📚 Topic Summary
In linear algebra, a mapping (or transformation) between vector spaces is considered linear if it preserves vector addition and scalar multiplication. To verify linearity, you need to check two conditions: first, that the mapping of the sum of two vectors equals the sum of the mappings of the individual vectors; second, that the mapping of a scalar multiple of a vector equals the scalar multiple of the mapping of the vector. If both conditions hold, the mapping is linear; otherwise, it is not.
More formally, let $V$ and $W$ be vector spaces. A mapping $T: V \rightarrow W$ is linear if and only if:
- $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v} \in V$.
- $T(c\mathbf{u}) = cT(\mathbf{u})$ for all $\mathbf{u} \in V$ and all scalars $c$.
🧮 Part A: Vocabulary
Match the term with its correct definition:
| Term |
Definition |
| 1. Linear Mapping |
A. A function that preserves vector addition and scalar multiplication. |
| 2. Vector Space |
B. A set of objects that can be added together and multiplied by scalars. |
| 3. Scalar Multiplication |
C. Multiplying a vector by a scalar, resulting in another vector. |
| 4. Vector Addition |
D. Combining two vectors to produce a third vector. |
| 5. Transformation |
E. A function from one vector space to another. |
Answers:
1-A, 2-B, 3-C, 4-D, 5-E
✍️ Part B: Fill in the Blanks
A mapping $T: V \rightarrow W$ is ________ if it preserves ________ addition and ________ multiplication. This means that for any vectors $\mathbf{u}$ and $\mathbf{v}$ in $V$, $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$, and for any scalar $c$ and vector $\mathbf{u}$ in $V$, $T(c\mathbf{u}) = ________$. If these conditions hold, the mapping is considered ________.
Answers:
linear, vector, scalar, $cT(\mathbf{u})$, linear
🤔 Part C: Critical Thinking
Consider the mapping $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $T(x, y) = (x^2, y)$. Is this mapping linear? Explain why or why not, using the properties of linear mappings.