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  "body": "\n## Proof\n\n1. Let $S$ be the space of solutions on $I$ of the [[homogeneous linear equation|0EQZSAB9EC]]. For $x \\in I$, let $\\Phi_x(y) = \\left(y(x), y'(x), \\dots, y^{(n-1)}(x)\\right)$ for $y \\in S$.\n2. By the [[dimension theorem for the solution space|0GK68PZ6KH]], $\\Phi_x : S \\to \\R^n$ is a linear bijection for each $x \\in I$.\n3. For $x \\in I$, let $M(x)$ be the $n \\times n$ matrix whose column $j$ is $\\Phi_x(y_j)$. By the definition of the [[Wronskian|0N5NF1RGBP]], $W(x) = \\det M(x)$.\n4. For $c = (c_1, \\dots, c_n) \\in \\R^n$, let $y_c = \\sum_j c_j y_j$. By the [[superposition principle|0F2VBPXNP9]], $y_c \\in S$. By linearity of $\\Phi_x$, $\\Phi_x(y_c) = M(x)\\, c$.\n5. (a) implies (c). Assume (a), and suppose $W(x) = 0$ for some $x \\in I$. Then [[$M(x)$ is not invertible|03Q7RE2HXR]].\n6. The linear map $c \\mapsto M(x) c$ of $\\R^n$ is then not bijective. By [[the injective endomorphism theorem|0912EEEVCT]], it is not injective. So there is $c \\ne 0$ with $M(x) c = 0$.\n7. By step 4, $\\Phi_x(y_c) = 0$. Since $\\Phi_x$ is injective, $y_c = 0$ on $I$.\n8. So $\\sum_j c_j y_j = 0$ on $I$ with $c \\ne 0$. This contradicts the [[linear independence|0EVHJFTG9Z]] in (a). Hence $W(x) \\ne 0$ for each $x \\in I$.\n9. (c) implies (b). The interval $I$ is not empty, so take any $x_0 \\in I$.\n10. (b) implies (a). Let $W(x_0) \\ne 0$, and let $c \\in \\R^n$ with $\\sum_j c_j y_j = 0$ on $I$. So $y_c = 0$, and all its derivatives vanish on $I$, by [[linearity of differentiation|0M0TKFRX2A]].\n11. In particular $M(x_0)\\, c = \\Phi_{x_0}(y_c) = 0$, by step 4. Since $\\det M(x_0) = W(x_0) \\ne 0$, the matrix $M(x_0)$ is invertible.\n12. Hence $c = M(x_0)^{-1} M(x_0)\\, c = 0$. So $y_1, \\dots, y_n$ are linearly independent on $I$.\n13. The implications (a) to (c), (c) to (b) and (b) to (a) show that the three conditions are equivalent.\n\n## Idea\n\nThe columns of the Wronskian matrix at $x$ are the initial values of the solutions at $x$. A solution is determined by its initial values at any one point. So a dependence among the columns at one point is a dependence among the solutions on all of $I$.\n",
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