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Linear independence (view source)
Revision as of 15:56, September 11, 2017
, 15:56, September 11, 2017Created the page with an explanation of linear independence and some methods of testing it
A set [[vector]]s, <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.<ref>[http://mathworld.wolfram.com/LinearlyIndependent.html Linear Independence on mathworld.wolfram.com]</ref> 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.<ref>[http://mathonline.wikidot.com/sufficient-conditions-for-a-set-of-vectors-to-be-a-basis mathonline.wikidot.com]</ref>
==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]].<ref>[https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/lindep/lindep.html Testing for Linear dependence]</ref> 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.
==References==
{{reflist}}
==See also==
*[[Vector]]
*[[Linear algebra]]
[[Category:Mathematics]]
[[Category:Linear algebra]]
:<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.<ref>[http://mathonline.wikidot.com/sufficient-conditions-for-a-set-of-vectors-to-be-a-basis mathonline.wikidot.com]</ref>
==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]].<ref>[https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/lindep/lindep.html Testing for Linear dependence]</ref> 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.
==References==
{{reflist}}
==See also==
*[[Vector]]
*[[Linear algebra]]
[[Category:Mathematics]]
[[Category:Linear algebra]]