Relationship of span with linear independence I am given such a question, with 2 parts.
Assume that we have a set of vectors $\{u, v, w\}$ that is linearly independent and is a subset of $R^n$ space.
Part 1:
Now I am given another vector $x$ such a condition that $span\{u, v, w\} \ne span\{u, v, x\}$. Would $\{u, v, w, x\}$ be linearly independent?
My idea is that it is. Imagining a 3D space in my head, if the spans are not equal, that means that $w$ and $x$ are heading in different directions. Hence if put together, the set would still be linearly independent.
Is my line of thought correct?
Part 2:
Suppose now that $\{$A$u, $A$v, $A$w\}$ is also linearly independent, must A be invertible?
My instinctive answer is no, it need not be. I imagine the multiplication of A to the three vectors to be a uniform transformation applied to all 3, hence no matter what A is, the set will still remain linearly independent. Yet I am highly sceptical, and I know I'm not really competent in this topic.
Can anyone point out the flaws in my thinking, and perhaps come up with a more rigorous explanation?
Thank you!
 A: Part 1:  The answer is not necessarily for $n>3$ dimensions.  For example, $x$ could be a combination of $u,v$ so that $\mathbb{span} \{u,v,x\} = \mathbb{span}\{u,v\} \neq \mathbb{span} \{u,v,w\}$.  In three dimensions, $x$ must depend on $u,v$ because $\mathbb{span}\{u,v,w\} = \mathbb R^3$.
Part 2:  If $\mathbf A : \mathbb{R} ^n \to \mathbb{R}^n$ then the answer is no if $n>3$ dimensions and yes if $n=3$.  For $n>3$ $\mathbf A$ can map independent vectors other than $u,v,w$ to zero, so cannot be invertible.  But when $n=3$ the $\mathbb {span} \{u,v,w\} = \mathbb R^3$, so that $\mathbb A$ has no null space and is therefore invertible.
A: Part 1: yes, you are correct. If $span\{u,v,w\}\neq span\{u,v,x\}$, as $u,v\in span\{u,v,w\}$, then the conclusion is that $x\notin span\{u,v,w\}$ and that means that if the family $\{u,v,w\}$ is linearly indepentent, then the family $\{u,v,w,x\}$ is linearly independent.
Part 2: it doesn't have to be independent. $A$ can even be a non-square matrix, sending $(0,0,1), (0,1,0), (1,0,0)$ to independent vectors in $\mathbb{R}^4$, with the obvious implication that $A$ is not invertible
