17Nov2010.  Lina Class-D  [This note uses mixed Maple/Math notation.]

Our goal is to compute an 
	Orthogonal-basis
for the (lefthand action) nullspace of matrix G.
  

> G := <<1 | 2 | 1 | 0 | 1>,<3 | 6 | 0 | -3 | 0>>;
    Computing   R := RREF(G)  uses two row-subtractions, and a row-scaling:

          [1    2    1     0    1]        [1    2    0    -1    0]
      G = [3    6    0    -3    0],   R = [0    0    1     1    1]

    Back-substitution gives a basis {V1,V2,V3} for Null(R), ie. for Null(G).

                            V1 := [-2, 1,  0, 0, 0]^t
                            V2 := [ 1, 0, -1, 1, 0]^t
                            V3 := [ 0, 0, -1, 0, 1]^t

[Aside:  V1 is already orthogonal to V3, so we could leave them alone and
just let B2 be the part of V2 orthogonal to Span(V1,V3).]

    But let's not get that clever, and just apply vanilla Gram-Schmidt.
    Since V1 isn't the zero-vector, we set B1 := V1 or rather B1 := -V1, 
since we want the leftmost entry to be positive.

> B1  :=  -V1  =  [2, -1, 0, 0, 0]^t

    Now project V2 on the line-through-B1:
> ProjOnB1ofV2  :=  (IP(B1,V2) / IP(B1,B1)) * B1
                 =  [4/5, -2/5, 0, 0, 0]^t

    Hence this subtraction gives Orth_B1(V2):
> scaledB2  :=  V2 - ProjOnB1ofV2  =  [1/5, 2/5, -1, 1, 0]^t

    Let's choose a B2 with integer-entries:
> B2  :=  5 * scaledB2  =  [1, 2, -5, 5, 0]^t

    We now want the part of V3 that is orthogonal to Span(B1,B2), so we 
seek ProjOnB1ofV3 and ProjOnB2ofV3.  We observe that V3 is ortho to B1, 
since
		B1 =  [2, -1,  0, 0, 0]^t  and
		V3 =  [0,  0, -1, 0, 1]^t .
Thus  ProjOnB1ofV3 =  [0,  0,  0, 0, 0]^t .

    Next
> ProjOnB2ofV3 := (IP(B2,V3)/IP(B2,B2)) * B2  = [1/11, 2/11, -5/11, 5/11, 0]^t

  Subtracting off the projections yields a vector ortho to Span(B1,B2):

> scaledB3 := V3 - ProjOnB1ofV3 - ProjOnB2ofV3 = [-1/11, -2/11, -6/11, -5/11, 1]^t

    Integer-normalizing the vector gives 
> B3  :=  -11 * scaledB3  =  [1, 2, 6, 5, -11]^t

    Let's check that {B1, B2, B3} forms an ortho-set:
>  IP(B1,B2) = 0, IP(B1,B3) = 0, IP(B2,B3) = 0.

    Let's verify that Span({V1, V2, V3}) includes Span({B1, B2, B3}).

> VVecsThenBVecs := Tpose(<V1,V2,V3,B1,B2,B3>);

                               [-2     1     0     2     1      1]
                               [ 1     0     0    -1     2      2]
              VVecsThenBVecs = [ 0    -1    -1     0    -5      6]
                               [ 0     1     0     0     5      5]
                               [ 0     0     1     0     0    -11]

> RREF(VVecsThenBVecs) ;
                        [1    0    0    -1    2      2]
                        [0    1    0     0    5      5]
                        [0    0    1     0    0    -11]
                        [0    0    0     0    0      0]
                        [0    0    0     0    0      0]

So all the pivots are in the {V1, V2, V3} part.  

    Now check the reverse inclusion:
> BVecsThenVVecs := Tpose(<B1,B2,B3,V1,V2,V3>);
                               [ 2     1      1    -2     1     0]
                               [-1     2      2     1     0     0]
              BVecsThenVVecs = [ 0    -5      6     0    -1    -1]
                               [ 0     5      5     0     1     0]
                               [ 0     0    -11     0     0     1]
    Drum roll, please:
> RREF(BVecsThenVVecs);

                      [1    0    0    -1    2/5     0  ]
                      [0    1    0     0    1/5    1/11]
                      [0    0    1     0     0    -1/11]
                      [0    0    0     0     0      0  ]
                      [0    0    0     0     0      0  ]

Now all the pivots are in the {B1, B2, B3} part.  Yea!

So our orthogonal-basis for LeftNull(G) is

    B1  =  [2, -1,  0, 0,   0]^t
    B2  =  [1,  2, -5, 5,   0]^t
    B3  =  [1,  2,  6, 5, -11]^t

			Ain't Science Wonderful!
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