Basis for the Kernel of T

Proved by: Not Applicable
References: Not Applicable
Justifications: Not Applicable

Specializations: Not Applicable
Generalizations: Not Applicable

!theorem of T.
Let be the number of nonpivotal columns of , and their positions. For each nonpivotal column, form the vector satisfying and such that its th entry is 1, and its th entries are all 0, for . The vectors form a basis of .

Example: Consider the matrix below, which describes a linear transformation from to .

The third and fourth columns of are nonpivotal, so and . The system has a unique solution for any values we choose of the third and fourth unknowns. In particular, there is a unique vector whose third entry is 1 and fourth entry is 0, such that . There is another, , whose fourth entry is 1 and third entry is 0, such that :

The first, second, and fifth entries of and correspond to the pivotal unknowns. We read their values from the first three rows of (remembering that a solution for is also a solution for ):

that is,



which gives



So for , where and , the first entry is , the second is and the fifth is ; the corresponding entries for are :

These two vectors form a basis of the kernel of .