---
tag: 5P604PACP9
kind: proof
name: Proof that the solution space of a homogeneous linear equation has dimension n
msc: [34A30]
proves: 0GK68PZ6KH
uses: [0EQZSAB9EC, 0F2VBPXNP9, 0M0TKFRX2A, 0TH9CPW3VS, 0A6QZ367GX, 0WS1A8E2P9, 0RFWW56Y22, 055VVSN00V, 025PKW8AMR, 0BJBYCGMWE, 0BFD3A6CRH, 0KCQ9Y3QKA, 05G2G0P224]
sources:
  - url: https://en.wikipedia.org/wiki/Linear_differential_equation
    locator: Linear differential operator
---

## Proof

We reduce to a first-order linear system and use existence and uniqueness for it.

1. The set $S$ consists of the solutions of the [[homogeneous linear equation|0EQZSAB9EC]] on $I$. The zero function lies in $S$. By the [[superposition principle|0F2VBPXNP9]], $S$ is closed under linear combinations. So $S$ is a subspace of the real vector space of functions $I \to \R$.
2. Fix $x_0 \in I$ and let $\Phi(y) = \left(y(x_0), y'(x_0), \dots, y^{(n-1)}(x_0)\right)$ for $y \in S$. By [[linearity of differentiation|0M0TKFRX2A]], as in the superposition principle, $\Phi : S \to \R^n$ is linear.
3. For $x \in I$, let $A(x)$ be the $n \times n$ matrix with entries $A_{k, k+1}(x) = 1$ for $k < n$, last row $\left(-a_0(x), \dots, -a_{n-1}(x)\right)$, and all other entries $0$. Each entry of $A$ is continuous on $I$.
4. Apply the [[reduction to a first-order system|0TH9CPW3VS]] with $U = I \times \R^n$ and $f(x, z) = -a_0(x) z_1 - \dots - a_{n-1}(x) z_n$. The system becomes the [[linear system|0A6QZ367GX]] $z' = A(x) z$.
5. By that reduction, $y \in S$ implies that $z = \left(y, \dots, y^{(n-1)}\right)$ solves $z' = A(x) z$ on $I$. Conversely, each solution $z$ on $I$ has $z_1 \in S$ and $z = \left(z_1, \dots, z_1^{(n-1)}\right)$.
6. Let $[c, d] \subseteq I$ be a closed bounded interval. The function $x \mapsto \left(\sum_{i, j} A_{ij}(x)^2\right)^{1/2}$ is continuous. By the [[extreme value theorem|0WS1A8E2P9]], it has a maximum $K$ on $[c, d]$.
7. Let $x \in [c, d]$ and $v \in \R^n$. Entry $i$ of $A(x) v$ is the dot product of row $i$ of $A(x)$ with $v$. The [[Cauchy–Schwarz inequality|0RFWW56Y22]] bounds its square by $\left(\sum_j A_{ij}(x)^2\right) \norm{v}^2$.
8. Summing over $i$ gives $\norm{A(x) v} \le K \norm{v}$. So $\norm{A(x) z - A(x) w} \le K \norm{z - w}$ for $x \in [c, d]$ and $z, w \in \R^n$, a [[Lipschitz condition|055VVSN00V]] in $z$.
9. Injectivity. Let $y \in S$ with $\Phi(y) = 0$, and let $z = \left(y, \dots, y^{(n-1)}\right)$. Then $z$ and the zero curve both solve $z' = A(x) z$ on $I$, and both vanish at $x_0$.
10. The field $(x, z) \mapsto A(x) z$ is continuous on the open set $D = I \times \R^n$. Each point $(x_1, z_1) \in D$ has the neighbourhood $(c, d) \times \R^n$, where $[c, d] \subseteq I$ and $c < x_1 < d$. By step 8, the field is Lipschitz in $z$ there.
11. By [[uniqueness of solutions for Lipschitz vector fields|025PKW8AMR]], $z = 0$ on $I$. So $y = z_1 = 0$. The kernel of $\Phi$ is trivial, so [[$\Phi$ is injective|0BJBYCGMWE]].
12. Surjectivity. Let $c \in \R^n$. Write $I = (\alpha, \beta)$ with $-\infty \le \alpha < \beta \le \infty$. Choose numbers $c_m$ decreasing to $\alpha$ and $d_m$ increasing to $\beta$, with $c_1 < x_0 < d_1$.
13. Let $J_m = [c_m, d_m] \subseteq I$. By step 8 with $[c, d] = J_m$, the field $A(x) z$ is continuous on $J_m \times \R^n$ and Lipschitz in $z$ on all of it.
14. By [[global existence for globally Lipschitz vector fields|0BFD3A6CRH]], there is exactly one solution $z_m : J_m \to \R^n$ of $z' = A(x) z$ with $z_m(x_0) = c$.
15. Let $m \le k$. The restriction of $z_k$ to $J_m$ is a solution on $J_m$ with value $c$ at $x_0$. By the uniqueness in step 14, it equals $z_m$.
16. For $x \in I$, let $z(x) = z_m(x)$ for any $m$ with $x \in J_m$. Such $m$ exists since $c_m \to \alpha$ and $d_m \to \beta$. By step 15, $z(x)$ does not depend on $m$.
17. Each $x \in I$ lies in some open interval $(c_m, d_m)$, and $z = z_m$ there. So $z$ is differentiable at $x$ with $z'(x) = A(x) z(x)$. Also $z(x_0) = c$.
18. By step 5, $y = z_1$ lies in $S$, and $\Phi(y) = z(x_0) = c$. So $\Phi$ is surjective.
19. Thus $\Phi$ is a linear bijection, and its inverse $\Phi^{-1}$ is linear. Let $e_1, \dots, e_n$ be the standard basis of $\R^n$, and let $b_k = \Phi^{-1}(e_k)$.
20. If $\sum_k \lambda_k b_k = 0$, applying $\Phi$ gives $\sum_k \lambda_k e_k = 0$, so each $\lambda_k = 0$. Each $y \in S$ equals $\Phi^{-1}(\Phi(y)) = \sum_k \Phi(y)_k\, b_k$.
21. So $(b_1, \dots, b_n)$ is a [[basis|0KCQ9Y3QKA]] of $S$. Hence $S$ is a real vector space of [[dimension|05G2G0P224]] $n$, and $\Phi$ is a linear bijection.

## Idea

A solution is fixed by its $n$ initial values $y(x_0), \dots, y^{(n-1)}(x_0)$, and every choice of these values occurs. Uniqueness gives injectivity and existence gives surjectivity. Existence on all of $I$ holds because the equation is linear: on each closed subinterval, the field is globally Lipschitz.