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 \(1\)
Exercise · ZW-RANK-NULLITY-THEOREM-007 · ★★☆ Intermediate
Choose one option.
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.
For a linear transformation \(T\) with finite-dimensional domain \(V\), the dimension of the kernel plus that of the image equals the dimension of \(V\).