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 \(\mathcal{X}=\{x_{1}, \ldots, x_{n}\}\) and \(\mathcal{Y}=\{y_{1}, \ldots, y_{m}\}\) 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 \(\mathcal{V}\) 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 \(\mathcal{S}\) of vectors \[y_{m}, x_{1}, \ldots, x_{n}\] Again we know that every vector is a linear combination of vectors of \(\mathcal{S}\) and that \(\mathcal{S}\) is linearly dependent. Reasoning just as before, we obtain a set \(\mathcal{S}^{\prime}\) of vectors \[y_{m}, x_{1}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\] 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 \(\mathcal{S}\) , 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 \(\mathcal{X}\) and \(\mathcal{Y}\) are bases (so that they each have both properties), then \(n \geq m\) and \(m \geq n\) . ◻