Linear independence
A set vectors, <math>\vec{v}_1, \vec{v}_2, ... \vec{v}_n</math> are said to be linearly independent if a linear combination of the vectors is zero if and only if all the coefficients of the vectors are zero.[1] In other words, the equation:
- <math>
a_1\vec{v}_1 + a_2\vec{v}_2 + ... + a_n\vec{v}_n = \sum^n_{i=1} a_i\vec{v}_i = \vec{0} </math>
only holds when <math>a_1, a_2,...a_n</math> all equal 0. If this equation is true for any other values of <math>a_1, a_2,...a_n</math>, then the vectors are said to be "linearly dependent". If a set of n vectors, each with dimension n, are linearly independent, then they form a basis for that space.[2]
Testing for Linear Independence
The simplest way to test for linear independence is to try to express one vector as a linear combination of the others. Consider the vectors:
- <math>
\vec{a}_1 \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \vec{a}_2 \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}, \vec{a}_3 \begin{pmatrix} 5 \\ 0 \\ 8 \end{pmatrix} </math>
These are not linearly independent as the third vector can be written as a linear combination of the other two:
- <math>
\vec{a}_3 = 2\vec{a}_1 + \vec{a}_2 </math>
Another method is to combine the vectors into a square matrix and calculates its determinant.[3] If the vectors are linearly dependent, the determinant will be zero. The vectors above can be combined into a matrix as the columns of the matrix.
- <math>
\begin{pmatrix} 1 & 3 & 5 \\ 2 & -1 & 0 \\ 3 & 2 & 8 \end{pmatrix} </math>
The order of the vectors in the matrix does not matter. The determinant can be found as:
- <math>
\begin{vmatrix} 1 & 3 & 5 \\ 2 & -1 & 0 \\ 3 & 2 & 8 \end{vmatrix} = -21 </math>
As the determinant is not zero, the vectors are not linearly dependent. This matrix method only works when the dimension of the vectors and the number of vectors are the same.