082nd_tut7sol - MATH1111/2008-09/Tutorial VII Solution 1...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH1111/2008-09/Tutorial VII Solution 1 Tutorial VII Suggested Solution 1. (a) Let B = MA where M is an × m matrix and A be m × n . i. Show that rank( B ) min(rank( M ) , rank( A )) where the notation min( a, b ) means the minimum of a and b . ii. If rank( A ) = rank( B ), show that N ( A ) = N ( B ). (b) Let A be a square matrix of order n . Show that A is invertible if and only if rank( A ) = n . Ans . (a) i. Let x N ( A ). Then A x = 0 B x = MA 0 = M 0 = 0 . i.e. x N ( B ), or N ( A ) N ( B ). By part (a), dim N ( A ) dim N ( B ). Since A is m × n , dim N ( A ) + rank( A ) = n . Similarly, B is × n , dim N ( B ) + rank( B ) = n . Using the inequality dim N ( A ) dim N ( B ), we conclude rank( B ) rank( A ). From B = MA , we have B T = A T M T . Apply the last result, we get rank( B T ) rank( M T ). Note that rank( B ) = rank( B T ). (Proof. By definition and Theorem 3.6.6, rank( B ) = dim r ( B ) = dim c ( B ) = dim r ( B T ) = rank( B T ).) Hence, rank( B ) rank( M ). To sum up, rank( B ) is less than or equal to both rank( A ) and rank( M ). Thus, rank( B ) min(rank( M ) , rank( A )).
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern