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.