Subspaces of R^3...

Date: 06/22/98 at 19:34:54
From: arnetta piper
Subject: Subspaces of R^3...

Prove that {[ a ]  | a + b + c + 0 } is a subspace of R^3.
            [ b ]
            [ c ]

Find a basis.  Determine the dimension of the subspace.

Date: 06/23/98 at 22:53:49
From: Doctor Mateo
Subject: Re: Subspaces of R^3...

Hello Arnetta,

You have some very interesting questions here.  I am going to answer 
them on the assumption that a + b + c + 0 is really a + b + c = 0.  
If this is not the case, please send the question back.

In the first part of your question we are being asked to prove that 
    _   _
   |  a  ||
  {|  b  || a + b + c = 0}  (let's call this W) is a subspace of R^3.  
   |_ c _|| 

 We know that W is not an empty subset of R^3 since 
       _   _ 
      |  0  |  
  0 = |  0  | belongs to W since 0 + 0 + 0 = 0.
      |_ 0 _|
 
****NOTE****  Let us agree that s_1 means s(*subscript*)1 and that 
              the underscore(_) means subscript.
              _      _                          _     _
             |   s_1  |                        |  t_1  |
 Now if  s = |   s_2  | belongs to W and  t  = |  t_2  | belongs to W, 
             |_  s_3 _|                        |_ t_3 _| 

     then it follows that s_1 + s_2 + s_3 = 0 , t_1 + t_2 + t_3 = 0
                     _             _
                    |   s_1 + t_1   |
     and   s + t =  |   s_2 + t_2   | belongs to W 
                    |_  s_3 + t_3  _|


     since (s_1 + t_1) + (s_2 + t_2) + (s_3 + t_3) =
           (s_1 + s_2 + s_3) + (t_1 + t_2 + t_3) = 0 + 0 = 0.
                                            _         _
                                           |   k*s_1   |
 Moreover, if k is any scalar, then  k*s = |   k*s_2   | belongs to W
                                           |_  k*s_3  _|

     since k*s_1 + k*s_2 + k*s_3 = k(s_1 + s_2 + s_3) = k(0) = 0.

What can you conclude since we have shown that W is closed under 
addition and scalar multiplication? Is W a subspace of R^3?

Now let us find a basis for W and the dimension of W.
                _   _
               |  a  ||
  Since W =   {|  b  || a + b + c = 0}  
               |_ c _|| 

  If we let a = p and  b = q, then we have c = -p-q.
  Each vector v in W can then be written as  
         _       _      _    _      _    _
        |   p     |    |   1  |    |   0  |
    W = |   q     | = p|   0  | + q|   1  |
        |_ -p-q  _|    |_ -1 _|    |_ -1 _| 

                             _     _               _     _ 
                            |   1   |             |   0   |
  Therefore if we let v_1 = |   0   |   and v_2 = |   1   |
                            |_ -1  _|             |_ -1  _|   

  then W = p*v_1 + q*v_2 so that W = Span{v_1, v_2}.
  
  Thus, v_1 and v_2 are not scalar multiples of one another and 
  therefore independent.
  
Now that we have shown the linear independence what can you deduce 
about the basis of W and the corresponding dimension of W ?

Hope that this helps.   

-Doctor Mateo,  The Math Forum