# Understanding the Replacement Theorem (Exchange Theorem)

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

• $S$ is not necessarily a subset of $B$. – Augustin May 26 '15 at 19:13
• @Augustin But is $S$ also a basis? – nTuply May 26 '15 at 19:14
• 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. – Augustin May 26 '15 at 19:16

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.
• Now it makes more sense. But what about $S$ and $Z$, how do these sets look like and how are they formed? – nTuply May 26 '15 at 19:53
• 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. – muaddib May 26 '15 at 19:56