# Thread: find the n transformation

1. ## find the n transformation

let $\displaystyle T\in\mathcal L(\mathbb R^3),$ such that $\displaystyle T(x,y,z)=(ux+y,uy+z,uz).$ Compute $\displaystyle T^n(x,y,z).$

2. obviously for each $\displaystyle 0\ne u\in\mathbb R.$

as for the problem, compute the matrix associated to the transformation (it's the canonical basis for $\displaystyle \mathbb R^3,$ so it's easy) and then compute the characteristic polynomial.

use the division algorithm and put $\displaystyle t^n=(u-\lambda)^3f(t)+at^2+bt+c,$ (1) where $\displaystyle (u-\lambda)^3$ is the characteristic polynomial and $\displaystyle at^2+bt+c$ is the remainder (its degree is one less than the characteristic polynomial), now you gotta compute $\displaystyle a,b,c.$

first put $\displaystyle t=u,$ and differentiate once and evaluate again at $\displaystyle t=u,$ differentiate again and evaluate at $\displaystyle t=u,$ this will be useful to find $\displaystyle a,b,c.$

once got those values, put them at (1) and then evaluate the associated matrix to the transformation; by the Cayley - Hamilton theorem, the matrix evaluated in the characteristic polynomial produces the null one so you'll end up (say the matrix is $\displaystyle A$) $\displaystyle A^n=aA^2+bA+cI_3.$

finally, explicitly find $\displaystyle A^n$ and multiply it by $\displaystyle \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right],$ and we're done!

3. Originally Posted by Morgan
let $\displaystyle T\in\mathcal L(\mathbb R^3),$ such that $\displaystyle T(x,y,z)=(ux+y,uy+z,uz).$ Compute $\displaystyle T^n(x,y,z).$
a very easy induction on $\displaystyle n$ will prove that: $\displaystyle T^n(x,y,z)=(u^nx + nu^{n-1}y+ \frac{n(n-1)}{2}u^{n-2}z, \ u^ny + nu^{n-1}z, \ u^n z).$

4. Originally Posted by Morgan
let $\displaystyle T\in\mathcal L(\mathbb R^3),$ such that $\displaystyle T(x,y,z)=(ux+y,uy+z,uz).$ Compute $\displaystyle T^n(x,y,z).$
Write T as a matrix and calculate the first few powers of T. You should see the pattern pretty quickly.

Or you could just calculate them directly from that formula, but I think it is easier to see with the matrix.