Question

For a linear transformation \(L:V\to W\) with \(\dim(V)=10\) and \(\dim(W)=8\), which one of the following options is correct?

• A. \(L\) is always injective
• B. \(L\) is always surjective
• C. \(\operatorname{Ker}(L)\) is at least 2-dimensional
• D. \(L\) is always invertible

Hint:

Use rank-nullity: nullity = dim(V) - rank(L), and rank(L) is capped by dim(W).

Answer 1

The Correct Option is C

Step 1: Apply the rank-nullity theorem to \(L:V\to W\). \[\dim(V) = \operatorname{rank}(L) + \operatorname{nullity}(L)\] Since \(\dim(V)=10\), this gives \(\operatorname{rank}(L) + \operatorname{nullity}(L) = 10\).

Step 2: Bound the rank using the codomain. The image of \(L\) is a subspace of \(W\), so \(\operatorname{rank}(L) \le \dim(W) = 8\).

Step 3: Bound the nullity from below. From \(\operatorname{nullity}(L) = 10 - \operatorname{rank}(L)\) and \(\operatorname{rank}(L)\le 8\), \[\operatorname{nullity}(L) \ge 10 - 8 = 2\] So \(\operatorname{Ker}(L)\) has dimension at least 2, no matter which such \(L\) is chosen. This also rules out the other options: since \(\dim(V) > \dim(W)\), \(L\) can never be injective, and since the kernel is forced to be nontrivial, \(L\) can never be invertible. Surjectivity is possible in some cases but is not guaranteed for every such \(L\).

\[\boxed{\operatorname{Ker}(L) \text{ is at least 2-dimensional}}\]