SOLUTION: Which of the following subsets is/are NOT a vector subspace of R3? W1={(x,x,x^2):x∈R} W2={(x,y,x+2y):x,y∈R} W3={(x,2x,2):x∈R} W4={(x,y,x):x,y∈R} W5={(x,y,x^2):x,y∈R

Algebra ->  Probability-and-statistics -> SOLUTION: Which of the following subsets is/are NOT a vector subspace of R3? W1={(x,x,x^2):x∈R} W2={(x,y,x+2y):x,y∈R} W3={(x,2x,2):x∈R} W4={(x,y,x):x,y∈R} W5={(x,y,x^2):x,y∈R      Log On


   

Question 1174472: Which of the following subsets is/are NOT a vector subspace of R3?
W1={(x,x,x^2):x∈R}
W2={(x,y,x+2y):x,y∈R}
W3={(x,2x,2):x∈R}
W4={(x,y,x):x,y∈R}
W5={(x,y,x^2):x,y∈R}
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let's analyze each subset to determine if it's a vector subspace of R3.
**Criteria for a Vector Subspace:**
A subset W of a vector space V is a subspace if it satisfies the following three conditions:
1. **Zero Vector:** The zero vector of V is in W.
2. **Closure under Addition:** If u and v are in W, then u + v is in W.
3. **Closure under Scalar Multiplication:** If u is in W and c is a scalar, then cu is in W.
**Analysis:**
* **W1 = {(x, x, x²): x ∈ R}**
* **Zero Vector:** If x = 0, we get (0, 0, 0), so the zero vector is in W1.
* **Closure under Addition:** Let u = (x1, x1, x1²) and v = (x2, x2, x2²). Then u + v = (x1 + x2, x1 + x2, x1² + x2²). For u + v to be in W1, the third component must be (x1 + x2)². However, x1² + x2² ≠ (x1 + x2)² unless x1 or x2 is 0. Therefore, W1 is not closed under addition.
* **Conclusion:** W1 is NOT a vector subspace.
* **W2 = {(x, y, x + 2y): x, y ∈ R}**
* **Zero Vector:** If x = 0 and y = 0, we get (0, 0, 0), so the zero vector is in W2.
* **Closure under Addition:** Let u = (x1, y1, x1 + 2y1) and v = (x2, y2, x2 + 2y2). Then u + v = (x1 + x2, y1 + y2, x1 + 2y1 + x2 + 2y2) = (x1 + x2, y1 + y2, (x1 + x2) + 2(y1 + y2)). Thus, u + v is in W2.
* **Closure under Scalar Multiplication:** Let u = (x, y, x + 2y) and c be a scalar. Then cu = (cx, cy, c(x + 2y)) = (cx, cy, cx + 2cy). Thus, cu is in W2.
* **Conclusion:** W2 is a vector subspace.
* **W3 = {(x, 2x, 2): x ∈ R}**
* **Zero Vector:** The zero vector is (0, 0, 0). If we set x = 0, we get (0, 0, 2), which is not the zero vector. Therefore, W3 does not contain the zero vector.
* **Conclusion:** W3 is NOT a vector subspace.
* **W4 = {(x, y, x): x, y ∈ R}**
* **Zero Vector:** If x = 0 and y = 0, we get (0, 0, 0), so the zero vector is in W4.
* **Closure under Addition:** Let u = (x1, y1, x1) and v = (x2, y2, x2). Then u + v = (x1 + x2, y1 + y2, x1 + x2). Thus, u + v is in W4.
* **Closure under Scalar Multiplication:** Let u = (x, y, x) and c be a scalar. Then cu = (cx, cy, cx). Thus, cu is in W4.
* **Conclusion:** W4 is a vector subspace.
* **W5 = {(x, y, x²): x, y ∈ R}**
* **Zero Vector:** If x = 0 and y = 0, we get (0, 0, 0), so the zero vector is in W5.
* **Closure under Addition:** Let u = (x1, y1, x1²) and v = (x2, y2, x2²). Then u + v = (x1 + x2, y1 + y2, x1² + x2²). For u + v to be in W5, the third component must be (x1 + x2)². However, x1² + x2² ≠ (x1 + x2)² unless x1 or x2 is 0. Therefore, W5 is not closed under addition.
* **Conclusion:** W5 is NOT a vector subspace.
**Therefore, the subsets that are NOT vector subspaces of R3 are W1, W3, and W5.**