linear algebraterminology
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.
It doesn't work. See Keith Conrad's note (namely page 16). Here is a relevant screenshot.
Yes, the proof is fine. You need not believe that you can find such linearly independent vectors. At each step, there either still exists another vector in $\mathcal B$ that is independent of what you already have (and then you can proceed), or there does not exist such a vector (and the process terminates). If you already know that a linearly independent set cannot have more than $\dim V$ elements, the process is doomed to terminate with some number $n\le \dim V$ vectors picked that way. By revisiting the termination condition, those $n$ vectors span $V$, hence must be a basis.
Best Answer
Yes. The following three terms are equivalent (for a vector space!):
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,
$$x_1=\frac{-\alpha_2}{\alpha_1}x_2+\cdots+\frac{-\alpha_m}{\alpha_1}x_m$$
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.