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Given that $\{v_1,v_2\}$, $\{v_2,v_3\}$ are linearly independent in $\mathbb{R}^n$, for $n\geq3$, where $v_1$, $v_2$ and $v_3$ are distinct, is $\{v_1,v_2,v_3\}$ linearly independent?

My thought so far is that if $v_1=(1,1,0)$, $v_2=(1,0,0)$ and $v_3=(0,1,0)$, $\{v_1,v_2\}$, $\{v_2,v_3\}$ are linearly independent but $\{v_1,v_2,v_3\}$ isn't linearly independent.

But I fear that $v_1$, $v_2$ and $v_3$ aren't distinct enough. Is my example enough to answer the question?

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  • $\begingroup$ What do you mean by "distinct enough"? They are distinct, and that's enough for this question. $\endgroup$ – Colescu May 21 '17 at 11:53
  • $\begingroup$ I'm just overthinking the question, Thank you for the assurance $\endgroup$ – TheMaui999 May 21 '17 at 11:56
  • $\begingroup$ That's a perfect counterexample. There's no such thing as "distinct enough". They are distinct. $\endgroup$ – Tristan Batchler May 21 '17 at 12:23
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Let $v_3:=v_1\ldots$ $\mbox{ }$

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No. For example you can take $v_1=<1,0,0>, \quad v_2 =<0,1,0> \quad$ and $v_3=<2,0,0>$.

These vectors are distinct, but $2 v_1=v_3$ and therefore $\{ v_1,v_2,v_3\}$ is not linearly independent.

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Let $v_1=i+j$, $v_2=i-j$ and $v_3=-2i$.

Vectors $v_1,v_2$ and $v_3$ are clearly distinct.

The sets of vectors $\{v_1,v_2\}$ and $\{v_2,v_3\}$ are linearly independent. But, $v_3=-v_1-v_2$, $v_2=v_1+v_3$ and $v_1=v_2-v_3$.

A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the others;

if no vector in the set can be written in this way, then the vectors are said to be linearly independent.


Hence the set of vectors $\{v_1,v_2,v_3\}$ are linearly dependent.

So, you got a counter example.

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