I am so lost on this question I am not sure even where to start. I am not looking for an answer but more of a turn in the right direction.
Find all the eigenvalues and eigenfunctions of the following linear transformation: $T(f(x)) = f(-x)$ from $\mathbf{P}_2$ to $\mathbf{P}_2$. IS T diagonalizable?
All I know for sure is that the standard basis is $1$, $x$, $x^2$, and if this is an eigenbasis for $T$ then $T$ is diagonalizable.
Thanks for any assistance!
Best Answer
An eigenvector of a linear tranformation $T\colon\mathbf{V}\to\mathbf{V}$ is a vector $\mathbf{v}$, $\mathbf{v}\neq\mathbf{0}$, such that there is a scalar $\lambda$ for which $T(\mathbf{v})=\lambda \mathbf{v}$. The scalar $\lambda$ is called an eigenvalue. $T$ is diagonalizable if and only if you can find a basis for $\mathbf{V}$ in which every vector in the basis is an eigenvector. This does not mean that any basis will consist of eigenvectors. It means that there is a way to pick a basis so that every element of the basis is an eigenvector.
So, here, $\mathbf{V}=\mathbf{P}_2$. Your function $T$ is defined by the given formula. For example, if $p(x) = x^2 - 3x + 1$, then $p(-x) = (-x)^2 -3(-x) + 1 = x^2 +3x + 1$, so $T(p) = x^2+3x+1$.
You are looking for eigenvalues and eigenvectors. If you already know about the characteristic polynomial, then you should figure out the characteristic polynomial of $T$ (say, using the standard basis to find a coordinate matrix for $T$) and proceed along those lines.
If you don't know about the characteristic polynomial, not all is lost. Suppose $p(x) = ax^2+bx+c$ is an eigenvector. That means that $a$, $b$, and $c$ are not all zero, and there is a scalar $\lambda$ such that $T(p)=\lambda p$. Since $T(p(x)) = ax^2 -bx +c$ and $\lambda p = \lambda ax^2 + \lambda bx + \lambda c$, the equation says that $$ax^2 - bx + c = T(p) = \lambda p = \lambda ax^2 +\lambda bx + \lambda c.$$ For this to happen, you must have: \begin{align*} a &= \lambda a\\ -b &= \lambda b\\ c &=\lambda c \end{align*} and $a$, $b$, and $c$ not all zero. You want to find all values of $\lambda$ for which there are soluctions $a,b,c$ not all zero. This is not terribly hard to do. Try that, and we can take it from there; edit your question if you get stuck to account for how far you have managed to go then.