2}au + \frac{1}{3}bv$
$x = au + bv \implies Cx = C(au +bv) = Cau + Cbv$
Thus
$Cau + Cbv = \frac{1}{2}au + \frac{1}{3}bv$
$\implies a(C-\frac{1}{2}I)u + b(C-\frac{1}{3}I)v = 0$
$\implies a(C-\frac{1}{2}I)u.u^T + b(C-\frac{1}{3}I)v.u^T = 0.u^T = 0$
$\implies a(C-\frac{1}{2}I) = 0$ , since u&v are orthanormal.
By the same reason we also have $ b(C-\frac{1}{3}I) = 0$
Therefore either $a = 0$ or $b = 0$
WLOG choose $b = 0$, then
$Cx = \frac{1}{2}au$, and $x = au$, and it is easy to show
$C^nx = \frac{1}{2^n}au$
