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Thread: Finding eigenvalues

  1. #1
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    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 $\displaystyle a_{0}v+a_{1}Tv+a_{2}T^{2}v=0$.
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  2. #2
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    Quote Originally Posted by zebra2147 View Post
    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

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