Final answer:
To prove the isomorphism, we need to show that there exists an injective homomorphism from each algebra to gl(2,R). For the 1-dimensional algebra, we can define a linear map that sends its basis element to any non-zero matrix in gl(2,R). For one of the 2-dimensional algebras, we can define a linear map that sends its basis elements to specific matrices. Similarly, we can define a linear map for the other 2-dimensional algebra. All these maps are injective and preserve the Lie bracket, making the algebras isomorphic to Lie subalgebras of gl(2,R).
Step-by-step explanation:
To prove that the 1-dimensional Lie algebra and the two 2-dimensional Lie algebras are isomorphic to Lie subalgebras of gl(2,R), we need to show that there exists an injective homomorphism from each of these algebras to gl(2,R). Let's go through each part:
a) For the 1-dimensional algebra, since it has only one basis element, we can define a linear map that sends this element to any non-zero matrix in gl(2,R). This map is injective and preserves the Lie bracket, so it is a Lie algebra homomorphism. Thus, the 1-dimensional algebra is isomorphic to a Lie subalgebra of gl(2,R).
b) For one of the 2-dimensional algebras, consider the basis elements [x] and [y]. We can define a linear map that sends [x] to a specific matrix in gl(2,R) and [y] to another specific matrix. This map is injective and preserves the Lie bracket, so it is a Lie algebra homomorphism. Therefore, this 2-dimensional algebra is isomorphic to a Lie subalgebra of gl(2,R).
c) Similarly, for the other 2-dimensional algebra with basis elements [u] and [v], we can define a linear map that sends [u] and [v] to specific matrices in gl(2,R). This map is injective and preserves the Lie bracket, making it a Lie algebra homomorphism. Hence, this 2-dimensional algebra is isomorphic to a Lie subalgebra of gl(2,R).