Difference between revisions of "Dependence MA265F11Walther" - Rhea

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 Linear Dependence

Definition

The vectors v₁,v₂,...,v_k in a vector space V are said to be linearly dependent if there exist constans a₁,a₂,...,a_k not all zero, such that

(1)

Otherwise, v₁,v₂,...,v_k are called linearly independent. That is, v1,v2,...,vk are linearly independent if, whenever a1,v1+a2v2+...+akvk=0,

                                     a₁= a₂=...=a_k=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) $

This there is a nontrivial solution

$ \left(\begin{array}{cccc}k\\-2k\\k\end{array}\right) $       k is not equal to 0

so the vectors are linearly dependent.

Example 2

Are the vectors

$ v1=\left(\begin{array}{cccc}1&0&1&2\end{array}\right) $

$ v2=\left(\begin{array}{cccc}0&1&1&2\end{array}\right) $

and

$ v3=\left(\begin{array}{cccc}1&1&1&3\end{array}\right) $     in R₄

linearly dependent or linearly independent?