How to find eigenvalues of this linear transformation?  I have a practice problem on linear transformations that I'd like help on. I have to find the eigenvalues and eigenvectors of 
$$T(ax^2 + bx + c) = bx^2 + cx .$$
So I know the general idea is that you solve $T(v) = \lambda v$, so from there I can get the expression:
$$bx^2 + cx = \lambda ax^2 + \lambda bx + \lambda c ,$$ which implies that
$$
\begin{align*}
b &= \lambda a \\
c &= \lambda b \\
0 &= \lambda c
\end{align*} $$
but I'm stuck there. I know you need one of $a, b, c$ to be non-zero, but that only works if $\lambda = 0$. Even after I find the eigenvalues, how exactly do I find eigenvectors, since there's no matrix given?
Thanks in advance. 
Edit: corrected my claim to state that at least one of $a, b, c$ should be non-zero. 
 A: You only need $a$, $b$, and $c$ to not all be zero, but some of them can be zero.
Start with $\lambda c=0$, because it gives you the most information: either $\lambda=0$ or $c=0$.
If $\lambda=0$, then $c$ has to be zero (so that $c=\lambda b$ holds), and $b$ has to be zero (so that $b=\lambda a$ holds), but $a$ seems like it can be anything. That suggests trying polynomials of the form $ax^2$ with $a\neq 0$ and verifying that they are eigenvectors for $\lambda=0$.
If $\lambda\neq 0$, then $\lambda c = 0$ means that we must have $c=0$. Then $c=0=\lambda b$ tells you that $b$ must be zero. And then...
Once you have an eigenvalue $\lambda$, you find the eigenvectors by solving $T(v)=\lambda v$, $v\neq\mathbf{0}$.
Let me also add that if you are more comfortable with matrices, you can always compute a matrix for $T$: first pick a basis for your vector space; for example, $1$, $x$, and $x^2$. Then compute the value of $T$ at the basis, and express the answers in terms of the basis:
$$\begin{align*}
T(1) &= x\\
&= 0(1) + 1(x) + 0(x^2).\\
T(x) &= x^2\\
&= 0(1) + 0(x) + 1(x^2).\\
T(x^2) &= 0\\
&= 0(1) + 0(x) + 0(x^2).
\end{align*}$$
These give you the columns of the matrix of $T$ relative to the basis $\beta=[1,x,x^2]$, so that $T$ is represented (with respect to this basis) by
$$\left(\begin{array}{ccc}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{array}\right).$$
Finding the eigenvalues and eigenvectors is now straightforward; you will obtain your answers in terms of coordinate vectors relative to the basis $\beta$, which you can then translate back into polynomials written the usual way.
