I know minimal spanning set is independent .
But how do i prove that a spanning set which is linearly independent i.e a basis is minimal spanning set ?
I tried to prove by contradiction by assuming that a set which is smaller and different than the basis set is a minimal set . But this way prove seems to be very lengthy.
is it possible that a smaller set containing different independent vectors (other than those is S) is minimal spanning vector ?
[Math] basis is minimal spanning set
linear algebravector-spaces
Related Solutions
Yes. The following three terms are equivalent (for a vector space!):
- A linearly independent spanning set.
- A minimal spanning set.
- 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,
$$x_1=\frac{-\alpha_2}{\alpha_1}x_2+\cdots+\frac{-\alpha_m}{\alpha_1}x_m$$
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.
It doesn't work. See Keith Conrad's note (namely page 16). Here is a relevant screenshot.
Best Answer
Let $S$ be your basis. If it was not a minimal spanning set, there would be a $s\in S$ such that $\bigl\langle S\setminus\{s\}\bigr\rangle$ is the whole space. Then, $s$ can be written as a linear combination $\alpha_1s_1+\cdots+\alpha_ns_n$ of elements of $S\setminus\{s\}$. That is, we have$$1.s-\alpha_1s_1-\cdots-\alpha_ns_n=0,$$ which is impossible, since $S$ is linearly independent.