MATH125 Lecture Notes - Lecture 32: Diagonalizable Matrix, Invertible Matrix

• School: University of Alberta
• Department: Mathematics
• Course: MATH125
• Professor: Nikita

Document Summary

Let p be such that p 1ap = b: det b = det(p 1ap ) = (det p ) 1(det a)(det p ) = det a, b(p 1a 1p ) = (p 1ap )(p 1a 1p ) = i. P 1a 1p is the inverse of b: we will use the following. For any two matrices a, b, one has rank(ab) rank b and rank(ab) rank a. Since the columns of ab are linear combinations of the columns of a, the column space of ab is contained in the column space of a. Therefore rank(ab) = dim col(ab) dim col(a) = rank a. With a help of transpose, the second inequality follows from the rst one: rank(ab) = rank(ab)t = rank(b t at ) rank b t = rank b. (cid:3) Returning to our similar matrices a and b, by the above lemma, we have: rank b = rank(p 1ap ) rank(ap ) rank a.