Degree is a coordinate

Take a polynomial and slide its variable: replace $x$ by $ax + b$. Nothing about the graph is destroyed, it is stretched by $a$ and slid by $b$, and it seems obvious that a quintic stays a quintic. It is obvious. But the reason is worth extracting, because the same reason will fail spectacularly in a later article when we allow ourselves to divide.

The map $T_{a,b} : p(x) \mapsto p(ax+b)$ is linear in $p$. Two polynomials substituted and then added give the same thing as added and then substituted, so $T_{a,b}$ has a matrix on the space of polynomials of degree at most $n$. Getting it is one application of the binomial theorem. Write $p = \sum_j c_j x^j$; then

$p(ax+b) = \sum_j c_j (ax+b)^j = \sum_j c_j \sum_{k=0}^{j} \binom{j}{k} a^k b^{\,j-k} x^k,$

so the coefficient of $x^k$ in the image is $\sum_{j \geq k} \binom{j}{k} a^k b^{\,j-k} c_j$, and the matrix is

$M(a,b)_{kj} = \binom{j}{k}\, a^k b^{\,j-k}.$

Everything below the diagonal is zero, because a $j$-th power cannot produce an $x^k$ with $k > j$. Every entry on the diagonal is $\binom{k}{k} a^k b^0 = a^k$.

interactive · p(x) ↦ p(ax + b) on the monomial basisopen ↗

Interactive demo: p(x) ↦ p(ax + b) on the monomial basis. Requires JavaScript to run — the source is at /interactive/funcspace/substitution.html.

Drag $a$ and $b$ and watch three things move together. The bottom-right entry, boxed, is $a^5$. It is the only thing the leading coefficient of the image depends on: the top coefficient of $p(ax+b)$ is $a^5 c_5$, with no contribution from any other $c_j$ and no contribution from $b$ at all. The determinant is the product of the diagonal, $a^{0}a^{1}\cdots a^{n} = a^{n(n+1)/2}$, which for $n = 5$ is $a^{15}$. And the coefficient bars redistribute wildly while the highest one refuses to vanish.

So the statement “an affine substitution preserves degree” is the statement “a triangular matrix with no zero on its diagonal is invertible”, plus the observation that the diagonal here is $a^k$ and vanishes only when $a$ does. Push $a$ to zero in the demo. The matrix collapses to a single row, its rank drops to one, the determinant drops to zero, and the quintic becomes the constant $p(b)$. That is not degree preservation failing under stress. It is the map no longer being a change of variable, because $x \mapsto b$ is not a change of anything.

The triangularity says more than degree preservation. A triangular matrix preserves the whole flag of subspaces $\mathcal{P}_0 \subset \mathcal{P}_1 \subset \cdots \subset \mathcal{P}_n$: if $p$ has degree at most 3, so does $p(ax+b)$, for every $a$ and $b$ at once. Degree is which step of the flag you first appear on, and $T_{a,b}$ moves nobody between steps. Sliding and stretching the variable cannot smuggle a cubic into the quadratics.

There is a decomposition hiding in the matrix that is worth naming, because the next two articles are about its two halves. Scaling and shifting compose:

$T_{a,b} = T_{a,0} \circ T_{1,b}, \qquad M(a,b) = M(a,0)\, M(1,b),$

since applying $T_{1,b}$ and then $T_{a,0}$ sends $p(x)$ to $p(x+b)$ to $p(ax+b)$. The first factor $M(a,0)$ is diagonal, $\operatorname{diag}(1, a, a^2, \ldots, a^n)$: the monomials are its eigenvectors and $a^k$ are its eigenvalues, which is a fact we will lean on hard when we get to log paper. The second factor $M(1,b)$ is Pascal’s triangle, ones on the diagonal, everything interesting strictly above it. It is unipotent: the identity plus something that is nothing but a nilpotent shadow, and that shadow is the derivative.

A diagonalizable part and a unipotent part, multiplied together. Anyone who has met the Jordan decomposition will recognize the shape of it, and it is not a coincidence that it appears here: the maps $x \mapsto ax + b$ with $a \neq 0$ form a group under composition — the affine group of the line — and $M$ is a representation of that group on $\mathcal{P}_n$. The scalings are the semisimple part, the translations are the unipotent part, and the semidirect product structure of the group is visible in the matrices.

The geometric reading is about roots. If $r$ is a root of $p$, then $p(a r' + b) = 0$ exactly when $ar' + b = r$, so the roots of $p(ax+b)$ are the roots of $p$ pulled back by the same affine map, $r' = (r - b)/a$. There are $n$ of them counted with multiplicity, before and after, in the complex plane. Nothing is created, nothing is destroyed, everything is slid and stretched. The affine maps of the line are exactly the transformations that permute the roots of every polynomial without ever losing one — and the reason they cannot lose one, which will only become clear once we have somewhere for a root to be lost to, is that they fix the point at infinity.

The affine maps in this article are the same affine maps that drive the fractal designer, where each control point is one contractive map $x \mapsto Ax + b$ of the plane rather than the line. Same group, one dimension up, and the same triangular bookkeeping. There it is doing something more dramatic than preserving a degree; see what an affine IFS actually is.