Let B be an n times n matrix If Rank BT n show that there i

Let B be an n times n matrix. If Rank B^T = n, show that there is an n times n matrix C such that CB^2 = 1.

Solution

If B is a n×n matrix and has rank n, then it has n pivot columns (and n pivot rows). This implies that we can row reduce B to an upper triangular form with pivots along the diagonal. The determinant is the product of these elements on the diagonal. Since the pivots are non-zero, hence their product is non-zero. Thus det(B) 0. This implies that B is invertible. Then ( B²)^-1 =( B.B)^-1 = B^-1.B^-1 = C (say). Since B^-1 is a n x n matrix, hence C is also a n x n matrix. Also CB² = (B^-1.B^-1)(B²) = (B^-1.B^-1)(B.B) = B^-1(B^-1B)B ( as matrix multiplication is associative) or, CB² = B^-1(I)B = B^-1(IB) = B^-1B = I.