VIPSolutionsGive step-by-step solution with explanation and final answer:[25 pts] Determine whether the given set and operations define a real vector space. (a) V is the set of real vectors (x1, zs, 23,0) in R* with the standard vector addition and scalar multiplication. (b) V is the set of real vectors (v1, x2, 23,1) in R? where vector addi- tion and scalar multiplication are given by: (21.22.23. 1) + (y1, 42, 43, 1) = (21 + yr, 22 + yo, 23 + y3. 1) c- (x1, 9,23, 1) = (cx, ca, cz, 1) for all c € R.
Question:
Give step-by-step solution with explanation and final answer:
[25 pts] Determine whether the given set and operations define a real
vector space.
(a) V is the set of real vectors (x1, zs, 23,0) in R* with the standard
vector addition and scalar multiplication.
(b) V is the set of real vectors (v1, x2, 23,1) in R? where vector addi-
tion and scalar multiplication are given by:
(21.22.23. 1) + (y1, 42, 43, 1) = (21 + yr, 22 + yo, 23 + y3. 1)
c- (x1, 9,23, 1) = (cx, ca, cz, 1)
for all c € R.
This is AI generated Response.
Asked by: fffffff
Created at: 2025-05-30 14:56:55
Asked by: fffffff
Created at: 2025-05-30 14:56:55
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Let's analyze each part step by step:
---
### (a) \( V \) is the set of real vectors \((x_1, x_2, x_3, 0)\) in \(\mathbb{R}^4\) with standard vector addition and scalar multiplication.
#### **Step 1: Closure under addition**
Let \(\mathbf{u} = (u_1, u_2, u_3, 0)\) and \(\mathbf{v} = (v_1, v_2, v_3, 0)\) in \(V\).
\[
\mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3, 0 + 0) = (u_1 + v_1, u_2 + v_2, u_3 + v_3, 0)
\]
This is still in \(V\).
#### **Step 2: Closure under scalar multiplication**
Let \(c \in \mathbb{R}\),
\[
c \cdot \mathbf{u} = (c u_1, c u_2, c u_3, c \cdot 0) = (c u_1, c u_2, c u_3, 0)
\]
Still in \(V\).
#### **Step 3: Zero vector exists**
The zero vector is \((0, 0, 0, 0)\), which is in \(V\).
#### **Step 4: Additive inverses exist**
For any \((x_1, x_2, x_3, 0)\), the inverse is \((-x_1, -x_2, -x_3, 0)\), which is in \(V\).
#### **Step 5: Other vector space axioms**
The set inherits associativity, commutativity, distributivity, etc., from \(\mathbb{R}^4\).
**Conclusion for (a):**
\[
\boxed{\text{Yes, this is a real vector space.}}
\]
---
### (b) \( V \) is the set of real vectors \((x_1, x_2, x_3, 1)\) in \(\mathbb{R}^4\) with the given operations.
#### Operations:
- \[(x_1, x_2, x_3, 1) + (y_1, y_2, y_3, 1) = (x_1 + y_1, x_2 + y_2, x_3 + y_3, 1)\]
- \(c \cdot (x_1, x_2, x_3, 1) = (c x_1, c x_2, c x_3, 1)\)
#### **Step 1: Closure under addition**
Let \(\mathbf{u} = (u_1, u_2, u_3, 1)\) and \(\mathbf{v} = (v_1, v_2, v_3, 1)\),
\[
\mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3, 1)
\]
This is in \(V\).
#### **Step 2: Closure under scalar multiplication**
Let \(c \in \mathbb{R}\),
\[
c \cdot (x_1, x_2, x_3, 1) = (c x_1, c x_2, c x_3, 1)
\]
The last coordinate is always 1, **regardless of \(c\)**.
#### **Step 3: Zero vector**
The zero vector should be \((0, 0, 0, 0)\), but in this set, every vector is of the form \((x_1, x_2, x_3, 1)\). So, there is **no zero vector** in \(V\).
#### **Step 4: Additive inverses**
The additive inverse of \((x_1, x_2, x_3, 1)\) would be \((-x_1, -x_2, -x_3, 1)\), but \((x_1, x_2, x_3, 1) + (-x_1, -x_2, -x_3, 1) = (0, 0, 0, 1)\), **not** the zero vector.
#### **Step 5: Scalar multiplication issue**
For \(c = 0\):
\[
0 \cdot (x_1, x_2, x_3, 1) = (0, 0, 0, 1)
\]
It **does not** produce the zero vector.
**Conclusion for (b):**
\[
\boxed{\text{No, this is not a real vector space.}}
\]
**Reason:** There is no zero vector, and scalar multiplication does not produce the zero vector. Additive inverses do not exist either.
---
## **Final Answers**
**(a)** Yes, it is a real vector space.
**(b)** No, it is not a real vector space.
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