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Basis for a Vector Space in R^3


Date: 11/25/98 at 13:49:55
From: Noura
Subject: Basis for a vector space.

Hi Dr. Math:

I have several questions about bases.

First, which of the following are bases for R^3?
 
   a) (2,-3,1), (4,1,1), (0,-7,1) 
   b) (1,6,4), (2,4,-1), (-1,2,5)

I know that both of them are not bases because it is written in my 
book. The second one is because it is linearly dependent, and so it is 
not a basis. I don't know why the first one is not a basis. 

Second, I need to determine the dimension of and a basis for the 
solution space of the system:

    X1 - 3X2 + X3  = 0
   2X1 - 6X2 + 2X3 = 0
   3X1 - 9X2 + 3X3 = 0

I know the dimension is 2 and the basis is: {(3,1,0), (-1,0,1)} How do 
you find it?

Third, I need to find a standard basis vector that can be added to the 
set {v1,v2} to produce a basis for R^3, where:

   v1 = (1,-1,0), v2 = (3,1,-2)

Thanks!


Date: 11/25/98 at 14:56:21
From: Doctor Anthony
Subject: Re: Basis for a vector space.

For the first question, part (a), the vectors are not bases if you can 
express one of them in terms of the other two, i.e. they are linearly 
dependent. Note:

   p[ 2] + q[4] + r[ 0] = 0
    [-3]    [1]    [-7]
    [ 1]    [1]    [ 1]

This gives 3 homogeneous equations in 3 unknowns, p, q and r, and if 
they are not consistent, the vectors are independent. If they are 
consistent one vector is a linear combination of the other two and so 
the vectors could not form a basis for R^3.
  
The condition for valid solutions of p, q, r other than the trivial 
ones p = q = r = 0 is that the determinant shown below equals zero.

       |2   4    0|
   M = |-3  1   -7| = 0  (if true then vectors are linearly dependent)
       |1   1    1|  

Since, Det(M) = 2 - 28 + 0 - 0 + 14 + 12 = 0, the vectors are linearly 
dependent and DO NOT form a basis for R^3.

For part (b), by the same reasoning as before, we must check the value 
of the determinant shown below:

       |1   6   4|   
   M = |2   4  -1| = 20 + 6 + 16 + 16 + 2 - 60 = 0 
       |-1  2   5|

again the vectors are linearly dependent and DO NOT form a basis for 
R^3.

For your second question, notice the 3 equations are all the same 
equation. If these are coordinates then X1 - 2X2 + X3 = 0 is the 
equation of a plane and so the dimension is 2. Any two lines chosen at 
random in this plane can be used as base vectors.

For example X1 = 2, X2 = 1 and X3 = 1 will satisfy the equation, so one 
base vector is (2, 1, 1). A second is (1, 1, 2), but there are an 
infinity of pairs (as long as they are not parallel) that can be used 
as base vectors.

Finally, for the third problem, any vector not in the plane of v1 and 
v2 could form the third vector of the basis. For example the vector 
perpendicular to both these is:

   |i    j    k|   
   |1   -1    0| = i(2) - j(-2) + k(4) =  i + j + 2k 
   |3    1   -2|   

So the third vector could be (1, 1, 2).

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Linear Algebra
High School Linear Algebra

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