1

Assume that, $\vec{v} \in \mathbb R^n$. Is the set $(\vec{v}, 2\vec{v} , − 5\vec{v} )$ linearly independent or dependent?

How can we deduce from this independence or dependence? I know the theorem which states that if a set has more vectors than there are entries in each vector, then the set is linearly dependent. Seem though that here we cannot use that nor the other definition where the only solution would be the trivial one?

user
  • 154,566
  • All three are in a one-dimensional subspace, so any two of them are linearly dependent. – Martin R Nov 28 '19 at 21:29
  • 1
    @Daniel: Sorry, may I ask what book in linear algebra you are reading? Hardly can I understand your confusion if I don't know what definitions you have learned. The three vectors $v,2v,-5v$ are all multiples of the vector $v$, which should immediately tells you that the set ${v,2v,-5v}$ is linearly dependent. But, of course, if one is taught in a different way, such observation might not be straightforward. –  Nov 29 '19 at 00:01

2 Answers2

1

By definition they are dependent indeed

$$(1)\cdot (\vec{v})+ (2)\cdot (2\vec{v})+(1)\cdot (− 5\vec{v}) =\vec 0$$

and more in general

$$\{c_1 \vec v, c_2 \vec v,\ldots,c_k \vec v\}$$

are always linearly dependent since

$$c_2\cdot c_1 \vec v-c_1\cdot c_2 \vec v +0 \cdot c_3\vec v+\ldots+0\cdot c_k \vec v=\vec 0$$

J. W. Tanner
  • 60,406
user
  • 154,566
1

You said

If a set has more vectors than there are entries in each vector, then the set is linearly dependent.

This is perhaps better stated as

A set of $n+1$ vectors in an $n$-dimensional vector space is linearly dependent.

Then you can conclude that any two of the three vectors $(\vec{v}, 2\vec{v} , − 5\vec{v} )$ are linearly dependent, because they lie all in the one-dimensional subspace generated by $\vec v$.

Martin R
  • 113,040
  • I can't believe that this is not covered by some more genral answer already given here on MSE. According to some new guide lines, now we need also to close already "covered" questions. – user Nov 28 '19 at 21:46
  • Possible duplicate. – user Nov 28 '19 at 21:47
  • Another one. – user Nov 28 '19 at 21:48
  • @user: If you think that the question has been asked and answered before then you are welcome vote to close as a duplicate (instead of posting an answer). – Martin R Nov 28 '19 at 21:49
  • Why do you suppose I think so? According to your curriculum, I expect you should vote to close question like this as a duplicate instead giving an answer. – user Nov 28 '19 at 21:53