Does the polynomial derivative matrix have complex eigenvalues and what do they mean?

Question

I was watching this video, which describes a matrix that applies the derivative to the space of polynomials. It's pretty clear that this matrix has no real eigenvalues since

$$\frac{d}{dx}kx^n \ne \lambda x^n$$

for any $k \ne 0$.

What if one tries to solve for complex eigenvalues and eigenvectors? Do these exist, and if so, how can we reason about them intuitively?

Answer 1

Actually the differentation operator does have an eigenvalue ($\lambda = 0$). Remember that if a linear operator $T$ is not invertible, then $\lambda = 0$ is an eigenvalue of $T$ and the corresponding eigenspace is $\ker T$. In the case of the differentiation operator, the eigenspace is the space of constant polynomials.

If you consider polynomials with complex coefficients, nothing changes because the rules for differentiating polynomials over $\mathbb{R}$ also hold for polynomials over $\mathbb{C}$.

Answer 2

You can reason about complex eigenvalues as rotations in $R^2$, but there are no non-trivial eigenvalues for any $n \times n$ "derivative matrix" as $\det{(A - \lambda I)}$ would just be $(-\lambda)^{n}$ (since $A - \lambda I$ is upper triangular) setting that equal to $0$ gives $\lambda = 0$.

Answer 3

One natural generalization is to the space of entire functions (i.e. functions analytic on $\mathbb C$, thus represented by power series with radius of convergence $\infty$). Here all complex numbers $\lambda$ are eigenvalues, with eigenvector $f(z) = e^{\lambda z}$. Of course these are not polynomials except in the case $\lambda = 0$.