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To determine if \(\mathbb{R}^2\) is a linear space over \(\mathbb{R}\) with the given scalar multiplications, we need to check if it satisfies the axioms of a vector space.

### Axiom Check for the First Scalar Multiplication:
\[ \lambda \cdot (x, y) = (0, \lambda y), \lambda \in \mathbb{R} \]

1. **Closure under scalar multiplication**: For any \(\lambda \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\), \(\lambda \cdot (x, y) = (0, \lambda y) \in \mathbb{R}^2\). This is satisfied.

2. **Associativity of scalar multiplication**: For \(\lambda, \mu \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\):
   \[ \lambda \cdot (\mu \cdot (x, y)) = \lambda \cdot (0, \mu y) = (0, \lambda (\mu y)) = (0, (\lambda \mu) y) \]
   \[ (\lambda \mu) \cdot (x, y) = (0, (\lambda \mu) y) \]
   This is satisfied.

3. **Identity element of scalar multiplication**: For \(1 \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\):
   \[ 1 \cdot (x, y) = (0, 1y) = (0, y) \neq (x, y) \]
   This is not satisfied.

Since the identity element axiom is not satisfied, \((\mathbb{R}^2, \lambda \cdot (x, y)) = (0, \lambda y)\) is not a vector space.

### Axiom Check for the Second Scalar Multiplication:
\[ \lambda \cdot (x, y) = (\lambda x, 1), \lambda \in \mathbb{R} \]

1. **Closure under scalar multiplication**: For any \(\lambda \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\), \(\lambda \cdot (x, y) = (\lambda x, 1) \in \mathbb{R}^2\). This is satisfied.

2. **Associativity of scalar multiplication**: For \(\lambda, \mu \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\):
   \[ \lambda \cdot (\mu \cdot (x, y)) = \lambda \cdot (\mu x, 1) = (\lambda (\mu x), 1) = ((\lambda \mu) x, 1) \]
   \[ (\lambda \mu) \cdot (x, y) = ((\lambda \mu) x, 1) \]
   This is satisfied.

3. **Identity element of scalar multiplication**: For \(1 \in \mathbb{R}\) and \((x, y) \in \mathbb{R}^2\):
   \[ 1 \cdot (x, y) = (1 \cdot x, 1) = (x, 1) \neq (x, y) \]
   This is not satisfied.

Since the identity element axiom is not satisfied, \((\mathbb{R}^2, \lambda \cdot (x, y)) = (\lambda x, 1)\) is not a vector space.

### Conclusion:
Neither of the scalar multiplications given define a vector space over \(\mathbb{R}\) on \(\mathbb{R}^2\) because they both fail to satisfy the identity element axiom.

Would you like more details or have any other questions? Here are some related questions you might find interesting:

1. What are the axioms of a vector space?
2. How to verify if a given set and operations form a vector space?
3. Can you provide examples of valid scalar multiplications for \(\mathbb{R}^2\)?
4. How does the failure of a single axiom affect the structure of a vector space?
5. What are some practical applications of vector spaces in real-world problems?

**Tip:** When verifying vector space properties, always ensure that all axioms are satisfied, as failing to meet even one axiom disqualifies the structure from being a vector space.

Explore the concept of vector spaces and scalar multiplication in \(\mathbb{R}^2\) with a detailed analysis of two given scalar multiplications. This problem involves verifying the axioms of vector spaces, including closure, associativity, and the identity element. Learn why neither scalar multiplication defines a vector space due to the failure to satisfy the identity element axiom. Suitable for advanced undergraduate students and those interested in abstract algebra concepts.

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