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

Represent \(L\) by an \(8\times 10\) matrix \(A\) once bases are fixed for \(V\) and \(W\) (8 rows for the target dimension, 10 columns for the source dimension).

The rank of \(A\) can be at most the smaller of its number of rows and columns, so \[\operatorname{rank}(A) \le \min(8,10) = 8\]

The kernel of \(L\) corresponds to the null space of \(A\), and the dimension theorem for matrices gives \[\dim(\text{null space of } A) = 10 - \operatorname{rank}(A)\]

Since \(\operatorname{rank}(A)\) can be at most 8, \[\dim(\text{null space}) \ge 10 - 8 = 2\]

This holds for every choice of \(A\) representing such a map, so the kernel always contains at least a 2-dimensional subspace. Checking the alternatives: injectivity would need the null space to be \(\{0\}\), which is impossible here; invertibility needs both injectivity and surjectivity, also impossible; surjectivity is only possible in special cases, not guaranteed for every such \(L\).

\[\boxed{\dim(\operatorname{Ker}(L)) \ge 2 \text{ always}}\]