3.2: Linear Independence
- Page ID
- 96151
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A set of vectors, \(\{u_1,\: u_2,\cdots , u_n\}\), are said to be linearly independent if for any scalars \(c_1,\: c_2,\cdots , c_n\), the equation
\[c_1\text{u}_1+c_2\text{u}_2+\cdots +c_n\text{u}_n=0\nonumber \]
has only the solution \(c_1 = c_2 =\cdots = c_n = 0\). What this means is that one is unable to write any of the vectors \(\text{u}_1,\: \text{u}_2,\cdots , \text{u}_n\) as a linear combination of any of the other vectors. For instance, if there was a solution to the above equation with \(c_1\neq 0\), then we could solve that equation for \(\text{u}_1\) in terms of the other vectors with nonzero coefficients.
As an example consider whether the following three three-by-one column vectors are linearly independent:
\[\text{u}=\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad\text{v}=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad\text{w}=\left(\begin{array}{c}2\\3\\0\end{array}\right).\nonumber \]
Indeed, they are not linearly independent, that is, they are linearly dependent, because \(\text{w}\) can be written in terms of \(\text{u}\) and \(\text{v}\). In fact, \(\text{w} = 2\text{u} + 3\text{v}\). Now consider the three three-by-one column vectors given by
\[\text{u}=\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad\text{v}=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad\text{w}=\left(\begin{array}{c}0\\0\\1\end{array}\right).\nonumber \]
These three vectors are linearly independent because you cannot write any one of these vectors as a linear combination of the other two. If we go back to our definition of linear independence, we can see that the equation
\[a\text{u}+b\text{v}+c\text{w}=\left(\begin{array}{c}a\\b\\c\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)\nonumber \]
has as its only solution \(a=b=c=0\).