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A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the other vectors. For example, the vectors

\begin{bmatrix}1\\0\\0\end{bmatrix},
\begin{bmatrix}0\\1\\0\end{bmatrix},
\begin{bmatrix}0\\0\\1\end{bmatrix}

are linearly independent. If vectors are linearly independent, they form the basis for a vector space. If the zero vector is in a set of vectors, they cannot be linearly independent, since zero times any vector is the zero vector.

Determining linear independence

Linear independence or lack thereof can be determined by creating a matrix with the columns being the vectors in question and setting it equal to zero. If there is at least one parameter, the vectors are dependent. Because of this, any set with more vectors than there are components to each vector will not be independent. For example,

\begin{align}
&\begin{bmatrix}5\\3\\1\end{bmatrix},
\begin{bmatrix}1\\-1\\-3\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix}\\
&\begin{bmatrix}5&1&1\\3&-1&0\\1&-3&-1\end{bmatrix}
\to\begin{bmatrix}1&0&\frac18\\0&1&\frac38\\0&0&0\end{bmatrix}
\end{align}

Since this matrix will have a non-leading variable, the vectors are not linearly independent.

Another way to test for linear independence if there are n n×1 vectors is to take the determinant of the matrix withe the vectors as columns. If the determinant is not zero, the vectors are independent. Using the same vectors,

\begin{vmatrix}5&1&1\\3&-1&0\\1&-3&-1\end{vmatrix}=0

So these vectors are not linearly independent. On the other hand, the vectors

\begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}3\\0\end{bmatrix}

are linearly independent since

\begin{vmatrix}2&3\\1&0\end{vmatrix}=-3\ne0

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