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What methods are most commonly used to determine whether a set of vectors is linearly dependent?

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Construct a matrix with one column corresponding to one vector, then compute the rank of the matrix. – chaohuang Sep 27 '12 at 8:57
@chaohuang How can the rank of a matrix be used to determine whether the matrix is linearly dependent? – Anderson Green Dec 4 '12 at 5:53
Do you know the definition of the rank? – chaohuang Dec 4 '12 at 16:20
I do now. A simple definition is given here: – Anderson Green Dec 4 '12 at 21:16
Then the rank tells you the number of linear independent vectors – chaohuang Dec 4 '12 at 21:55
up vote 3 down vote accepted

It's equivalent to show $Ax=0$ with $x\neq 0$ is solvable, where $A$ is the matrix of your vectors, each column corresponding to a vector. This can be shown by many methods, one being Gaussian elimination

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You want solvable, not unsolvable. The OP is asking when the vectors are dependent. Also $x\neq 0$. – Alex Becker Sep 26 '12 at 18:18
@Alex Becker: thanks for the correction – Alex R. Sep 26 '12 at 18:19
Also note my edited comment. – Alex Becker Sep 26 '12 at 18:19

If you have $\{v_1,...,v_n\}\in\mathbb{R}^n$, i.e $n$ vectors in $\mathbb{R}^n$ or generally in some field $\mathbb{F}^n$ then the determinant of the matrix $A=(v_1,...,v_n)$ is a very good and easy way to find out. If $det(A)=0$ then $A$ is singular which is equivalent to the colums of $A$ being linearly dependent.

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The OP didn't say the number of vectors was $n$. – Stefan Smith Sep 28 '12 at 0:08

Just by using the definition of linearly depended and solving the corresponding system of linear equations(with any method you want, usually by Gaussian elimination).

For example to deside if the vectors $v_1=(1,2,3,4),v_2=(2,1,3,4),v_3=(1,2,4,8),v_4=(0,3,4,8)$ are linearly depended you will do:

If $r_1,r_2,r_3,r_4 \in \mathbb{F}$($\mathbb{F}$ being your field) such that $r_1v_1+r_2v_2+r_3v_3+r_4v_4=(0,0,0,0)$ then this is equivalent to $(r_1+2r_2+r_3,2r_1+r_2+2r_3+3r_4,3r_1+3r_2+4r_3+4r_4,4r_1+4r_2+8r_3+8r_4)=(0,0,0,0)$ witch is equivalent to the system of linear equations

\begin{split} r_1&+2r_2+r_3\:\:\:\:\:\;\;\;\;\;\;\;=0\\ 2r_1&+\;\;r_2+2r_3+3r_4=0\\ 3r_1&+3r_2+4r_3+4r_4=0\\ 4r_1&+4r_2+8r_3+8r_4=0 \ . \end{split}

If you solve this system (say, by Gaussian elimination) you will find that it has non zero solution (for example $r_1=1,r_2=-1,r_3=1,r_4=-1$) so


and your vectors are linearly depented.

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One might be able to exhibit an explicit linear combination yielding $0$.

Or one might be able to show that the vectors are all in a vector space of dimension smaller than the given family of vectors.

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