Defining Row Space and Column Space
For any m × n matrix A:
- The Row Space is the span (set of all linear combinations) of the row vectors of A. It is a subspace of Rn.
- The Column Space (or Range) is the span of the column vectors of A. It is a subspace of Rm.
When we perform Gauss-Jordan elimination to get the RREF, we use elementary row operations. Let's explore why these operations treat the row space and column space differently.
Why Row Space is Preserved
Every elementary row operation is reversible and involves either swapping rows, scaling a row, or adding a multiple of one row to another. This means that:
- Every row in the new matrix is a linear combination of the rows of the old matrix. Thus, the new row space is contained within the old row space.
- Because operations are reversible, every row in the old matrix is also a linear combination of the rows of the new matrix. Thus, the old row space is contained within the new row space.
Therefore, the row space of A and the row space of RREF(A) are **identical**. Row operations do not change the row space.
Why Column Space Changes
In contrast, row operations work across columns, modifying individual column coordinates. The columns of the RREF matrix usually point in completely different directions compared to the columns of the original matrix. For example, consider:
| 1 | 2 |
| 2 | 4 |
The column space of this matrix is the span of the vector [1, 2]T. When reduced to RREF, we get:
| 1 | 2 |
| 0 | 0 |
The column space of the RREF is the span of the vector [1, 0]T. These two vectors point in different directions and span different lines in R2. The column space has clearly changed.
The Core Connection: Column Dependencies
Although the column space changes, row operations preserve the linear dependency relationships between columns. If the second column of A is twice the first column, the second column of RREF(A) will also be twice the first column. This is why we can use RREF to find a basis for the column space of A: we identify which columns contain pivots in RREF, and then select the corresponding columns from the original matrix A to form the basis.