Exercise · ZW-RANK-NULLITY-THEOREM-007 · ★★☆ Intermediate

Can there be an injective linear map \(T:\mathbb{R}^3\to\mathbb{R}^2\)?

Choose one option.

Show answer
  • D. No: the image has dimension at most \(2\), so the kernel has dimension at least \(1\)

Explanation

Since \(\dim\operatorname{Im}T\le2\), the nullity \(3-\dim\operatorname{Im}T\) is at least \(1\), and a nontrivial kernel rules out injectivity. The kernel always contains zero; injective means it contains only zero.

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Review the concept: Rank–nullity theorem

For a linear transformation \(T\) with finite-dimensional domain \(V\), the dimension of the kernel plus that of the image equals the dimension of \(V\).

\[\dim V=\dim\ker T+\dim\operatorname{Im}T\]

More practice problems: rank–nullity theorem

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