In my Linear Algebra script I have this passage here:
For $b_1,\ldots,b_n\in\mathbb{R}^n$ with $A = (b_1,\ldots,b_n)\in\mathbb{R}^{n\times n}$ the following statements are equivalent:
- $b_1,\ldots,b_n$ is a basis of $\mathbb{R}^n$
- $b_1,\ldots,b_n$ is a generator (German: Erzeugendensystem) of $\mathbb{R}^n$
- $b_1,\ldots,b_n$ are linear independent
- $A\in \text{GL}_n(\mathbb{R})$ is invertible
But if a basis is previously defined as $b_1,\ldots,b_n$ being (1) a generator and (2) linear independent, how is one of these conditions 1 or 2 sufficient to tell the other? Isn't $$\text{span}\{b_1,\ldots,b_n,b_{n+1}\} = \mathbb{R}^n \quad \text{with}\quad b_{n+1} = \lambda\cdot b_n,\, \lambda\in\mathbb{R}$$ also a generator of $\mathbb{R}^n$ without $b_1,\ldots,b_n,b_{n+1}$ being linear independent? From this equivalence here $b_1,\ldots,b_n,b_{n+1}$ are a generator of $\mathbb{R}^n$, therefore a basis of $\mathbb{R}^n$ and also linear independent – which they aren't.