KindiakMath

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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