I read the following in my textbook:

Find as small a set of vectors that span the row space of $A$ as you can. Such a set is called a minimal spanning set.

Is this terminology synonymous with the basis of the vector space? A basis is also made up of the largest set of linearly independent vectors that span a vector space.

  • $\begingroup$ They are the same. Also, as you noted, you could refer to the basis as the maximal linearly independent set. $\endgroup$ – tylerc0816 Sep 26 '13 at 1:35

Yes. The following three terms are equivalent (for a vector space!):

  1. A linearly independent spanning set.
  2. A minimal spanning set.
  3. A maximal linearly independent set.

The first obviously implies the second and third. To see that 2. implies 1., suppose that if $\{x_1,\ldots,x_m\}$ is a minimal spanning set, but not a basis. Then, for some constants $\alpha_1,\ldots,\alpha_m$, not all zero, we have that

$$\displaystyle \alpha_1 x_1+\cdots+\alpha_m x_m=0$$

So, assume that $\alpha_1\ne 0$. Then,


Thus, $\{x_2,\ldots,x_m\}$ is a spanning set (why?) and thus this contradicts minimality.

You and try to prove that 3 implies 1.

  • 1
    $\begingroup$ $\alpha_1 \neq 0$ is a bit of an unnecessary condition, no? Just consider the convention that: $\text{span}() = \{0\}$. $\endgroup$ – St Vincent Sep 26 '13 at 2:21
  • $\begingroup$ @StVincent I don't follow. The zero space has an empty set as a spanning set, but in that case $\alpha_1$ doesn't even make sense! $\endgroup$ – Alex Youcis Sep 26 '13 at 2:23
  • $\begingroup$ If $\alpha_1 = 0$ then surely it can be linearly dependent in a list of length $1$, and as a result, can be thrown out, and still have span the set (even though it is empty) that still spans $0$. $\endgroup$ – St Vincent Sep 26 '13 at 2:26
  • $\begingroup$ Wait, sorry I was confusing $\alpha_1$ with a vector for some reason, very silly. See this post: math.stackexchange.com/questions/492989/linear-dependence-lemma $\endgroup$ – St Vincent Sep 26 '13 at 2:28
  • 1
    $\begingroup$ You should be a bit more careful with your proof of "(2) implies (1)" since a minimal spanning set may not be finite. $\endgroup$ – user21820 Apr 26 '15 at 11:05

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