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]);

A := matrix([[2, 3], [0, 0]])

B := matrix([[0, 1], [1, 0]])

C := matrix([[0, 2], [0, 0]])

> nullspace(A);nullspace(B); kernel(C);

{vector([-3/2, 1])}

{}

{vector([1, 0])}

Remember, there is no difference between nullspace and kernel, and Maple is aware of that.

In response to either command, Maple gives a basis for the nullspace.

The dimensions of the null spaces of matrices A and C are 1. This confirms the dimension theorem:

dim(col space) + dim(null space) = dimension(input vector s.

Since the input dimension is all these cases is 2, and we calculated earlier that the column spaces of these matrices had dimension 1, the null spaces must have dimension 1 as well.

(Recall that the dimension of the column space of a matrix is called its rank , and the dimension of the nullspace is called its nullity .)

Maple gave us empty braces for the basis of the nullspace of matrix B. That is because the null space of B must have dimension 0. This means that there is only one vector in the nullspace, namely the zero vector.

Comparisons with the column spaces . Although the dimensions of the column space and the nullspace are related, the spaces themselves don't have any special relationship. For instance, for matrix A, the basis vectors for the column and null spaces are

[Maple Metafile]

which don't have much to do with each other. For matrix C, the nullspace and the column space are actually the same! It is also possible for the nullspace and the column space to be perpendicular. Example:

> G := matrix(2,2,[1,1,1,1]); colspace(G); nullspace(G);

G := matrix([[1, 1], [1, 1]])

{vector([1, 1])}

{vector([-1, 1])}

The solutions "by hand."

Matrix A . Finding the null space is the same as finding the set of solutions to

A x = 0 .

We can do this by row reduction. For matrix A, the set of solutions works out to be all solutions of the form

[Maple Metafile]

which is the same as the line spanned by the basis vector

[Maple Metafile]

Matrix B . The determinant of this square matrix is not zero. Therefore the matrix is invertible, therefore the only solution to B x = 0 is the obvious one, x = 0 .

Matrix C . This is also a simple row-reduction problem.

Ready for bigger matrices? Read on.