A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • one year ago

Please explain how to find basis for set of 2x3 matrices whose null space contains (2, 1, 1). This is a question from Recitation video in http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/matrix-spaces-rank-1-small-world-graphs/ It's not very well explained in the video.

  • This Question is Open
  1. JoshDanziger23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Lokeshh, if A is a (2,3) matrix with u=(2,1,1) in its null space then it must be that both rows of A are orthogonal to u. Any two independent vectors v and w orthogonal to u will span a plane perpendicular to u, and the recitation lecturer suggests the two vectors v=(1,0,-2) and w=(0,1,-1) are natural choices. As long as the rows of A are linear combinations of v and w then we will have Au=0; and because v and w span the whole plane, linear combinations of v and w will generate all vectors orthogonal to u. So we can make a basis for all (2,3) matrices satisfying Au=0 by using the basis [1,0,-2; 0,0,0], [0,1,-1; 0,0,0], [0,0,0; 1,0,-2], [0,0,0; 0,1,-1]. Does that help? Josh.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.