Proving the Isomorphism between {00y xy€R} and R2

Understanding isomorphisms is crucial in linear algebra and abstract algebra. An isomorphism between two algebraic structures is a bijective homomorphism that preserves the structure. In the context of vector spaces, two vector spaces are isomorphic if there exists a bijective linear transformation between them. This article will walk you through the steps to prove that the set {00y xy€R} (where 00y and xy are elements of the set of real numbers, R) is isomorphic to R2.

Defining an Isomorphism

The definition of an isomorphism between two vector spaces V and W can be formally stated as follows:

A function T: V rarr; W is an isomorphism if and only if:

T is a linear transformation (i.e., T(v w) T(v) T(w) and T(c*v) c*T(v) for all v, w ∈ V and c ∈ R) T is bijective (i.e., T is both injective and surjective)

Our goal is to show that the set {00y xy€R} is isomorphic to R2. Let's start by defining the set and the transformation.

Defining the Set V and the Transformation T

We define the set V {00y xy€R}, where 00y and xy are real numbers. We also define the transformation T: V rarr; R2 by the mapping T00y xy.

To establish that T is well-defined, we need to show that it is indeed a function from V to R2. This step is straightforward since the mapping is explicitly defined.

We need to show that T is a linear transformation. Let's consider the two properties of a linear transformation:

Additivity: T((00y1 00y2), (xy1 xy2)) T(00y1, xy1) T(00y2, xy2) Homogeneity: T(c * (00y, xy)) c * T(00y, xy) for any scalar c ∈ R

Let's prove each property:

Additivity: T((00y1 00y2), (xy1 xy2)) (xy1 xy2) T(00y1, xy1) T(00y2, xy2) (xy1) (xy2) Thus, T((00y1 00y2), (xy1 xy2)) T(00y1, xy1) T(00y2, xy2) Homogeneity: T(c * (00y, xy)) c * (xy) T(00y, xy) (xy) Thus, T(c * (00y, xy)) c * T(00y, xy)

Showing that T is Bijective

Now that we have established that T is a linear transformation, we need to show that T is bijective. For T to be bijective, it must be both injective (one-to-one) and surjective (onto).

Injectivity (One-to-One)

To show that T is injective, assume that T(00y1, xy1) T(00y2, xy2). This implies:

(xy1) (xy2)

Since (xy1) (xy2), it follows that 00y1 00y2. Thus, T is injective.

Surjectivity (Onto)

To show that T is surjective, we need to demonstrate that for every element (a, b) ∈ R2, there exists an element (00y, xy) ∈ V such that T(00y, xy) (a, b).

Given (a, b) ∈ R2, we can choose 00y a and xy b. Then:

T(a, b) (b)

Thus, T is surjective.

Since T is both injective and surjective, it is bijective.

Conclusion

We have shown that the set {00y xy€R} with the transformation T: V → R2 defined by T00y xy is a linear transformation and is bijective. Therefore, {00y xy€R} is isomorphic to R2 through the isomorphism T.

Understanding this concept is important for deeper dives into linear algebra and applications in physics, engineering, and computer science.