Examples of Linear Maps

Differentiation and integration as linear maps underlie the entire theory of linear differential equations: writing $D$ for the derivative operator, an equation like $y'' - 3y' + 2y = 0$ becomes $(D^2 - 3D + 2I)(y) = 0$, turning solving the ODE into finding the kernel of a linear map — precisely the technique used across engineering and physics. In quantum mechanics, physical observables are modeled as linear operators on Hilbert spaces of wavefunctions, with differentiation operators representing momentum. In signal processing and cryptography, shift maps on sequences model delay lines and linear feedback shift registers, the latter foundational to stream ciphers. In computer graphics and computer vision, projection maps (often orthogonal projections rather than oblique ones) are used constantly to render 3D scenes onto 2D image planes, and to reduce data dimensionality via projection onto subspaces. In statistics, projection matrices are the essential tool in least-squares regression, projecting an observed data vector onto the column space of a design matrix. All of these depend, at bottom, on the fact — established here and generalized in linear maps determined by their action on a basis — that once a handful of canonical examples are shown to be linear, an entire structural theory (rank-nullity, invertibility, matrix representation) becomes available to analyze them.

Ask the tutor: Can you show me how to build the matrix of the differentiation map $D$ on $\mathcal{P}_3(\mathbb{R})$ from first principles, using the basis $1, x, x^2, x^3$?

A linear map is a function that "commutes" with the two basic vector-space operations: it does not matter whether you add vectors first and then apply the map, or apply the map first and then add the results — the answer is the same. The same holds for scaling. This single requirement is extraordinarily restrictive: most functions one writes down (squaring, taking absolute value, adding a constant) fail it instantly. What makes calculus and geometry rich sources of linear maps is that several of their most basic operations do satisfy it, essentially because differentiation and integration are themselves defined through limits of sums and scalar multiples, and geometric projection is defined through a linear decomposition of space. It helps to picture the vector space of polynomials of degree at most $3$, with basis $1, x, x^2, x^3$, as a $4$-dimensional coordinate space with axes labeled by these basis vectors. Differentiation moves each axis one slot to the left and rescales it, mapping $x^3 \mapsto 3x^2$, $x^2 \mapsto 2x$, and so on. Integration (with the convention of integrating from $0$) moves each axis one slot to the right and rescales oppositely. This "shift-and-scale" picture is exactly analogous to the sequence-shift map, which simply moves coordinates without rescaling — the same underlying mechanism, dressed differently. Projection is the geometric heart of the story: decompose space into two complementary directions, keep the component along one direction, and discard the component along the other. The picture below shows this in the plane, projecting onto the $x$-axis along the direction of a line of slope $1$.

Example 1: Differentiation and integration on $\mathcal{P}_3(\mathbb{R})$, tracked through an explicit non-symmetric polynomial. Let $p(x) = 4x^3 - 3x^2 + x - 7 \in \mathcal{P}_3(\mathbb{R})$, and let $D: \mathcal{P}_3(\mathbb{R}) \to \mathcal{P}_2(\mathbb{R})$ be differentiation and $J: \mathcal{P}_2(\mathbb{R}) \to \mathcal{P}_3(\mathbb{R})$ be integration from $0$. Compute $D(p)$, then $J(D(p))$, and compare to $p$. Step 1 — differentiate termwise:

Step 2 — integrate $D(p)$ from $0$:

Comparing to the original $p(x) = 4x^3-3x^2+x-7$, we find

so $J \circ D$ is not the identity map: it recovers $p$ only up to the loss of the constant term $-7$. This is expected because $D$ kills constants (its kernel is exactly $\mathcal{P}_0(\mathbb{R})$, the constants), so information about the constant term of $p$ cannot survive differentiation; integrating back in cannot restore what was destroyed. Concretely this also verifies that $D$ is not injective: $D(p) = D(p-7)$ but $p \neq p - 7$. By contrast, compute $D(J(q))$ for $q(x) = 2x^2 - 5x + 3 \in \mathcal{P}_2(\mathbb{R})$:

Here $D(J(q)) = q$ exactly: integrating first and then differentiating does recover the original polynomial, because $J$ introduces no ambiguity ($J$ is injective, as it always produces a polynomial with zero constant term uniquely determined by $q$), and differentiation exactly undoes the antiderivative operation by the Fundamental Theorem of Calculus. This asymmetry, $D\circ J = I$ but $J \circ D \neq I$, is a first hint of the difference between one-sided inverses and true inverses, to be developed once invertibility of linear maps is formally treated.

Example 2: A non-trivial oblique projection on $\mathbb{R}^2$, computed on an explicit vector. Let $U = \operatorname{span}\{(1,2)\}$ and $W' = \operatorname{span}\{(3,1)\}$. First check $\mathbb{R}^2 = U \oplus W'$: the vectors $(1,2)$ and $(3,1)$ are linearly independent (neither is a scalar multiple of the other, since $1/3 \neq 2/1$), and two linearly independent vectors in a $2$-dimensional space span it, so $U + W' = \mathbb{R}^2$ and $U \cap W' = \{0\}$ (an element of both would have to be a multiple of $(1,2)$ and of $(3,1)$ simultaneously, forcing it to be $0$ by the independence just noted). Hence the direct sum decomposition holds and the oblique projection $P$ onto $U$ along $W'$ is well defined. Take $v = (7, 5)$. We must write $v = a(1,2) + b(3,1)$ for scalars $a,b$, i.e., solve

From the second equation, $b = 5 - 2a$. Substituting into the first:

Then $b = 5 - 2\cdot\frac{8}{5} = 5 - \frac{16}{5} = \frac{9}{5}$. Check: $a(1,2)+b(3,1) = \frac{8}{5}(1,2) + \frac{9}{5}(3,1) = \left(\frac{8}{5}+\frac{27}{5},\ \frac{16}{5}+\frac{9}{5}\right) = \left(\frac{35}{5}, \frac{25}{5}\right) = (7,5) = v$. ✓. So the decomposition is $v = \underbrace{\tfrac{8}{5}(1,2)}_{\in U} + \underbrace{\tfrac{9}{5}(3,1)}_{\in W'}$, and by definition

As a sanity check on idempotence, apply $P$ again: since $P(v) \in U$, its own decomposition is $P(v) = P(v) + 0$, so $P(P(v)) = P(v)$, matching $P^2 = P$ proved in general above. Notice also that this projection is not the familiar orthogonal projection (which would use the standard inner product and project along the perpendicular direction); it is genuinely oblique, projecting along $(3,1)$, illustrating that "projection" in the linear-algebra sense depends on the choice of complementary subspace, not merely on $U$.