A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^3$ is represented by a $3 \times 5$ real matrix $A$. If the column space of $A$ has dimension 3 (full row rank), the dimension of the null space (kernel) of $A$ is:

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles By the Rank-Nullity Theorem (Dimension Theorem for Linear Transformations): $$\text{Rank}(A) + \text{Nullity}(A) = n$$ where: - $A$ is an $m \times n$ matrix representing linear map $T: \mathbb{R}^n \to \mathbb{R}^m$ - $n$ is the dimension of the domain (number of columns of $A$) - $\text{Rank}(A) = \dim(\text{Col}(A))$ is the dimension of the column space (range of $T$) - $\text{Nullity}(A) = \dim(\ker(A))$ is the dimension of the null space. ### 2. Analytical Formulation & Calculation Given: - Matrix size: $3 \times 5 \implies n = 5$ columns (domain dimension is 5) - Column space dimension: $\text{Rank}(A) = 3$ Applying Rank-Nullity Theorem: $$3 + \text{Nullity}(A) = 5$$ $$\text{Nullity}(A) = 5 - 3 = 2$$ ### 3. Distractor Forensics - Candidate using number of rows $m=3$ gets $3 - 3 = 0$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Remember $n$ in the Rank-Nullity Theorem is always the number of *columns* (dimension of domain), not rows. - **Standard Reference:** Strang, G., *Linear Algebra and Its Applications*, Chapter 3.

← Prev Question Next Question →