📂Linear Algebra

Every n-dimensional Real Vector Space is Isomorphic to R^n

Definition¹

Let $V$ and $W$ be called a vector space. If there exists an invertible (bijective) linear transformation $T : V \to W$, then $V$ and $W$ are said to be isomorphic. $T$ is called an isomorphism.

Theorem

Every $n$-dimentional real vector space is isomorphic to $\mathbb{R}^{n}$.

Explanation

Another way to express the theorem is as follows.

“A $\mathbb{R}$-vector space $V$ being isomorphic to $\mathbb{R}^{n}$” is equivalent to “being $\dim{V}=n$”.

Proof²

Let $V$ be a $n$-dimensional real vector space. Then, the proof is completed by showing the existence of a one-to-one and onto linear transformation $T$ that satisfies the following.

$$T : V \to \mathbb{R}^{n}$$

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Let’s call $S = \left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n} \right\}$ a basis of $V$. Then, for every $\mathbf{v} \in V$, there exists a unique linear combination of bases as follows.

$$\mathbf{v} = k_{1}\mathbf{v}_{1} + k_{2}\mathbf{v}_{2} + \cdots + k_{n}\mathbf{v}_{n},\quad k_{i}\in \mathbb{R}$$

Now, define the transformation $T$ as follows.

$$T(\mathbf{v}) = (k_{1}, k_{2}, \dots, k_{n})$$

• Part 1. $T$ is linear

  Let’s say $\mathbf{v}, \mathbf{u} \in V$ is expressed as follows.

  $$\begin{equation} \mathbf{v}=k_{1}\mathbf{v}_{1} + k_{2}\mathbf{v}_{2} + \cdots + k_{n}\mathbf{v}_{n} \quad \text{and} \quad \mathbf{u}=d_{1}\mathbf{v}_{1} + d_{2}\mathbf{v}_{2} + \cdots + d_{n}\mathbf{v}_{n} \end{equation}$$

  And let’s call it $c\in \mathbb{R}$. Then, by the following, $T$ is linear.

  $$\begin{align*} T(\mathbf{v} + c\mathbf{u}) &= T\left( (k_{1}+ cd_{1})\mathbf{v}_{1} + (k_{2}+cd_{2})\mathbf{v}_{2} + \cdots + (k_{n}+cd_{n})\mathbf{v}_{n} \right) \\ &= \left( k_{1}+ cd_{1}, k_{2}+cd_{2}, \dots, k_{n}+cd_{n}\right) \\ &= \left( k_{1}, k_{2}, \dots, k_{n}\right) + c\left(d_{1}, d_{2}, \dots, d_{n}\right) \\ &= T(\mathbf{v}) + cT(\mathbf{u}) \end{align*}$$

• Part 2. $T$ is one-to-one.

  If $\mathbf{v}, \mathbf{u}$ satisfies $(1)$, and let’s call it $\mathbf{v} \ne \mathbf{u}$. Then, for at least one $i$, $k_{i}\ne d_{i}$ must hold. Therefore,

  $$\left( k_{1}, k_{2}, \dots, k_{n}\right) = T(\mathbf{v}) \ne T(\mathbf{u}) = \left(d_{1}, d_{2}, \dots, d_{n}\right)$$

• Part 3. $T$ is onto

  Let’s call it $\mathbf{x} = (x_{1}, x_{2}, \dots, x_{n})\in \mathbb{R}^{n}$. Then, since $V$ is the set of all linear combinations of $\mathbf{v}_{i}$s, there exists a $\mathbf{v} \in V$ that satisfies the following.

  $$\mathbf{v} = x_{1}\mathbf{v}_{1} + x_{2}\mathbf{v}_{2} + \cdots + x_{n}\mathbf{v}_{n}$$

  Therefore, every $T$ is onto.

See Also

• Other Proofs

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