📚 Quick Study Guide
To determine if a subset $W$ of $R^n$ is a subspace, you must verify the following three axioms:
- ➕ Closure under addition: If $\mathbf{u}$ and $\mathbf{v}$ are in $W$, then $\mathbf{u} + \mathbf{v}$ must also be in $W$.
- 🔢 Closure under scalar multiplication: If $\mathbf{u}$ is in $W$ and $c$ is any scalar, then $c\mathbf{u}$ must also be in $W$.
- 📍 Contains the zero vector: The zero vector, $\mathbf{0}$, must be in $W$.
Practice Quiz
- Which of the following is NOT a requirement for a subset $W$ of $R^n$ to be a subspace?
- Contains the zero vector.
- Closed under addition.
- Closed under scalar multiplication.
- Contains the identity vector.
- Let $W = \{(x, y) \in R^2 : x = y\}$. Is $W$ a subspace of $R^2$?
- No, because it doesn't contain the zero vector.
- No, because it's not closed under scalar multiplication.
- Yes, it satisfies all three axioms.
- No, because it's not closed under addition.
- If $W$ is a subspace of $R^n$, and $\mathbf{u}, \mathbf{v} \in W$, which of the following must also be in $W$?
- $\mathbf{u} - \mathbf{v}$
- $\mathbf{u} \times \mathbf{v}$
- $\frac{\mathbf{u}}{\mathbf{v}}$
- $\mathbf{u} \cdot \mathbf{v}$
- Let $W = \{(x, y) \in R^2 : x^2 + y^2 = 1\}$. Is $W$ a subspace of $R^2$?
- Yes, because it contains the zero vector.
- Yes, because it's closed under addition.
- No, because it doesn't contain the zero vector and is not closed under scalar multiplication.
- Yes, because it's closed under scalar multiplication.
- Which of the following subsets of $R^3$ is a subspace?
- $\left\{(x, y, z) : x + y + z = 1\right\}$
- $\left\{(x, y, z) : x = y = z\right\}$
- $\left\{(x, y, z) : x^2 + y^2 + z^2 = 0\right\}$
- $\left\{(x, y, z) : xyz = 0\right\}$
- If $W$ is a subspace, and $\mathbf{u} \in W$, which of the following is always true?
- $-\mathbf{u} \notin W$
- $-\mathbf{u} \in W$
- $\mathbf{u} = \mathbf{0}$
- $\mathbf{u}$ is not a vector.
- Let $W = \{(x, y, z) \in R^3 : z = 0\}$. Is $W$ a subspace of $R^3$?
- No, because it does not contain the zero vector.
- No, because it's not closed under scalar multiplication.
- Yes, it satisfies all three axioms.
- No, because it's not closed under addition.
Click to see Answers
- D
- C
- A
- C
- B
- B
- C