Let be an -dimensional vector space with basis and let be an -dimensional vector space with basis . Let be a linear transformation, and let . (That is, the th column of is the coordinates of in terms of the ordered basis . Let be the reduced row echelon form of .
- Prove that the dimension of is the row rank (i.e. number of nonzero rows) of . Prove moreover that the columns of corresponding to the pivotal columns of form a basis of .
- Note that the kernel of is precisely the solution of the equation . We saw how to solve such equations in this previous exercise. Prove that is injective if and only if has row rank . If is not injective, construct a basis for .
- Write as a column matrix. Then . In particular, the image of is the span of the columns of . In particular, as we showed previously, the columns of corresponding to the pivotal columns of form a basis of . In particular, the dimension of this image is the row rank of .
- Note that is injective if and only if has dimension , if and only if (by part 1) has row rank . We showed previously how to construct the solutions of ; specifically, for each nonpivotal column of , we fix one corresponding entry in to be 1 and the other (nonpivotal) entries 0. The other entries of are then determined uniquely. These vectors are certainly linearly independent, and generate .