# Any minimal spanning set must be the maximal independent set?

By definition:

1. A minimal spanning set is a spanning set such that no proper subset thereof is spanning.
2. A maximal independent set is an independent set such that no set that contains it properly is independent.

For example in $\mathbb{R}^2$, $$\{ \begin{bmatrix}1 \\0 \end{bmatrix}, \begin{bmatrix}0 \\1 \end{bmatrix} \}$$

is both 1. and 2.

Of course, $$\{ \begin{bmatrix}1 \\1 \end{bmatrix}, \begin{bmatrix}1 \\-1 \end{bmatrix} \}$$
is also both 1. and 2.

My question is that any minimal spanning set must at the same time be a maximal independent set? If not, any example?

Let $S$ be a minimal spanning set. Then $S$ is linearly independent, otherwise one of its elements, say $v$, would be a linear combination of the elements in $S\setminus\{v\}$; but this would imply $S\setminus\{v\}$ is still a spanning set, contradicting minimality.
If $w\notin S$, then $w$ is a linear combination of the elements of $S$, so $S\cup\{w\}$ is not linearly independent. Thus $S$ is a maximal linearly independent set.