🔗 Steinitz Exchange Lemma

Let $\\{v_1, \\ldots, v_m\\}$ span $V$ and let $\\{w_1, \\ldots, w_n\\}$ be linearly independent in $V$. Then $n \\leq m$, and after reordering, $\\{w_1, \\ldots, w_n, v_{n+1}, \\ldots, v_m\\}$ spans $V$.

Proof: We proceed by induction on $n$.

**Base case ($n = 1$):** Since $w_1 \\neq \\mathbf{0}$ and the $v_i$ span $V$, we can write $w_1 = \\sum a_i v_i$ with some $a_j \\neq 0$. After reordering, say $a_1 \\neq 0$. Then $v_1 \\in \\text{span}(w_1, v_2, \\ldots, v_m)$, so $\\{w_1, v_2, \\ldots, v_m\\}$ ...

From: Advanced Linear Algebra

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