Let $G = \{e_1, e_2, e_3\}$ (standard basis) and $L = \{(1, 1, 0)\}$. Then $m = 1$, $n = 3$, and we can replace one element of $G$. For instance:

$\text{span}\{(1, 1, 0), e_2, e_3\} = \mathbb{R}^3$

since $(1, 1, 0), e_2, e_3$ are linearly independent. We replaced $e_1$ (we could also have replaced $e_2$, but not $e_3$ since $(1,1,0)$ has $0$ in the third component and we would lose coverage of that direction).

$G = \{(1, 0), (0, 1)\}$ spans $\mathbb{R}^2$ and $L = \{(1, 1), (1, -1)\}$ is linearly independent with $|L| = 2 = |G|$. The theorem says $m \leq n$ ($2 \leq 2$) and we can replace all of $G$ by $L$:

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

The set $H$ is empty.

$G = \{1, x, x^2\}$ spans $P_2(\mathbb{R})$ with $|G| = 3$. The set $L = \{1 + x, 1 - x\}$ is linearly independent ($|L| = 2$). The theorem guarantees we can keep one element of $G$:

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

Indeed: $1 = \tfrac{1}{2}(1+x) + \tfrac{1}{2}(1-x)$ and $x = \tfrac{1}{2}(1+x) - \tfrac{1}{2}(1-x)$.

In $\mathbb{R}^3$, we cannot have 4 linearly independent vectors. If $\{v_1, v_2, v_3, v_4\}$ were independent and $G = \{e_1, e_2, e_3\}$, the theorem gives $4 \leq 3$, a contradiction.

The theorem requires $G$ to span $V$. Consider $V = \mathbb{R}^3$, $G = \{(1, 0, 0), (0, 1, 0)\}$ (not spanning), and $L = \{(0, 0, 1)\}$. Here $L$ is independent and $|L| = 1 \leq 2 = |G|$, but we cannot replace an element of $G$ and still span $V$ (since $G$ did not span $V$ to begin with).

The theorem is also known as the Steinitz Exchange Lemma, after Ernst Steinitz who proved it in 1913. The "exchange" viewpoint: if $v$ is a vector outside $\text{span}(S)$, we can add $v$ to $S$ and remove some element of $S$ while preserving the span.

The subset $H$ in the theorem is generally not unique. In $\mathbb{R}^3$ with $G = \{e_1, e_2, e_3\}$ and $L = \{(1, 1, 1)\}$, we can take $H = \{e_2, e_3\}$ or $H = \{e_1, e_3\}$ or $H = \{e_1, e_2\}$. All three choices work.

In practice, the replacement is done one vector at a time. Given $G = \{g_1, g_2, g_3\}$ spanning $\mathbb{R}^3$ and $v_1 = 2g_1 + 3g_2 + g_3$:

Since the coefficient of $g_1$ is nonzero, we can solve for $g_1$: $g_1 = \tfrac{1}{2}(v_1 - 3g_2 - g_3)$. So $\text{span}\{v_1, g_2, g_3\} = \text{span}\{g_1, g_2, g_3\}$.

A linearly independent set $L$ in a finite-dimensional space $V$ is a basis if and only if $|L| = \dim V$. This follows from the Replacement Theorem: if $|L| = \dim V = n$ and $G$ is any basis, then $H = \varnothing$, so $\text{span}(L) = V$.

A spanning set $G$ for $V$ is a basis if and only if $|G| = \dim V$. If $|G| = n = \dim V$ and $G$ were dependent, removing a vector gives a spanning set of size $n-1$. But any basis has $n$ elements, contradicting that it can be extended from a subset of this smaller spanning set.

By the Replacement Theorem, the subspaces of $\mathbb{R}^4$ have dimensions $0, 1, 2, 3$, or $4$:

• $\dim = 0$: only $\{0\}$
• $\dim = 1$: lines through the origin
• $\dim = 2$: planes through the origin
• $\dim = 3$: hyperplanes $\{x \mid a \cdot x = 0\}$
• $\dim = 4$: $\mathbb{R}^4$ itself

In infinite-dimensional spaces, the Replacement Theorem still holds for finite independent sets, but the full theory of dimension requires more care (Zorn's lemma for the existence of bases).