📂Linear Algebra

Homomorphism

Definition¹

For two vector spaces $V, W$, if there exists an invertible linear transformation $T : V \to W$, then $V$ is said to be isomorphic to $W$, and is denoted as follows.

$$V \cong W$$

Furthermore, $T$ is called an isomorphism.

Explanation

By the equivalence condition of being invertible, saying $T$ is an isomorphism means that $T$ is a bijective function. Therefore, if there exists a bijective function $T : V \to W$, then $V, W$ is isomorphic.

That $V, W$ is isomorphic means that there is virtually no difference between $V$ and $W$.

Theorem

Let $V, W$ be a finite-dimensional vector space. The necessary and sufficient condition for $V$ and $W$ to be isomorphic is that $\dim (V) = \dim (W)$ holds.

Corollary

Let $V$ be a vector space. The necessary and sufficient condition for $V$ to be isomorphic to $\mathbb{R}^{n}$ is that $\dim (V) = n$ holds.

Proof

$(\Longrightarrow)$

Assume that $T : V \to W$ is an isomorphism. Then, $T$ is invertible, and by the properties of invertible linear transformations,

$$\dim (V) = \dim (W)$$

$(\Longleftarrow)$

Assuming $\dim (V) = \dim (W)$, let $\beta = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{n} \right\}, \gamma = \left\{ \mathbf{w}_{1}, \dots, \mathbf{w}_{n} \right\}$ be the basis for $V, W$, respectively. Then, the following linear transformation exists between finite-dimensional vector spaces.

$$T : V \to W \quad \text{ by } \quad T(\mathbf{v}_{i}) = \mathbf{w}_{i}$$

Moreover, then $T(\beta) = \gamma$ is true, and since $T(\beta)$ generates $R(T)$,

$$R(T) = \span (T(\beta)) = \span (\gamma) = W$$

Therefore, $T$ is surjective. Then, since we assumed $\dim (V) = \dim (W)$, $T$ is also injective. Thus, there exists a bijective function $T : V \to W$, and $V$ and $W$ are isomorphic.

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