$G = \{e_1, e_2, e_3\}$ spans $\mathbb{R}^3$, $n = 3$. Let $L = \{v_1, v_2\}$ with $v_1 = (1, 1, 0)$, $v_2 = (0, 1, 1)$.

Step 1 ($m = 1$): We write $v_1 = 1 \cdot e_1 + 1 \cdot e_2 + 0 \cdot e_3$. The coefficient of $e_1$ is $1 \neq 0$, so replace $e_1$:

$\mathbb{R}^3 = \text{span}\{v_1, e_2, e_3\}.$

Step 2 ($m = 2$): We write $v_2 = (0, 1, 1)$ in terms of $\{v_1, e_2, e_3\}$:

$v_2 = 0 \cdot v_1 + 1 \cdot e_2 + 1 \cdot e_3.$

We need a nonzero coefficient on $e_2$ or $e_3$. The coefficient of $e_2$ is $1 \neq 0$, so replace $e_2$:

$\mathbb{R}^3 = \text{span}\{v_1, v_2, e_3\}.$

Result: $H = \{e_3\}$, $|H| = 1 = 3 - 2$.

$G = \{1, x, x^2\}$ spans $P_2(\mathbb{R})$, $n = 3$. Let $L = \{1 + x + x^2\}$, so $m = 1$.

$v_1 = 1 + x + x^2 = 1 \cdot 1 + 1 \cdot x + 1 \cdot x^2$. All coefficients are nonzero. Choose to replace $1$:

$P_2(\mathbb{R}) = \text{span}\{1 + x + x^2, x, x^2\}.$

Indeed: $1 = (1 + x + x^2) - x - x^2$.

$H = \{x, x^2\}$, $|H| = 2 = 3 - 1$.

Try to find 4 independent vectors in $\mathbb{R}^3$. Say $v_1, v_2, v_3, v_4$ are independent. After replacing 3 vectors from $G = \{e_1, e_2, e_3\}$:

$\mathbb{R}^3 = \text{span}\{v_1, v_2, v_3\}.$

Then $v_4 \in \text{span}\{v_1, v_2, v_3\}$, contradicting independence of $\{v_1, v_2, v_3, v_4\}$.

In the proof, when $v_m = \sum a_i v_i + \sum b_j g_j$, we can replace any $g_j$ with $b_j \neq 0$. For $v_1 = (1, 2, 3)$ and $G = \{e_1, e_2, e_3\}$:

$v_1 = 1 \cdot e_1 + 2 \cdot e_2 + 3 \cdot e_3.$

All three coefficients are nonzero, so we could replace any one of $e_1$, $e_2$, or $e_3$. Each choice gives a valid spanning set.

Unlike many existence proofs in mathematics, the Replacement Theorem's proof is fully constructive: it provides an explicit algorithm to find $H$. Process $v_1, v_2, \ldots, v_m$ one at a time, each time replacing one element of $G$ based on a nonzero coefficient. This algorithm is essentially Gaussian elimination.

If $\beta_1$ and $\beta_2$ are two bases for $V$, then $\beta_1$ is independent and $\beta_2$ is spanning, so $|\beta_1| \leq |\beta_2|$ by the Replacement Theorem. By symmetry, $|\beta_2| \leq |\beta_1|$. Hence $|\beta_1| = |\beta_2|$.

This is the most important consequence: the number $\dim V$ does not depend on the choice of basis.

An independent set $L$ with $|L| = \dim V$ is a basis. The Replacement Theorem gives $H \subseteq G$ with $|H| = n - n = 0$, so $\text{span}(L) = V$.

A spanning set $G$ with $|G| = \dim V$ is a basis. If $G$ were dependent, we could remove an element and still span, giving a spanning set of size $n - 1 < n$. But any independent subset (e.g., a basis) has $n$ elements, contradicting $n \leq n - 1$.

If $L = \{v_1, \ldots, v_m\}$ is independent and $m < n = \dim V$, then there exist $g_1, \ldots, g_{n-m} \in G$ such that $\{v_1, \ldots, v_m, g_1, \ldots, g_{n-m}\}$ is a basis for $V$.

The Replacement Theorem can be seen as a statement about the lattice of subspaces: any maximal chain from $\{0\}$ to $V$ (a composition series) has the same length, $\dim V$. This is a special case of the Jordan-Holder theorem in group theory.

In infinite-dimensional spaces, the analogous theorem holds with Zorn's lemma replacing induction. Any linearly independent set can be extended to a (Hamel) basis, and all bases have the same cardinality.

If $W \subseteq V$ is a subspace of a finite-dimensional space $V$, then $\dim W \leq \dim V$. Any basis of $W$ is an independent set in $V$, and by the Replacement Theorem, its size cannot exceed $|G| = \dim V$. Moreover, $\dim W = \dim V$ implies $W = V$ (extend a basis of $W$ -- but no extension is needed since it already has $\dim V$ vectors, so it spans $V$).