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I've read on the internet this definition of Steinitz exchange lemma:

Steinitz exchange lemma

Let $S = \{v_1, \ldots , v_m\}$ satisfy $\mathrm{Span}(S) = V$ and let $T = \{w_1, \ldots , w_k\}$ be linearly independent.

Then

  1. $k \le m$ and

  2. after possibly reordering the set $S$, we have

    $\mathrm{Span}(\{w_1, \ldots , w_k, v_{k+1}, \ldots , v_m\}) = V$

However,my teacher gave us a slightly different definition: she said that $\{w_1,\ldots,w_k,v_{k+1},\ldots,v_m\}$ doesn't only span $V$, but is also a basis. Is that system really linearly independent or did she make a mistake?

egreg
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Andrei Florea
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1 Answers1

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It is basically true, if $S$ is a basis:

Suppose $(v_1,\dots, v_i, \dots,v_n)$ is basis of $V$ and $w = \lambda_1 v_1 + \dots + \lambda_n v_n$, where some $\lambda_i \neq 0$. Then $(v_1,\dots, v_{i-1}, w,\dots, v_{i+1}, \dots, v_n)$ is a basis of $V$.

Of course you can use this fact again and again to "replace" multiple vectors, like so:

Suppose the set $B$ is a basis of $V$ and $A$ is a linearly independent subset of $V$. Then $|A|\leq |B|$ and there is a subset $B'$ of $B$, s.t. $B'\cup A$ is basis of $V$.

(I avoided listing the elements of $A$ and $B$, so that I not have to talk about "reordering" elements)

Stefan Perko
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