2 by 2 examples.

> A := matrix(2,2,[2,3,0,0]); B := matrix(2,2,[0,1,1,0]); C := matrix(2,2,[0,2,0,0]);

> colspace(A);

What Maple is providing is a basis for the column space. For the first example, we could just have chosen the first column as the basis, but Maple has chosen a simpler basis by making the first component 1 rather than 2.

We can almost always choose the first column of a matrix as one of the basis vectors (the only exception is when the first column is the zero vector). Then move on to the second column. If it is linearly independent of the first one, we can also use it as a basis vector. For matrix A, however, the second column is a multiple of the first one. So the two columns are not linearly independent. The column space is one-dimensional.

> colspace(B);

The two columns of B are not proportional, so they are independent. The column space of B is the whole two-dimensional space R2.

> colspace(C);

In this case, the first column is the zero vector, so we cannot use it in the basis; the zero vector is not linearly independent of any other vector. So we move on to the second column. Maple has again chosen to simplify the basis by arranging for as many 1's and 0's as possible. This is optional.

The solutions "by hand."

Matrix A . The first column is not the zero vector, so we put it into the "bucket" as part of the basis. We then check the second column. Problem: The second column is just a multiple of the first column, so not independent. Therefore a basis is

This answer would be perfectly correct. You might prefer to simplify by dividing by 2, as Maple did.

Matrix B . The determinant of this square matrix is not zero. Therefore the matrix is invertible, therefore the column space is the entire two-dimensional space R^2. Any two non-proportional vectors will serve as a basis. The simplest choice is

but

would also be correct. The first choise has two advantages; not only are the basis vectors simpler, they are also orthogonal (= perpendicular).

Matrix C . We proceed as for matrix A, but when we look at the first column, it is the zero vector, which is not allowed to be part of the basis. Therefore we move on to the second colums and choose it, or any multiple of it.

Ready for bigger matrices? Read on.