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 $= au\frac{1}{2^{k+1}} + bu\frac{1}{3^{k+1}}$
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 $\pmatrix{ 1/2 &amp; 0 \\ 0 &amp;1/3}$.

Therefore, the matrix of $C^n$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2^n &amp; 0 \\ 0 &amp;1/3^n}$. 



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.
$Cx = \frac{1}{2}au + \frac{1}{3}bv$

$x = au + bv \implies Cx = C(au +bv) = Cau + Cbv$

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&amp;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

$C^nx = \frac{1}{2^n}au$