A Shocking Trichotomy

Problem 1. Let $V$ be a vector space over $\mathbb R$ . Prove that if $V$ has at least one nonzero element, then $V$ has infinitely many elements.

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Solution. Suppose $V$ contains some nonzero $v$ . Since $V$ is a vector space over $\mathbb R$ , $V$ contains $tv$ for any $t \in \mathbb R$ . Each $t$ yields a unique $tv$ since $v \neq 0$ . Thus, $V$ contains an infinite number of elements given by the injection $\mathbb R \to V, t \mapsto tv$ .

Problem 2. Let $V, W$ be a vector spaces over $\mathbb R$ , and $T : V \to W$ be a linear transformation. For any $y \in W$ , prove that the equation $T(x) = y$ either has zero solutions, or one solution, or infinitely many solutions.

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Solution. If the equation $T(x) = y$ has no solutions, then we are done. Otherwise, it has at least one solution $u$ , which yields $T(u) = y$ .

If the equation has at most one solution, then we are done. Otherwise, the equation has at least one other solution $v \neq u$ , which yields $T(v) = y$ . Since $T$ is linear,

$T(v-u) = T(v) - T(u) = y - y = 0.$

For any $t \in \mathbb R$ , consider the vector $x_t := u + t(v-u)$ . Each $t$ yields a unique $x_t$ since $v-u \neq 0$ . On the other hand

$\begin{aligned} T(x_t) &= T(u + t(v-u)) \\ &= T(u) + t \cdot T(v-u) \\ &= y + t \cdot 0 \\ &= y + 0 = y. \end{aligned}$

Thus, the equation $T(x)=y$ has infinitely many solutions defined by $x_t = u + t(v-u)$ .

—Joel Kindiak, 30 Nov 24, 0112H

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Differential Polynomials – KindiakMath

April 20, 2025 at 8:12 pm

[…] It turns out that if we can find just one function such that , then we can find all functions such that . This approach isn’t unique to differential polynomials, but applies to any linear transformation. […]

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A Shocking Trichotomy

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