Isomorphic Finite Dimensional Normed Linear Spaces

# Isomorphic Finite Dimensional Normed Linear Spaces

Recall from the Equivalence of Norms in a Finite-Dimensional Linear Space page that if $X$ is a finite-dimensional normed linear space then any two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on $X$ are equivalent.

We will now prove an important result which states that any two finite-dimensional normed linear spaces of the same dimension are isomorphic to one another.

Theorem 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be finite-dimensional normed linear spaces with $\mathrm{dim} (X) = \mathrm{dim} (Y)$. Then $X$ and $Y$ are isomorphic. |

**Proof:**Let $X$ be a finite-dimensional normed linear space with $\mathrm{dim} (X) = n$. We will show that $X$ is isomorphic to $\mathbb{R}^n$. Since $X$ is finite-dimensional with $\mathrm{dim} (X) = n$ there exists a basis $\{ e_1, e_2, ..., e_n \} \subseteq X$ of $X$. Define a function $T : X \to \mathbb{C}^n$ for each element $x \in X$ with $\displaystyle{x = \sum_{k=1}^{n} a_ke_k}$ by:

\begin{align} \quad T(x) = (a_1, a_2, ..., a_n) \end{align}

- We show that $T : X \to \mathbb{R}^n$ is a linear map. Let $x, y \in X$ with $\displaystyle{x = \sum_{k=1}^{n} a_ke_k}$ and $\displaystyle{y = \sum_{k=1}^{n} b_ke_k}$ and let $\lambda \in \mathbb{R}$. Then:

\begin{align} \quad T(x + y) = (a_1 + b_1, a_2 + b_2, ..., a_n + b_n) = (a_1, a_2, ..., a_n) + (b_1, b_2, ..., b_n) = T(x) + T(y) \end{align}

(3)
\begin{align} \quad T(\lambda x) = (\lambda a_1, \lambda a_2, ..., \lambda a_n) = \lambda (a_1, a_2, ..., a_n) = \lambda T(x) \end{align}

- So indeed $T$ is a linear map. We now show that $T$ is bijective. Suppose that $T(x) = T(y)$. Then $(a_1, a_2, ..., a_n) = (b_1, b_2, ..., b_n)$ which implies that $x = y$. So $T$ is injective. And for every $(x_1, x_2, ..., x_n) \in \mathbb{r}^n$ by setting $z = x_1e_1 + x_2e_2 + ... x_ne_n \in X$ we have that $T(z) = (x_1, x_2, ..., x_n)$. So $T$ is surjective.

- Now observe that for all $x \in X$ that:

\begin{align} \quad \| T(x) \|_Y = \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} = \| x \|_{*} \end{align}

- Therefore:

\begin{align} \quad 1 \| x \|_* = \| T(x) \|_Y = 1 \| x \|_* \end{align}

- But $\| x \|_*$ is equivalent to the norm on $X$. So there exists $C, D \in \mathbb{R}$, $C, D > 0$ such that $C \| x \| \leq \| x \|_* \leq D \| x \|$ for all $x \in X$. Hence:

\begin{align} \quad C \| x \|_X \leq \| T(x) \|_Y \leq D \| x \|_X \quad \forall x \in X \end{align}

- So $X$ and $\mathbb{R}^n$ are isomorphic. Since $\dim (Y) = n$ as well, $Y$ and $\mathbb{R}^n$ are isomorphic. Hence $X$ and $Y$ are isomorphic. $\blacksquare$