Difference between revisions of "Dependence MA265F11Walther" - Rhea

Revision as of 19:58, 7 December 2011

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,...,v_k are linearly independent if, whenever a₁v₁+ a₂v₂+...+a_kv_k=0

                                     a₁= a₂=...=a_k=0

If S = {v₁,v₂,...,v_k} ,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,...,v_k, Equation (1) always holds if we choose all the scalars a₁,a₂,...,a_k, 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 (1) yields a homogeneous linear system of equations. it is always consistent, since a₁=a₂=...=a_k=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?

Solution

We form Equation (1),

a₁v₁+a₂v₂+a₃v₃=0

and solve for a₁,a₂,a₃.

The resulting homogeneous system is

a₁ + a₃ = 0

a₂ + a₃ = 0

a₁ + a₂ + a₃ = 0

2a₁ + 2a₂ + 3a₃ = 0

The corresponding augmented matrix is

$ \left[\begin{array}{cccc}1&0&1&|0\\0&1&1&|0\\1&1&1&|0\\2&2&3&|0\end{array}\right] $

and its reduced row echelon form is

$ \left[\begin{array}{cccc}1&0&0&|0\\0&1&0&|0\\0&0&1&|0\\0&0&0&|0\end{array}\right] $

Thus the only solution is the trivial solution

a₁ = a₂ = a₃ = 0

so the vectors are linearly independent.

Let {S = v₁,v₂,...,v_n}

be a set of n vectors in Rⁿ (<span class="texhtml" />Rn). Let A be the matrix whose columns (rows) are the elements of S. Then S is linearly independent if and only if det(A) is not equal to 0.

Proof

We shall prove the result for columns only; the proof for rows is analogous.

Suppose that S is linearly independent. Then it follows that the reduced row echelon form of A is I_n<span class="texhtml" />. Thus, A is row equivalent to I_n, and hence det(A) is not equal to 0. Conversely, if det(A) is not equal to 0, then A is row equivalent to I_n. Now assume that the rows of A are linearly dependent. Then it follows that the reduced row echelon form of A has a zero row, which contradicts the earlier conclusion that A is row equivalent to I_n. Hence, rows of <span class="texhtml" />A are linearly independent.

Example 3

Is