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Let $V$ be a vector space over a field $k$, and let $T:V\rightarrow V$ be linear, and let $f\in k[x]$. Suppose that $\lambda\in k$ is an eigenvalue of $T$ and let $v\in V$ be a corresponding eigenvector. Show that $f(T)v=f(\lambda)v$.

I'm not sure where to start. All I know is that $Tv=\lambda v$. I was thinking it might be easier to show that $f(T-\lambda I)=0$, but I still don't know where to start. Any help is appreciated!

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    $\begingroup$ Since $Av=\lambda v$, what can you say about $A^2 v$? $A^3 v$? $\endgroup$
    – Dylan
    Nov 18, 2015 at 21:59
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    $\begingroup$ Since f is a polynomial with coefficients in $k$, we can write $f(x) = \sum_{i=0}^n a_i x^i $ with $n = deg (f)$ and each $a_i$ in k. Along with the fact that $Tv=\lambda v $, this should give you the required equation. $\endgroup$
    – H1ghfiv3
    Nov 18, 2015 at 22:01
  • $\begingroup$ Ok so:$\\$ $f(T)v=\sum_{i=0}^na_i(T)^iv\\$ $=a_0T^0v+a_1T^1v+a_2T^2v+\dots\\$ $=a_0\lambda^0v+a_1\lambda^1v+a_2\lambda Tv+\dots\\$ $=a_0\lambda^0v+a_1\lambda^1v+a_2\lambda^2v+\dots\\$ $=\sum_{i=0}^na_i\lambda^iv\\$ $=f(\lambda)v$ Is there a way I can write this while keeping summation notation, I don't really like the look of expanding out the summation... $\endgroup$ Nov 18, 2015 at 22:22
  • $\begingroup$ You don't need to expand the sum, in particular not if you've shown that $\lambda^i v=T^i v $ beforehand. $\endgroup$
    – H1ghfiv3
    Nov 19, 2015 at 7:25

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