Linear operators with upper triangular matrix representations Can I have an example of a linear operator having an upper triangular (not just diagonal) matrix representation for a given basis? I am having tough time understanding upper triangular matrices related to operator matrices. 
Suggestions for any text with examples would be helpful.
 A: In fact, for any linear transformation $T$ from a finite-dimensional vector space $\Bbb V$ (over the field, say, $\Bbb F$) to itself, there is a basis with respect to which its matrix representation $[T]$ is upper-triangular, provided that we allow extensions of the base field. More precisely:
If $\Bbb F$ is algebraically closed, there is a basis of $\Bbb V$ with respect to which the matrix representation $[T]$ is in Jordan normal form, and in particular upper-triangular.
In fact: Any $n \times n$ complex matrix $A$ is unitarily triangulizable, that is, there is a unitary matrix $Q$ (i.e., a matrix such that $Q^* Q = I$) and an upper triangularize matrix $U$ such that $A = Q U Q^{-1}$. This is called a Schur decomposition of the matrix $A$.
Here's an important concrete example:

Example For a fixed $k$ let $P_k$ denote the space of (real or complex) polynomials (in $x$) of degree $\leq k$. Then, with respect to the standard basis $\{1, x, \ldots, x^k\}$ of $P_k$, the derivative operator $$\frac{d}{dx} : P_k \to P_k$$ is upper triangular. In fact, since differentiation decreases the degree of a polynomial, $\frac{d}{dx}$ is actually strictly upper triangular (its diagonal entries are all zero) and nilpotent of degree $k + 1$. (I encourage you to verify these claims in, say, the case $k = 3$ by computin)

