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posted by  knowledge-seeker on 3/20/2008 2:25:01 AM  |  status: Live  

LINEAR ALGEBRA

Course Textbook Chapter Problem
Linear Algebra Linear Algebra and Its Applications 3rd by Lay, Stade 4.3 - 4.4 N/A
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posted by jks777 on 3/20/2008 2:45:02 AM  |  status: Live
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knowledge-seeker's comment:
"thank you so much, that was very helpful and clear, but the only column that doesn't have a 0 on diag is c2 and i don't see where your final nulA entries came from. Could you clarify please? Thank you so much."
Response Details:
To find the null space of this matrix A, first it is transformed into a row echelon form matrix.
 
  1   1    2    -1
  2   3    6    -2
-2   1     2     2
  0  -2   -4    0
 
The first step is to multiply row 1 by -2 and add to row 2.  Then multiply row 1 by 2 and add to row 3.
 
  1   1    2    -1
  0   1    2     0
  0   3    6     0
  0  -2   -4    0
 
Next is to multiply row 2 by -3 and add to row 3.  Then multiply row 2 by 2 and add to row 4.
 
  1   1    2    -1
  0   1    2     0
  0   0    0     0
  0   0    0     0
 
Now we look at the diagonal and the number of components that do not have a 0 (which are two of them.  That means the columns that do not have a 0 on its diagonal will be the set of vectors spanning the nullspace of A.
 
null(A) = {(1, 2, -2, 0), (1, 3, 1, -2)}
 
Oracle
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posted by jks777 on 3/20/2008 3:09:20 AM  |  status: Live
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knowledge-seeker's comment:
"awesome now i got it"
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From the row echelon form of matrix A,
 
  1   1    2    -1
  0   1    2     0
  0   0    0     0
  0   0    0     0
 
The diagonal elements are shown in red.  From this diagonal, the first and second columns have a 1 and the third and fourth columns have a 0.  This means that when we determine the basis for this matrix, we will need to use the first and second columns of vectors from the original matrix.  These first and second column vectors from the original matrix will make the basis for the nullspace of A.
 
    Vector 1  Vector 2
            ↓       ↓ 
A =
 
Since the vectors are now found, these two vectors become the nullspace of A.
 
null(A) = {(1, 2, -2, 0), (1, 3, 1, -2)}
 
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posted by knowledge-seeker on 3/20/2008 3:09:59 AM  |  status: Live
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GOT IT!  Thanks so much, sorry, brain is getting worn down because its so late.  I really appreciate your help.  :)
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posted by Xico on 3/20/2008 8:27:40 AM  |  status: Live
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knowledge-seeker's comment:
"Thank you very much, don't worry your English was just fine. I'm gonna sit down and see if I understand everything here and get back to you if I have a question. I really appreciate your time :)"
Response Details:
you have the null space wrong.
 
after the Gauss-Jordan elemination method, you have:
 
 
 
the columns with pivots ( elements in the diagonal ) form not the null space of  A but the base of the subspace...
 
to know the null space you have to solve the rest of equation:
 
so by the matrix you have  that x3 and x4 are free variables, so write x1 and x2 in function of x3 and x4 wiil give you a basis for the null space, so:
 
and
 
<=> x1 = -x2 - 2x3 + x4  and x2 = - 2x3
<=> x1 = -(-2x3) - 2x3 + x4 and x2 = - 2x3
so x1 = x4  and  x2 = - 2x3
 
so the vectors that form the null space is:
 
{ ( 1, 0, 0, 1 ) , ( 0, -2, 1, 0 ) } 
Sorry about my english, but if you understand, please rate...
And if you have any doubt about the resolution don't hesitate to send me a message..
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posted by knowledge-seeker on 5/12/2008 9:19:24 PM  |  status: Live
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Response Details:
Any chance you might be able to help me with the following?  No one seems to be responding:

http://answerboard.cramster.com/advanced-math-topic-5-260269-0.aspx
 
 
Thank you very much.  :)
 
p.s. -- how have you been?
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