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I'm learning about Basis and Spans and now that's I've figured out what these are, I'm trying to understand the Replacement Theorem(also called the Exchange Theorem).

The definition goes like this:

Let $V$ be a vector space and let $B = \{v_{1}, v_{2}, ..., v_{n}\}$ be a basis for V with n elements. Choose and integer $m\leq n$ and let $S = \{w_1, w_2, ..., w_m\}$ be a finite set of linearly independent vectors. Then, there is a set $Z\subset B$ containing exactly $n-m$ elements such that $span({Z\cup S}) = V$

Now this is what I understand from the above definition. $V$ is a vector space and $B$ being the basis of $V$ is the set of linearly independent vectors such that for any vector, for example, $u_1 \in V$, we have: $u_1 = \sum u_1v_i$ where $i$ is the number of elements in $B$. If, now, I make a set $S$ with $n-m$ or $n=m$ elements then I'll have the remaining vectors $w_i$ in my set $Z$. These vectors are also linearly independent.

Finally, set $span(\{Z \cup S\}) = V$. Now that's my problem:

Why it is not $span(\{Z\cup S\}) = B$ since $S$ is a set of linearly independent vectors(which is a basis) and if I'm making a set $Z$ with $n-m$ elements(which is from $B$ as well) ?

It would really help if an example is given in the explanation, let's say we have a vector space $V = \{u_1, u_2, u_3, u_4\}$.

Thanks

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  • $\begingroup$ $S$ is not necessarily a subset of $B$. $\endgroup$ – Augustin May 26 '15 at 19:13
  • $\begingroup$ @Augustin But is $S$ also a basis? $\endgroup$ – nTuply May 26 '15 at 19:14
  • $\begingroup$ No, because there may not be enough vectors in $S$ for it to be a basis. The only thing you know about $S$ is that its vectors are linearly independant. Then the theorem states that you can pick vectors in $B$ (which is a basis) and add them to $S$ in order to form a (new) basis. $\endgroup$ – Augustin May 26 '15 at 19:16
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Perhaps I've misunderstood the issue but there seems to be some confusion about the difference between a basis, a vector space, and the span operation. A basis is a set of $n$ vectors. The vector space equals the set of all linear combinations of elements of a basis. The span of a set of vectors is the linear combination of all the elements in that set.
$$span(B) = \{\sum_i c_i v_i\}$$ where $c_i$ are arbitrary scalars.

You would not say that the span of a set of vectors is equal to an $n$ point set as in $span(Z \cup S) = B$. Instead you might ask if it equals the entire vector space $V$ or a subspace.

So for example. The vector space $\mathbb{R}^2$ has a basis $v_1, v_2$ which could be the $(1, 0)$ and $(0, 1)$ vectors. The elements of that space are vectors $(c_1, c_2)$ where $c_1, c_2$ are real numbers. The basis is a two element set $\{v_1, v_2\}$. The span of, for example, the one element set $\{v_1\}$ is all the vectors of the form $(c_1, 0)$ where $c_1$ is a real number. That is an infinite set.

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  • $\begingroup$ Yes, my misunderstanding is with the Basis and Vector space. Could you please elaborate your answer with a simple example which will aid my understanding? Thanks :) $\endgroup$ – nTuply May 26 '15 at 19:30
  • $\begingroup$ Sure, I updated the answer with examples. $\endgroup$ – muaddib May 26 '15 at 19:38
  • $\begingroup$ Now it makes more sense. But what about $S$ and $Z$, how do these sets look like and how are they formed? $\endgroup$ – nTuply May 26 '15 at 19:53
  • $\begingroup$ As you wrote above, they are both sets with a finite number of vectors. $S$ has $m$ elements and $Z$ has $n-m$ elements. The theorem does not offer a construction of $Z$ (tells you how it's formed). It simply tells you one such set exists with $n-m$ elements. $\endgroup$ – muaddib May 26 '15 at 19:56
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    $\begingroup$ That's correct, neither of them are a basis, they are just sets of vectors. However, their union is a basis. $\endgroup$ – muaddib May 26 '15 at 20:01

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