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Basis, Dimension, and Orthonormality


Date: 06/28/98 at 09:41:28
From: Naomi
Subject: Linear algebra

Hi! I have two questions:

1) How would you find a basis of the kernel, a basis of the image 
   and determine the dimension of each for this matrix?  The 
   matrix is:  

      1  2  1  2  1
      1  2  2  1  2
      2  4  3  3  3
      0  0  1 -1 -1

2) Are the following 3 vectors linearly dependent?  How can 
   you decide?  

      1     1      1
      1     2      4
      1     3      7
      1     4      10


Date: 06/28/98 at 16:47:22
From: Doctor Anthony
Subject: Re: Linear algebra

Problem 1:

The kernel is the vector subspace that maps into the zero vector. It 
represents the number of dimensions lost in the transformation. So:

   Dimension of range = dimension of domain - dimension of kernel

The rank of the matrix is the number of linearly independent rows and 
columns, and the dimension of the range is the rank of the matrix.

By row operations (which I shall leave for you to do) you partition 
the matrix into the form (known as the canonical form):

   [1   0    0 | a   b]
   |0   1    0 | c   d]
   |0   0    1 | e   f]      
   [-----------|------]
   [0   0    0 | 0   0]     

Assume that this is how it works out, and we can see that in this 
example the number of independent rows (or columns) is 3, and so the 
rank is 3.

The basis of the kernel will be the vectors:

   [ a]         [ b]
   | c|   and   | d]
   | e|         | f|
   |-1|         | 0|
   [ 0]         [-1] 

Problem 2:

If the vectors are linearly dependent, one of the vectors can be 
expressed as a linear combination of the other two. So, can you 
express vector 3 as a linear combination of vectors 1 and 2?
If so, we have:

    1        [1]      [1]
    4    =  p|1|  +  q|2|
    7        |1|      |3|
   10        [1]      [4]

From this we require:

   p +  q = 1
   p + 2q = 4
   p + 3q = 7
   p + 4q = 10

Solving the first two gives: q = 3, p = -2.

Put this into the third equation: -2 + 9 = 7, which is correct,
and into the fourth equation: -2 + 12 = 10, which is correct.

So the third vector is (-2) times the first vector plus 3 times the 
second vector. It follows that the vectors are linearly dependent.

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   


Date: 07/03/98 at 12:37:19
From: Naomi
Subject: Linear algebra

Hi, I have another question. How would you perform the Gram-Schmidt 
process on:

  2   3   5
  0   4   6
  0   0   7


Date: 07/03/98 at 18:57:50
From: Doctor Anthony
Subject: Re: Linear algebra

If M is the space R3 generated by {2,0,0}, {3,4,0}, {5,6,7}, we must 
find an orthonormal basis for M.

The inner product (or dot product) of two vectors x = {x1,x2,x3} 
and y = {y1,y2,y3} is defined to be:

   x.y = x1 * y1 + x2 * y2 + x3 * y3

Note that if two vectors are orthonormal, their dot product is 0. We 
first construct an orthogonal set {w1,w2,w3} and will later reduce 
these to unit vectors {e1,e2,e3}, which will be the orthonormal basis.

Take w1 = {2,0,0}.

Let u2 = {3,4,0} - s{2,0,0}

       = {3-2s, 4, 0}

To find s, we use the fact that we want u2.w1 to be 0. So:

   u2.w1 = 2(3-2s) + 0 + 0 = 0  if s = 3/2

Thus:

   u2 = {0,4,0}

We take w2 = k * u2. We need to choose k so that elements of w2 are 
integers. This is satisfied with k = 1. Then:

   w2 = {0,4,0}

Now let:

   u3 = {5,6,7} - s * w1 - t * w2

   u3 = {5,6,7} - s{2,0,0} - t{0,4,0}

   u3 = {5-2s, 6-4t, 7}

Again we use the fact that we want u3.w1 to be 0 and u3.w2 to be 0 to 
find s and t:

  u3.w1 = 2(5-2s) + 0 + 0 = 0  if  s = 5/2

and

  u3.w2 = 0 + 4(6-4t) + 0 = 0  if  t = 3/2

So u3 = {5,6,7} - (5/2){2,0,0} - (3/2){0,4,0}

      = {0,0,7}

Taking w3 = u3 (because u3 consists of integers), we get:

   w3 = {0,0,7}

So our orthogonal basis {w1,w2,w3} is given by:

   w1 = {2,0,0}

   w2 = {0,4,0}

   w3 = {0,0,7}

And the orthonormal basis is:

   e1 = {1,0,0}

   e2 = {0,1,0}

   e3 = {0,0,1}   

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
College Linear Algebra

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