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== Is &nbsp; &nbsp;&nbsp;<math>S=\left[\begin{array}{cccc}1&2&3\end{array}\right]</math>&nbsp;<math>,\left[\begin{array}{cccc}0&1&2\end{array}\right]</math>&nbsp;<math>,\left[\begin{array}{cccc}3&0&-1\end{array}\right]</math> ==
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== Is &nbsp; &nbsp;&nbsp;<math>S=\left[\begin{array}{cccc}1&2&3\end{array}\right]</math>&nbsp;<math>,\left[\begin{array}{cccc}0&1&2\end{array}\right]</math>&nbsp;<math>,\left[\begin{array}{cccc}3&0&-1\end{array}\right]</math> ==
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a linearly independent set of vectors in&nbsp;<span class="texhtml">''R''</span><sup><span class="texhtml">3</span></sup><span class="texhtml">?</span>
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&lt;span class="texhtml" /&gt;
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== Solution  ==
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We form the matrix&nbsp;<span class="texhtml">''A''</span>&nbsp;whose rows are the vectors in&nbsp;<span class="texhtml">''S''</span>:
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<br>
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<math>A=\left[\begin{array}{cccc}1&2&3\\0&1&2\\3&0&-1\end{array}\right]</math>
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<br>
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Since det(<span class="texhtml">''A''</span>) = 2, we conclude that&nbsp;<span class="texhtml">''S''</span>&nbsp;is linearly independent.
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Revision as of 20:51, 7 December 2011


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 a1v1+ 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 (1) 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 + a2a3 = 0

2a1 + 2a2 + 2a3 = 0

a1a3 = 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,...,vn}

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 In<span class="texhtml" />. Thus, A is row equivalent to In, and hence det(A) is not equal to 0. Conversely, if det(A) is not equal to 0, then A is row equivalent to In. 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 In. Hence, rows of <span class="texhtml" />A are linearly independent.



Example 3


Is     $ S=\left[\begin{array}{cccc}1&2&3\end{array}\right] $ $ ,\left[\begin{array}{cccc}0&1&2\end{array}\right] $ $ ,\left[\begin{array}{cccc}3&0&-1\end{array}\right] $

a linearly independent set of vectors in R3?

<span class="texhtml" />

Solution

We form the matrix A whose rows are the vectors in S:


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


Since det(A) = 2, we conclude that S is linearly independent.


Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett