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- Feb 5, 2012

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Here's another question that I solved. Let me know if you see any mistakes or if you have any other comments. Thanks very much.

**Problem:**

Prove that the eigenvector \(v\) of \(f:V\rightarrow V\) over a field \(F\), with eigenvalue \(\lambda\), is an eigenvector of \(P(f)\) where \(P(t)\in F[t]\). What is the eigenvalue of \(v\) with respect to \(P(f)\)?

**My Solution:**

Let \(A\) be the matrix representation of the linear transformation \(f\). Then we can write, \(Av=\lambda v\). Now let, \(P(t)=a_0+a_1 t+a_2 t^2+\cdots+a_n t^n\) where \(a_i\in F\,\forall\,i\). Then,

\[P(A)=a_0+a_1 A+a_2 A^2+\cdots+a_n A^n\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 (Av)+a_2 (A^2 v)+\cdots+a_n (A^n v)\]

\[\Rightarrow P(A)(v)=a_0 v+a_1 \lambda v+a_2 \lambda^2 v+\cdots+a_n \lambda^n v\]

\[\therefore P(A)(v)=(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n )(v)\]

Hence \(v\) is an eigenvector for \(P(f)\) and the corresponding eigenvalue is, \(a_0 +a_1 \lambda +a_2 \lambda^2 +\cdots+a_n \lambda^n\).