Column Space or Range of A

A basis for a space is a set of vectors that completely determine the set. That is to say, that each vector in the set may be written uniquely as a linear combination of the vectors in the basis. For example, the set { [1,0,0], [0,1,0], [0,0,1]} is a basis for R^3. The dimension of a space is the number of elements in a basis. The dimension of R^3 is 3 (naturally!). This is written dim(R^3)=3.

Since Gauss-Jordan elimination works on rows and we are interested in R(A), linear combinations of columns, we should look at Transpose[A]. The rows of Transpose[A] are the columns of A. If Gauss-Jordan elimination eliminates a row of Transpose[A], that means that the corresponding column of A was "extraneous" as far as linear combinations are concerned. Therefore, a basis for R(A) is found by the following procedure:

1. Let M= RowReduce[Transpose[A]]. (If M=0, then R(A)={0}.)
2. If M is not 0, then look at the submatrix of M consisting of nonzero rows.
3. The nonzero columns of the transpose of this submatrix form the standard basis for R(A).

The nonzero columns of Transpose[GJSF of Transpose[A]] form the standard basis for Range of A. Remember, the Range of Column Space is the space of all linear combinations of the columns of A. Finding the transpose of A, doing row elimination, and then transposing back will do the trick to eliminate the unnecessary columns. Let's try some.

In[32]:=

  a={{1,1,1,7},{1,-1,3,11},{2,-1,1,4}};
  MatrixForm[a]

Out[32]=

  1    1    1    7
  
  1    -1   3    11
  
  2    -1   1    4

In[33]:=

  m=RowReduce[Transpose[a]];
  MatrixForm[m]

Out[33]=

Answer:

Answer: ...

In[34]:=

  c={{1,0,-2,1},{2,1,4,-3},{5,2,6,-5}};
  MatrixForm[c]

Out[34]=

  1    0    -2   1
  
  2    1    4    -3
  
  5    2    6    -5

In[35]:=

  m=RowReduce[Transpose[c]];
  MatrixForm[m]

Out[35]=

Answer:

Answer: ...

Another example: Let L:R^4 to R^3 be defined by
L(Transpose[a1 a2 a3 a4])=Transpose[a1+a2 a3+a4 a1+a3]. Therefore the matrix of L is:

In[36]:=

  l={{1,1,0,0},{0,0,1,1},{1,0,1,0}};
  MatrixForm[l]

Out[36]=

  1   1   0   0
  
  0   0   1   1
  
  1   0   1   0

Don't believe me? Check it out!

In[37]:=

  l.{a1,a2,a3,a4}

Out[37]=

  {a1 + a2, a3 + a4, a1 + a3}

In[38]:=

  MatrixForm[Transpose[RowReduce[Transpose[l]]]]

Out[38]=

  1   0   0   0
  
  0   1   0   0
  
  0   0   1   0

So, a basis for Range of L is Transpose[1 0 0], Transpose[0 1 0], Transpose[0 0 1], that is, the range of L is all of R^3.

Let's try another one.

In[39]:=

  Clear[a]
  a={{1,2,1,2},{2,1,2,1},{3,2,3,2},{3,3,3,3},{5,3,5,3}};
  MatrixForm[a]

Out[39]=

  1   2   1   2
  
  2   1   2   1
  
  3   2   3   2
  
  3   3   3   3
  
  5   3   5   3

In[40]:=

  MatrixForm[Transpose[RowReduce[Transpose[a]]]]

Out[40]=

  
  
  1   0   0   0
  
  
  
  0   1   0   0
  
  1   4
  -   -
  3   3   0   0
  
  
  
  1   1   0   0
  
  1   7
  -   -
  3   3   0   0

Up to Bases for These Subspaces