The vectors you give for a basis of the range of E can't possibly be right because they are in . The operator E (multiplication by the matrix A with 2 rows and 3 columns) is from to . The range is a subspace of .

The "complicating point is that we are here thinking of the matrix, A, as the variable, and the vector v, as fixed: . Okay, since A is "variable", lets just write it as .

The simplest way to do this is to use the "rank-nullity" relation: if L is a linear transformation from U to V, then the "rank" of L (dimension of its range in V) plus the "nullity" of L (dimension of its null-space or "kernel" in U) is equal to the dimension of U.

Here, "U" is the set of all 2 by 3 real matrices and so has dimension 2(3)= 6. The null-space is the set of all matrices, M, such that Mv= 0.

Taking M as , we have . If M is in the null-space then we must have . That is, we must have a+ 2b+ 7c= 0 and d+ 2e+7f= 0. From those two equations, we get a= -2b- 7c and d= -2d- 7f. That is, b, c, d, and f can be any numbers whatsoever and we can calculate a and d from them. That tells us that the null-space of E is afourdimensional subspace of the 6 dimensional space of all such matrices. The range of E, then, is a 6-4= 2 dimensional subspace of . But is itself 2 dimensional! That means that E maps that set of matrices toallof [itex]R^2[/tex] and we can take as basis vectors of the range, any basis vectors for [itex]R^2[/itex], in particular, the "standard basis vectors", and .