# Every $n$-Dimensional Vector Space is Isomorphic to the Vector Space $\R^n$

## Problem 545

Let $V$ be a vector space over the field of real numbers $\R$.

Prove that if the dimension of $V$ is $n$, then $V$ is isomorphic to $\R^n$.

## Proof.

Since $V$ is an $n$-dimensional vector space, it has a basis

\[B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\},\]
where each $\mathbf{v}_i$ is a vector in $V$.

Define a map $T: V\to \R^n$ by sending each vector $\mathbf{v}\in V$ to its coordinate vector $[\mathbf{v}]_B$ with respect to the basis $B$.

More explicitly, if

\[\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_n \mathbf{v}_n \text{ with } c_1, \dots, c_n \in \R,\] then the coordinate vector with respect to $B$ is

\[[\mathbf{v}]_B=\begin{bmatrix}

c_1 \\

c_2 \\

\vdots \\

c_n

\end{bmatrix} \in \R^n.\] Then the map $T: V \to \R^n$ is defined by

\[T(\mathbf{v})=\begin{bmatrix}

c_1 \\

c_2 \\

\vdots \\

c_n

\end{bmatrix}.\]

It follows from the properties of the coordinate vectors that the map $T$ is a linear transformation.

We show that $T$ is bijective, hence an isomorphism.

### $T$ is injective.

To show that $T$ is injective, it suffices to show that the null space of $T$ is trivial: $\calN(T)=\{\mathbf{0}\}$.

(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for a proof of this fact.)

If $\mathbf{v}\in \calN(T)$, then we have

\[\mathbf{0}=T(\mathbf{v})=[\mathbf{v}]_B.\]
So the coordinate vector of $\mathbf{v}$ is zero, hence we have

\[\mathbf{v}=0\mathbf{v}_1+\cdots +0\mathbf{v}_n=\mathbf{0}.\]
Thus, $\calN(T)=\{\mathbf{0}\}$, and $T$ is injective.

### $T$ is surjective.

To show that $T$ is surjective, let

\[\mathbf{a}=\begin{bmatrix}

a_1 \\

a_2 \\

\vdots \\

a_n

\end{bmatrix}\]
be an arbitrary vector in $\R^n$.

Then consider the vector

\[\mathbf{v}:=a_1\mathbf{v}_1+\cdots+ a_n \mathbf{v}_n\]
in $V$.

Then it follows from the definition of the linear transformation $T$ that

\[T(\mathbf{v})=[\mathbf{v}]_B=\begin{bmatrix}

a_1 \\

a_2 \\

\vdots \\

a_n

\end{bmatrix}=\mathbf{a}.\]
Therefore $T$ is surjective.

In summary, $T: V \to \R^n$ is a bijective linear transformation, and hence $T$ is an isomorphism.

Thus, we conclude that $V$ and $\R^n$ are isomorphic.

## Generalization

We may use more general field $\F$ instead of the field of real numbers $\R$.

Then the general statement is as follows.

The proof is identical as above except that we replace $\R$ by $\F$ everywhere.

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