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<br> | <br> | ||
| − | '''Solution''' | + | '''Solution''' |
| − | <math>a1\left(\begin{array}{cccc}3\\2\\1\end{array}\right)</math><math>+a2\left(\begin{array}{cccc}1\\2\\0\end{array}\right)</math><math>+a3\left(\begin{array}{cccc}-1\\2\\-1\end{array}\right)</math><math>=\left(\begin{array}{cccc}0\\0\\0\end{array}\right)</math> | + | Forming Equation (1) |
| + | |||
| + | <math>a1\left(\begin{array}{cccc}3\\2\\1\end{array}\right)</math><math>+a2\left(\begin{array}{cccc}1\\2\\0\end{array}\right)</math><math>+a3\left(\begin{array}{cccc}-1\\2\\-1\end{array}\right)</math><math>=\left(\begin{array}{cccc}0\\0\\0\end{array}\right)</math> | ||
| + | |||
| + | we obtain the homogeneous system | ||
| + | |||
| + | <span class="texhtml">3''a''<sub>1</sub> + ''a''<sub>2</sub> − ''a''<sub>3</sub> = 0</span> | ||
| + | |||
| + | <span class="texhtml">2''a''<sub>1</sub> + 2''a''<sub>2</sub> + 2''a''<sub>3</sub> = 0</span> | ||
| + | |||
| + | <span class="texhtml">''a''<sub>1</sub> − ''a''<sub>3</sub> = 0</span> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | The corresponding augmented matrix is | ||
| + | |||
| + | <math>\left(\begin{array}{cccc}3&1&-1&|0\\2&2&2&|0\\1&0&-1&|0\end{array}\right)</math> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | Whose reduced row echelon form is | ||
| + | |||
| + | <math>\left(\begin{array}{cccc}1&0&-1&|0\\0&1&2&|0\\0&0&0&|0\end{array}\right)</math> | ||
Revision as of 15:41, 7 December 2011
Linear Dependence
Definition
The vectors v1,v2,...,vk in a vector space V are said to be linearly dependent if there exist constans a1,a2,...,ak not all zero, such that
(1)
Otherwise, v1,v2,...,vk are called linearly independent. That is, v1,v2,...,vk are linearly independent if, whenever a1,v1+a2v2+...+akvk=0,
a1= a2=...=ak=0
If S=[v1,v2,...,vk], then we also say that the set S is linearly dependent or linearly independent if the vectors have the corresponding property.
It should be emphasized that for any vectors v1,v2,...,vk, Equation (1) always holds if we choose all the scalars a1,a2,...,ak, equal to zero. The important point in this definition is whether it is possible to satisfy (1) with at least one of the scalars different from zero.
Remark: Definition is started for a finite set of vectors, but it also applies to an infinite set S of a vector space, using corresponding notation for infinite sums.
To determine whether a set of vectors is linearly independent or linearly dependent, we use Equation (1). Regardless of the form of the vectors, Equation (q) yields a homogeneous linear system of equations. it is always consistent, since a1=a2=...=ak=0 is a solution. However, the main idea from the Definition is whether there is a nontrivial solution.
Example 1
Determine whether the vectors
$ v1=\left(\begin{array}{cccc}3\\2\\1\end{array}\right) $
$ v2=\left(\begin{array}{cccc}1\\2\\0\end{array}\right) $
$ v3=\left(\begin{array}{cccc}-1\\-2\\-1\end{array}\right) $
are linearly independent.
Solution
Forming Equation (1)
$ a1\left(\begin{array}{cccc}3\\2\\1\end{array}\right) $$ +a2\left(\begin{array}{cccc}1\\2\\0\end{array}\right) $$ +a3\left(\begin{array}{cccc}-1\\2\\-1\end{array}\right) $$ =\left(\begin{array}{cccc}0\\0\\0\end{array}\right) $
we obtain the homogeneous system
3a1 + a2 − a3 = 0
2a1 + 2a2 + 2a3 = 0
a1 − a3 = 0
The corresponding augmented matrix is
$ \left(\begin{array}{cccc}3&1&-1&|0\\2&2&2&|0\\1&0&-1&|0\end{array}\right) $
Whose reduced row echelon form is
$ \left(\begin{array}{cccc}1&0&-1&|0\\0&1&2&|0\\0&0&0&|0\end{array}\right) $
