Recursive formulae involving a linear operator

Question

Given a basis $e_{1}$, $e_{2}$ in the plane, define the linear operator $F$ as $F(e_{1})=3e_{1}+e_{2}$ and $F(e_{2})=e_{2}$. Furthermore, define the sequence $u_{1},u_{2},\dots$ of vectors in the plane by letting $u_{1}=6e_{1}+2e_{2}$ and $u_{n+1}=F(u_{n})$. Give a formulae for $u_{n}$

This is what I done: $u_{2}=F(u_{1})=F(6e_{1}+2e_{2})=6F(e_{1})+2F(e_{2})=6(3e_{1}+e_{2})+2(e_{2})=18e_{1}+8_{2}$ Similarly, I obtain $u_{3}=48e_{1}+26e_{2}$ and $u_{4}=144e_{1}+74e_{2}$ Is this the right method, and would anyone be so kind as to help me determine the formulae from the results I have got?

Answer 1

So $F$ is given by the matrix $$A=\pmatrix{3&0\cr1&1\cr}$$ and you want a formula for $A^n(6,2)$. If you can find eigenvectors $v_1$ and $v_2$ of $A$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, and if you can find $c_1$ and $c_2$ such that $(6,2)=c_1v_1+c_2v_2$, then you will have $$A^r(6,2)=c_1\lambda_1^rv_1+c_2\lambda_2^rv_2$$

EDIT: As noted in the comments, we get $\lambda_1=3$ with $v_1=(2,1)$, and $\lambda_2=1$ with $v_2=(0,1)$, so we get $c_1=3$ and $c_2=-1$; then $$A^r(6,2)=(3)(3^r)(2,1)-(0,1)$$

Answer 2

Since $F$ is a linear transformation, and you are told what the transformation does to a basis of your space, you should be able to create the matrix that represents $F$ in the basis $\{e_1,e_2\}$. Let this matrix be $A$, Then: $$u_{n+1}=F(u_n)\iff u_{n+1}=Au_n.$$ Now your question is how to get powers of this matrix.