### Computation of Basis

##### 1. Basis of null space

In the next example, we find a basis of the null space of a matrix.

Example The null space of the matrix

 A = [ 1 3 2 0 1 ] -1 -1 -1 1 0 0 4 2 4 3 1 3 2 -2 0

consists of all the solutions of the homogeneous system Ax = 0. By the row operations in this example and this example, we have the reduced row echelon form.

 [ 1 0 1/2 0 -1/4 ] 0 1 1/2 0 1/4 0 0 0 1 1/2 0 0 0 0 0

Then we derive the solutions of the homogeneous system.

x1 = - (1/2)x3 - (1/4)x5
x2 = - (1/2)x3 - (1/4)x5
x3 is arbitrary
x4 = - (1/2)x5
x5 is arbitrary

Thus nulA consists of vectors of the form

 x = [ x1 ] = [ - (1/2)x3 - (1/4)x5 ] = x3[ -1/2 ] + x5[ -1/4 ] = x3u + x5v. x2 - (1/2)x3 - (1/4)x5 -1/2 -1/4 x3 x3 1 0 x4 - (1/2)x5 0 -1/2 x5 x5 0 1

Since x3 and x5 are arbitrary, we have nulA = span{u, v}. Moreover, the following argument shows that u and v are linearly independent.

x3u + x5v = 0
⇒ The solution x = 0
⇒ All coordinates of the solution x = 0
⇒ The 3rd and 5th coordinates x3 = x5 = 0.

Consequently, u and v form a basis of nulA.

The discussion in the example applies to the general case. Thus we have the following method for finding bases of the null spaces.

If the general solution of Ax = 0 is

x = c1v1 + c2v2 + ... + ckvk,

where c1, c2, ..., ck are the free variables. Then v1, v2, ..., vk form a basis of nulA.

The bases of kernels of linear transformations can be found in a similar way.

Example Consider the linear transformation

 T(X) = AX - XA: M(2, 2) → M(2, 2), where A = [ 1 2 ]. 3 4

Denote

 X = [ x y ]. z w

Then by looking at the four entries, T(X) = O is the same as

 3y - 2z = 0 2x + 3y - 2w = 0 - 3x - 3z + 3w = 0 - 3y + 2z = 0

The general solution is x = - z + w, y = (2/3)z, with z and w arbitrary. In the matrix form, the solution is

 X = [ - z + w (2/3)z ] = z[ -1 2/3 ] + w[ 1 0 ]. z w 1 0 0 1

Thus a basis of kernelT is given by the matrices

 [ -1 2/3 ], [ 1 0 ]. 1 0 0 1