1. ## Finding eigenvalues

The following is an exercise that my professor gave us. I am confident that I can solve it but I'm not sure how to interpret $\displaystyle T^2v$.

Exercise: Suppose the linear map $\displaystyle T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is given by $\displaystyle T(a,b)=(a+4b,3a+5b)$.
Let $\displaystyle v=(2,1)$. Find coefficients $\displaystyle a_0,a_1,a_2$ so that $\displaystyle a_{0}v+a_{1}Tv+a_{2}T^{2}v=0$.

2. Originally Posted by zebra2147
The following is an exercise that my professor gave us. I am confident that I can solve it but I'm not sure how to interpret $\displaystyle T^2v$.

Exercise: Suppose the linear map $\displaystyle T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is given by $\displaystyle T(a,b)=(a+4b,3a+5b)$.
Let $\displaystyle v=(2,1)$. Find coefficients $\displaystyle a_0,a_1,a_2$ so that $\displaystyle a_{0}v+a_{1}Tv+a_{2}T^{2}v=0$.
$\displaystyle T^2v$ means $\displaystyle T(T(v))$, so that

$\displaystyle T^2(a,b)=T(T(a,b))=T(a+4b,3a+5b)$
$\displaystyle =(a+4b+4(3a+5b),3(a+4b)+5(3a+5b))$
$\displaystyle =(13a+24b,18a+37b)$.