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Alright, so I've learned the three basic criteria that make a set of vectors linearly dependent. I'll go ahead and list them.

  1. More vectors than entries per vector, e.g.:

    $$\pmatrix{1\\2},\pmatrix{3\\4},\pmatrix{5\\6}$$

  2. All vectors are scalar multiples of each other, e.g.:

    $$\pmatrix{1\\-2\\-5},\pmatrix{-2\\4\\10},\pmatrix{3\\-6\\-15}$$

  3. Presence of a zero vector $\bar{0}$, e.g.:

    $$\pmatrix{1\\2},\pmatrix{0\\0}$$

Clearly determining linear dependence is not too tall a task considering we only have three tests to apply to our set.

However, rather than perform row operations and convert a corresponding matrix to reduced echelon form to determine linear independence, could we say that a set of vectors that fails to pass these three tests is necessarily linearly independent? If not, could someone provide a counterexample set that is dependent but meets none of the above criteria?

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    $\begingroup$ $(1,0,0),(0,1,0),(1,1,0)$. $\endgroup$
    – lulu
    Sep 19, 2016 at 0:30

2 Answers 2

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The criteria you mention are only special cases. The set $\left\{\pmatrix{1 \\ 0 \\ 0}, \pmatrix{0 \\ 1 \\ 1}, \pmatrix{1 \\ 1 \\ 1}\right\}$ is linearly dependent but doesn't satisfy any of your criteria.

The actual definition of linear independence is the following:

A set of vectors $\{\mathbf v_1, \dots, \mathbf v_n\}$ is linearly independent if $$a_1\mathbf v_1 + a_2\mathbf v_2 + \cdots + a_n\mathbf v_n = \mathbf 0 \implies a_1 = a_2 = \cdots = a_n = 0$$ where $a_1, \dots, a_n$ are scalars (regular numbers).
A set of vectors is linearly dependent if it is not linearly independent.

A necessary and sufficient criterion for linear independence is that none of the vectors are a linear combination of any (finite) subset of the others. The set I give above fails this test because the last vector is a linear combination of the first two.

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  • $\begingroup$ Ah, thank you. So do any other criteria exist that help easily determine the linear dependence of a set (like this one) that I haven't already listed? $\endgroup$
    – Lanier Freeman
    Sep 19, 2016 at 0:33
  • $\begingroup$ See my revised answer. $\endgroup$
    – user137731
    Sep 19, 2016 at 0:36
  • $\begingroup$ Ah, so if vector $x_3$ can be written as the vector addition of $x_1$ and $x_2$ then the set is dependent? $\endgroup$
    – Lanier Freeman
    Sep 19, 2016 at 0:37
  • $\begingroup$ Well a linear combination isn't just a sum -- it's a weighted sum. I could equally well replace the last vector in my set with $\pmatrix{3 \\ 2 \\ 2}$ and it'd still be linearly dependent because $$\pmatrix{3 \\ 2 \\ 2} = 3\pmatrix{1 \\ 0 \\ 0} + 2\pmatrix{0 \\ 1 \\ 1}$$ $\endgroup$
    – user137731
    Sep 19, 2016 at 0:38
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    $\begingroup$ Yes. If you can see that one of your vectors is a linear combination of the others then you'll immediately know that the set is linearly dependent. In practice it's not always easy to tell though so you go to Gauss-Jordan or the determinant or the exterior product or any of the other techniques to figure it out. $\endgroup$
    – user137731
    Sep 19, 2016 at 0:43
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A set of vectors vectors $\{a_1, \ldots, a_k\}$ is linearly independent if it is not linearly dependent, i.e.,
$$ \beta_1 a_1 + \ldots + \beta_k a_k = 0 $$ holds only when $\beta_1 = \cdots = \beta_k = 0$.

The criteria you specified are only sufficient.

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  • $\begingroup$ So how might I go about generating a more comprehensive approach to linear dependence? $\endgroup$
    – Lanier Freeman
    Sep 19, 2016 at 0:35
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    $\begingroup$ One (numerical) way to show a set of vectors is linearly independent or dependent is to use Gram-Schmidt. If the algorithm terminates with a zero-vector, then the set is dependent, otherwise it is independent. $\endgroup$
    – ahmedb
    Sep 19, 2016 at 0:38

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