📂Banach Space

Proof That Every Finite-Dimensional Normed Space Has a Basis

Theorem¹

Every finite-dimensional normed space has a basis.

Explanation

It may seem unfamiliar to establish the mere existence of a basis rather than a basis satisfying some particular condition, but in fact, the definition of a basis never stated that every vector space has a basis. Depending on how one defines finite dimensionality, it is also a fact so obvious that no separate proof is needed.

Proof

Strategy: Use finite dimensionality to construct a basis concretely.

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Since $(X, | \cdot | )$ is finite-dimensional, there exists $\left\{x_{1} , \dots , x_{n} \right\}$ satisfying $\text{span} \left\{ x_{1} , \dots , x_{n} \right\} = X$. Let $y_{1} : = x_{1}$. If $x_{2} \in \text{span} \left\{ y_{1} \right\}$, consider $x_{3}$. If $x_{2} \notin \text{span} \left\{ y_{1} \right\}$, set $y_{2} := x_{2}$. If $x_{3} \in \text{span} \left\{ y_{1}, y_{2} \right\}$, consider $x_{4}$. If $x_{3} \notin \text{span} \left\{ y_{1}, y_{2} \right\}$, set $y_{3} := x_{3}$. Defining $M = \left\{ y_{1} , \dots , y_{k} \right\}$ in this way, for $1 \le j \le k$

$$ y_{j} \notin \text{span} \left\{ y_{1} , \dots , y_{j-1} \right\} $$

Suppose $M$ is not linearly independent; then for some $\lambda_{j} \ne 0$

$$ \lambda_{1} y_{1} + \dots + \lambda_{k} y_{k} = 0 $$

Letting $j_{0}$ be the largest such $j$,

$$ y_{j_{0}} = - {{1} \over { \lambda_{j_{0}} }} \sum_{j < j_{0}} \lambda_{j} y_{j} - {{1} \over { \lambda_{j_{0}} }} \sum_{j > j_{0}} \lambda_{j} y_{j} = - {{1} \over { \lambda_{j_{0}} }} \sum_{j < j_{0}} \lambda_{j} y_{j} $$

Therefore $\displaystyle y_{j_{0}} = - {{1} \over { \lambda_{j_{0}} }} \sum_{j < j_{0}} \lambda_{j} y_{j} \in \text{span} \left\{ y_{1} , \dots , y_{j_{0}-1} \right\}$, which is a contradiction. Since $M \subset \left\{ x_{1} , \dots , x_{n} \right\}$ is linearly independent and satisfies $\text{span} M = X$, $M$ is a basis of $X$.

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Looking closely at the proof, one can see that $\left\{ x_{1} , \dots , x_{n} \right\}$ spans $X$ quite generously. Here, everything that obstructed linear independence was discarded, and only $M$ was taken, showing concretely that it is a basis.