Find eigenvalues for linear transformation $\mathbb R_2 \rightarrow \mathbb R_2$, $ p(x) \rightarrow x*p'(x) $. I have no idea on how to start this problem. Any help would be more then welcome.
 A: If you don't want to use a matrix representation (that require the choose of a basis in the vector space of polynomials), you can note that your linear transformation, for a polynomial $p(x)=ax^2+bx+c$ is:
$$
T(p(x))=x p'(x)=x(2ax+b)= 2ax^2+bx
$$
so the eigenvalues and eigenvectors are the real numbers $\lambda$ and the polynomials $q(x)$ such that:
$$
T(q(x))=\lambda q(x)
$$
that, for $q(x)=ax^2+bx+c$ becomes:
$$
2ax^2+bx=\lambda a x^2 +\lambda bx +\lambda c
$$
Now, using the identityt principle for polynomials, you can find that this is equivalent to the system
$$
\begin{cases}
a(\lambda -2)=0\\
b(\lambda-1)=0\\
c\lambda=0
\end{cases}
$$
that, for $a,b,c$ not all null, has solutions for $\lambda=2$, $\lambda=1$ or $\lambda=0$ . And substituting any of such eigenvalues you can find the corresponding eigenspaces ( or a single eigenvector in the eigenspace).
A: Your vector space has basis $\{1, x, x^2\}$.  Use this to get the matrix representation $A$ of your linear transformation by applying it to each basis vector, then writing the answer in terms of this very basis, and making the coordinates your columns.  Now find the eigenvalues of this matrix $A$ by the usual $\det(\lambda I - A)=0$ business. 
A: Here's some help, although not a complete answer: you probably know how to find eigenvalues of a matrix. (You can use Gaussian elimination, carefully, for instance). 
So if you could express your xformation as a matrix, you'd be in good shape. For that, you need a basis. A good choice is 
$$
1, x, x^2
$$
Consider the polynomial whose coordinates in this basis are $\pmatrix{1\\0\\0}$; that's $1 \cdot 1 + 0 \cdot x + 0 \cdot x^2$. 
What's the result of applying your transformation (let's call it $T$) to this vector? It's the $0$ polynomial, whose coordinates are $\pmatrix{0\\0\\0}$, so that's the first column of your matrix. 
Do the same thing with $\pmatrix{0\\1\\0}$ and $\pmatrix{0\\0\\1}$ to find the second and third columns. And then you have a matrix problem you can (I hope) solve. 
A: As an alternative to the above, solve the differential equation:
$$x\cdot \frac{dy}{dx}=\lambda \cdot y.$$
Almost certainly not what the exercise is supposed to test you on but a useful exercise nonetheless.
