Discussion
Let V and V' be finite dimensional vector spaces over K and let \(\{\sqrt{1}, \ldots, \sqrt{n}\}\) be a basis of V. If T \(\in\) L(V, V') and T is injective, then which of the following statements is FALSE ?
Answer: C
The false statement is dim V = dim V', option C.
The basis is printed with square-root signs (\(\sqrt{1}, \ldots, \sqrt{n}\)), which look like a typesetting slip for basis vectors. The argument below uses only that it is a basis of V with n elements and that T is injective.
Injective means \(\operatorname{Ker} T = \{0\}\), which is option D, so D is true. By rank-nullity:
\[ \begin{aligned} \dim V &= \dim \operatorname{Ker} T + \dim \operatorname{Im} T \\ &= \dim \operatorname{Im} T \le \dim V' \end{aligned} \]
so \(\dim V \le \dim V'\), option B, is true. Equality needs T to be onto as well, which is not given: the map \(x \mapsto (x, 0)\) from \(\mathbf{R}\) to \(\mathbf{R}^{2}\) is injective, yet 1 is not 2. So C can fail. An injective map takes a linearly independent set to a linearly independent set, so the images of the basis vectors stay independent and A is true.
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