# Proof involving eigenvectors/values of a linear map and polynomials.

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!

• Since $Av=\lambda v$, what can you say about $A^2 v$? $A^3 v$? Nov 18, 2015 at 21:59
• 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. Nov 18, 2015 at 22:01
• 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... Nov 18, 2015 at 22:22
• You don't need to expand the sum, in particular not if you've shown that $\lambda^i v=T^i v$ beforehand. Nov 19, 2015 at 7:25