Hey! Sign in to get help with your study questionsNew here? Join for free to post

urgent! matrix help!

Announcements Posted on
How often do you go out? 11-02-2016
How proud of your uni are you? Take our survey for the chance to win a holiday! 09-02-2016
  1. Offline

    ReputationRep:
    let A =

    i need to find a 5x5 matrix M with rank(M) = 2 that AM = 4X5 zero matrix.
    I thought of finding the inverse of A at first, but A is not a square matrix!!! i got stuck... help pls thanks!

    and for any 5X5 matrix B such that AB = 0 then rank B <= 2. how can i prove this?
  2. Offline

    ReputationRep:
    For the first bit, since rank(M) = 2 why don't you try:

    M = \begin{pmatrix} a & b & c & d & e \\f & g & h & i & j \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \end{pmatrix}?

    Then you could sub it in AM = 0 to find relations on a-j and just choose any old matrix that works.
  3. Offline

    ReputationRep:
    omg are you saying i have to multiply that all out and solve for it? is there a proper way of doing it?
  4. Offline

    ReputationRep:
    That is the proper way of doing it. Pick any two linearly independent solutions of the system Ax=0. Then let B be the matrix whose first two columns are the solutions, and fill the remaining columns with zeros.

    You can do the second bit easily if you know the following inequality. If A is an mxn matrix, and B is an nxp matrix, then
    rank(A) + rank(B) - n <= rank(AB)

    (Try to justify it by drawing a diagram or something, if you don't feel like proving it carefully.)

    So in this case,
    3 + rank(B) - 5 <= 0
    => rank(B) <= 2

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: March 28, 2007
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR

Student Money Week continues...

Find out which Q&As are happening today

Poll
Do you listen to podcasts?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.