# 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!

• 1) What about $x=v$? 2) If you work in $\Bbb R^3$ (or any 3dim vector space it holds true, otherwise it is false. Outside of $span{u,v,w}$ $A$ can do anything and need not be invertible... Oct 1 '20 at 9:14

## 2 Answers

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

• Isn't it safe to assume that a square matrix $A$ is assumed in part 2? Otherwise the question of invertibility is void. Oct 1 '20 at 9:35

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.