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 $\R^n$.

 
Add to solve later

Sponsored Links

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.

Click here if solved 99
Add to solve later

Sponsored Links