Help with Understanding Eigenvalues, Eigenvectors, and Mapping with Lamda and P

[terryfields]

this is exam revision not homework so feal free to help

let lamda be an eigenvalue of T, and let P be a polynomial with coefficients in F, define the linear mapping S=p(T) and show that p(lamda) is an eigenvalue of S

i know that an eigenvalue of T is a element vnot=0 such that T(v)=lamda v for some lamda in the field, and that the scalar lamda here is the eigenvalue, however i don't understand this question at all

so lamda is our eigenvalue meaning there must be a corresponding eigenvector v not equal to zero but what's p and what does this mapping show? please help

 

[statdad]

First,
If $$A$$ is the matrix and $$\lambda$$ the eigenvector, then $$A \vec x = \lambda \vec x$$ for some vector $$x$$. Note these facts.

$$\begin{align*}<br /> A^n \lambda & = A^{(n-1)} (A \vec x) \\<br /> & = A^{(n-2)} A (\lambda \vec x)\\<br /> & = \hdots \\<br /> & = \lambda^n \vec x<br /> \end{align*}$$

so $$\lambda^n$$ is an eigenvalue of $$A^n$$, with the same eigenvector

Next, if $$c$$ is any constant, then

$$(cA^n) \vec x = c\left(A^n \vec x\right) = c \lambda^n \vec x$$

so $$c\lambda^n$$ is an eigenvalue of $$cA^n$$

Finally, if $$a, b$$ are constants, and $$m, n$$ are integers, consider the
two-term polynomial $$p(s) = as^m + bs^n$$. The polynomial [tex]p(A) [tex]is<br /> <br /> $$p(A) = a A^m + bA^n$$<br /> <br /> which is a matrix the same size as $$A$$. The product $$p(A) \vec x$$ is<br /> <br /> $$\begin{align*}<br /> (a A^m + b A^n) \vec x & = (a A^m) \vec x + (b A^n) \vec x \\<br /> & = \left(a \lambda^m\right) \vec x + \left(b \lambda^n\right) \vec x\\<br /> & = \left(a \lambda^m + b \lambda^n \right) \, \vec x \\<br /> & = p(\lambda) \, \vec x<br /> \end{align*}$$<br /> <br /> That case does not have a constant term in the polynomial. If you have<br /> $$p(s) = as^m + bs^n + d$$<br /> <br /> where $$d$$ is a constant, the appropriate modification is<br /> <br /> $$p(A) = aA^m + bA^n + d I_n$$<br /> <br /> where $$I_n$$ is the identity matrix the same size as $$A$$. Again, it is easy to show that<br /> <br /> $$\begin{align*}<br /> p(A) \, \vec x & = \left(a A^m + b A^n + dI_n\right) \, \vec x\\<br /> & = \left(a \lambda^m + b \lambda^n + d\lambda\right) \, \vec x\\<br /> & = p(\lambda) \, \vec x<br /> \end{align*}$$<br /> <br /> so again $$p(\lambda)$$ is an eigenvalue of $$p(A)$$.<br /> <br /> The case for a general polynomial requires a little more notation but the steps are the same.<br /> <br /> The idea: if $$A$$ is the matrix for a linear operator, so is $$p(A)$$ for<br /> any polynomial $$p$$, and the eigenvalues behave ``as we expect them to''.[/tex][/tex]