1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is singular matrix is a subspace of vector space V?

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data

    S is a subset of vector space V,

    If V is an 2x2 matrix and S={A|A is singular},

    a)is S closed under addition?
    b) is S closed under scalar multiplication?


    2. Relevant equations
    S is a subspace of V if it is closed under addition and scalar multiplication.


    3. The attempt at a solution
    I tried to use the definition of sinularity. i.e. a matrix in not invertible. But could not decide if it was closed under addition and scalar multiplication.
     
  2. jcsd
  3. Oct 27, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Well, play around with examples of singular matrices. It's not too hard to find two that add up to an invertible one; in fact, you can find two singular matrices that add up to the identity.
     
  4. Oct 27, 2008 #3
    yeah, i can get an identity if i add [1 0; 0 0] and [0 0; 01]. So what does that insinuate? If we add two we get an invertible matrix, which implies that the solution to AX=B is unique. But how do I get to point where i can decide if they are closed under addition and scalar multiplication? Further help will be highly appreciated.

    thank you
     
  5. Oct 27, 2008 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    What does it mean for S to be closed under addition and scalar multiplication?
     
  6. Oct 27, 2008 #5
    if vec u and vec v are in S, vec u + vec v also should be in S (closed under addition)
    for any scalar r, r(vec u) = r*vec(u) and lies in S(closed under scalar multiplication)
     
  7. Oct 27, 2008 #6

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    OK, now think about what you did: you added two things in S and got the identity matrix. What does this tell you about S?
     
  8. Oct 27, 2008 #7
    the identity matrix belongs to S. So u and v are closed under addition. Right?
     
  9. Oct 27, 2008 #8

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Really -- does the identity matrix belong to S?
     
  10. Oct 27, 2008 #9
    ohh... so S is a singular, meaning that it is not invertible. The identity matrix is invertible, hence does not belong to S. I hope i got it right this time
     
  11. Oct 27, 2008 #10

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Yup. So S isn't closed under addition.

    Now, let's turn to scalar multiplication. Suppose we have a singular matrix A and we multiply it by a scalar r. Can rA ever be nonsingular (i.e. invertible)? Suppose it can be - what does this tell you about A?
     
  12. Oct 27, 2008 #11
    again, if we turn it into an invertible matrix, it would mean that A does not belong to S.
    But i did not get an invertible matrix by multiplying a singular matrix with a scalar. Say A = [1 0; 0 0], if we multiply by r, we get, rA= [r 0; 0 0], which is still singular, right?
     
  13. Oct 27, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    More succinctly, if det(A)= 0, then det(xA), for any scalar x, is what?
     
    Last edited: Oct 27, 2008
  14. Oct 27, 2008 #13
    we have not done determinant so far. But i believe in the above case, det(xA) should equal 0. I think so because det(xA) = x* det(A).
     
  15. Oct 27, 2008 #14

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is singular matrix is a subspace of vector space V?
Loading...