Name: Bryan Murray Account Name: btm6580 Attendance Sheet #: 29

Assignment #: 11 Problem #: 7.1.10

URL: http://acs.tamu.edu/~btm658b/m311p7_1_10.html

Unfinished business from Sec. 5.4: Prove that the third condition of Theorem 4 of that section is indeed equivalent to the other two.

Theorem 4: Let L:VV be a linear function from a vector space into itself, and let L be represented (with respect to some basis for V) by the (square) matrix A. Then the following conditions are equivalent:

1. L (or A) has the maximal rank possible (namely, the dimension of V).

2. A is invertible.

3. The determinant of A is not zero.

Since the first two conditions of Theorem 4 have been proved equivalent then it is only necessary to prove the third is equivalent to one of the first two. To prove that the third condition in equivalent to the first, one must prove that if the matrix is not full rank then the determinant is 0 and if the determinant is 0 then the matrix is not full rank. If the matrix is not full rank then row reduction to form an upper triangular matrix will result in a row that is all zeros. This would make an element of the main diagonal of the matrix 0, making the product of the elements of the main diagonal also 0. Therefore, since the determinant of this reduced matrix is the product of the main diagonal entries the determinant is 0. Therefore, if the matrix is not full rank then the determinant is 0. Conversely, if the matrix is full rank then the reduced matrix has a diagonal of all ones. Therefore, the original matrix has a nonzero determinant but the determinant is not necessarily equal to one.