How to solve this eigenvector eigenvalue problem? Let $L: P_1 \to P_1$ be the linear operator defined by $L(at+b)=-bt-a$. Find, if possible, a basis for $P_1$ with respect to which $L$ is represented by a diagonal matrix.
How about this?
 A: The vector space $P_1$ is the set of polynomials of degree at most $1$, and so there is a natural isomorphism $T: P_1 \to \mathbb R^2$ given by $T(at+b) = (a,b)$. Now you can easily come up with a matrix for $L$ with respect to the standard basis of $\mathbb R^2$, and diagonalize as usual, finding eigenvalues and eigenvectors.
A: Denote by $L_0$ the operator on $P_1$ defined by
$L_0(at + b) = bt + a; \tag{1}$
then
$L = -L_0. \tag{2}$
The operator $L_0$ has been thoroughly analyzed in my answer to the question https://math.stackexchange.com/questions/1211802/how-to-solve-this-eigenvalue-eigenvector-problem, where it is shown that the eigenvalues of $L_0$ are $\pm 1$ with eigenvectors $1 + t$ (for eigenvalue $1$) and $1 - t$ (for $1$).  Now use the fact that the eigenvalues of the negative of any operator $T$ are the negatives of the eigenvalues of $T$, with the same eigenvectors:  if
$Tv = \mu v, \tag{3}$
then
$(-T)v = -Tv = -(\mu v) = (-\mu) v. \tag{4}$
It thus follows that the eigenvalues of $L$ are also $\pm 1$, with the same eigenvectors; their roles however are reversed:  here the eigenvector associated with $1$ is not 
$1+ t$ but $1 - t$:
$L(1 - t) = -t + 1 = 1 - t; \tag{5}$
likewise $-1$ corresponds to $1 + t$:
$L(1 + t) = -t - 1 = -(1 + t).  \tag{6}$
The above in fact shows that the operator $L$ is diagonal in the basis $\{ 1 + t, 1 - t \}$ of $P_1$.
Diagonalizing an operator invariably involves finding its eigenvectors, which in turn requires calculating its eigenvectors.  Here the eigenvalues were found by solving the characteristic equation of $L_0$, and the associated eigenvalues by simple inspection (intelligent guessing). In more complicated examples, the vectors are usually found by solving a linear system.  Either way, the eigenbasis diagonalizes (when, that is, diagonalization is possible).
