Linear independence

A set vectors, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}_1, \vec{v}_2, ... \vec{v}_n} 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. In other words, the equation:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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} }

only holds when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_1, a_2,...a_n} all equal 0. If this equation is true for any other values of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_1, a_2,...a_n} , 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.

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:

${\vec {a}}_{1}{\begin{pmatrix}1\\2\\3\end{pmatrix}},{\vec {a}}_{2}{\begin{pmatrix}3\\-1\\2\end{pmatrix}},{\vec {a}}_{3}{\begin{pmatrix}5\\0\\8\end{pmatrix}}$

These are not linearly independent as the third vector can be written as a linear combination of the other two:

${\vec {a}}_{3}=2{\vec {a}}_{1}+{\vec {a}}_{2}$

Another method is to combine the vectors into a square matrix and calculates its determinant. 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.

${\begin{pmatrix}1&3&5\\2&-1&0\\3&2&8\end{pmatrix}}$

The order of the vectors in the matrix does not matter. The determinant can be found as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{vmatrix} 1 & 3 & 5 \\ 2 & -1 & 0 \\ 3 & 2 & 8 \end{vmatrix} = -21 }

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.

See also

• Vector
• Linear algebra