Dimension

Theorem 1. The number of elements in any basis of a finite-dimensional vector space $𝒱$ is the same as in any other basis.

Proof. The proof of this theorem is a slight refinement of the method used in Section: Linear combinations , and, incidentally, it proves something more than the theorem states. Let $𝒳 = {x_{1}, \dots, x_{n}}$ and $𝒴 = {y_{1}, \dots, y_{m}}$ be two finite sets of vectors, each with one of the two defining properties of a basis; i.e., we assume that every vector in $𝒱$ is a linear combination of the $x$ ’s (but not that the $x$ ’s are linearly independent), and we assume that the $y$ ’s are linearly independent (but not that every vector is a linear combination of them). We may apply the theorem of Section: Linear combinations , just as above, to the set $𝒮$ of vectors $y_{m}, x_{1}, \dots, x_{n}$ Again we know that every vector is a linear combination of vectors of $𝒮$ and that $𝒮$ is linearly dependent. Reasoning just as before, we obtain a set \mathcal{S}^{\prime} of vectors $y_{m}, x_{1}, \dots, x_{i - 1}, x_{i + 1}, \dots, x_{n}$ again with the property that every vector is a linear combination of vectors of \mathcal{S}^{\prime} . Now we write $y_{m - 1}$ in front of the vectors of \mathcal{S}^{\prime} and apply the same argument. Continuing in this way, we see that the $x$ ’s will not be exhausted before the $y$ ’s, since otherwise the remaining $y$ ’s would have to be linear combinations of the ones already incorporated into $𝒮$ , whereas we know that the $y$ ’s are linearly independent. In other words, after the argument has been applied $m$ times, we obtain a set with the same property the $x$ ’s had, and this set differs from the set of $x$ ’s in that $m$ of them are replaced by $y$ ’s. This seemingly innocent statement is what we are after; it implies that $n \geq m$ . Consequently if both $𝒳$ and $𝒴$ are bases (so that they each have both properties), then $n \geq m$ and $m \geq n$ . ◻

Definition 1. The dimension of a finite-dimensional vector space $𝒱$ is the number of elements in a basis of $𝒱$ .

Observe that since the empty set of vectors is a basis of the trivial space $𝒪$ , the definition implies that that space has dimension $0$ . At the same time the definition (together with the fact that we have already exhibited, in Section: Bases , one particular basis of $ℂ^{n}$ ) at last justifies our terminology and enables us to announce the pleasant result: $n$ -dimensional coordinate space is $n$ -dimensional. (Since the argument is the same for $ℝ^{n}$ and for $ℂ^{n}$ , the assertion is true in both the real case and the complex case.)

Our next result is a corollary of Theorem 1 (via the theorem of Section: Bases ).

Theorem 2. Every set of $n + 1$ vectors in an $n$ -dimensional vector space $𝒱$ is linearly dependent. A set of $n$ vectors in $𝒱$ is a basis if and only if it is linearly independent, or, alternatively, if and only if every vector in $𝒱$ is a linear combination of elements of the set.

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