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.