• If I have 3 vectors $\vec{v_1}, \vec{v_2}$ and $\vec{v_3},$ can they be linearly independent in $\mathbb{R^4}$?
  • If I have 5 vectors $\vec{v_1}, \vec{v_2},\vec{v_3},\vec{v_4}$ and $\vec{v_5},$ can they be linearly independent in $\mathbb{R^4}$?

What is generally said about the linear dependency of $k$ number of vectors in $n$ number of dimensions?

I formulated the two statements below by myself. Is the reasoning in them correct?

  1. I know that for the span of the vectors, it follows that if $n<k,$ then the vectors $\vec{v_1},...,\vec{v_k}$ can't span $\mathbb{R}^n$. For example 2 vectors can't span a space $\mathbb{R}^3$.
  2. If we have 4 vectors in $\mathbb{R^3}$ then that vector matrix would still have rank 3 and can still span $\mathbb{R^3}$.
  • 1
    $\begingroup$ In general, if you have any collection of more than $n$ vectors in $\Bbb R^n$ then these vectors will be linearly dependent. However, if you have a collection of at most $n$ vectors in $\Bbb R^n$, then they could either be linearly dependent or independent, but some further analysis is required. $\endgroup$ – Dave Oct 12 '17 at 21:02
  • Three vectors can be linearly independent in $\Bbb R^4$. Example: $v_1=(1, 0, 0, 0), v_2=(0, 1, 0, 0),$ and $v_3=(0, 0, 1, 0)$ are linearly independent vectors in $\Bbb R^4.$

  • You cannot have $5$ linearly independent vectors in $\Bbb R^4$. This is because $4$ vectors are necessary to span $\Bbb R^4$, and any additional vector will be a linear combination of those.

Statement one is correct if you switch $n$ and $k$ so that $k<n$. The example you give has $n=3$ and $k=2$. Aside from this, it is correct.

Statement two is correct.

  • $\begingroup$ Thanks for the answer. How can one see that, the vectors you've proposed, are linearly independent in without having to actually do the arithmetic and showing that the system of equations only has trivial solutions? $\endgroup$ – Parseval Oct 12 '17 at 20:59
  • $\begingroup$ @Parseval You can show pretty easily that every vector $av_1+bv_2+cv_3$ for $a,b,c\in\Bbb R$ is unique to show their linear independence. $\endgroup$ – Austin Weaver Oct 12 '17 at 21:02
  • $\begingroup$ By "show that every vector is unique", you mean solving $av_1+bv_2+cv_3=0$ and showing that the last row (after gaussian elimination) reads $0=0$, which means that the equation can only be solved by $a=b=c=0$? $\endgroup$ – Parseval Oct 12 '17 at 21:09
  • $\begingroup$ Yes. I was trying to show more intuitively why that works. To formally show that, you have to solve the system. $\endgroup$ – Austin Weaver Oct 12 '17 at 21:10
  • $\begingroup$ So how do you show it intuitively so easily? $\endgroup$ – Parseval Oct 12 '17 at 21:11

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