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Homework Help: 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

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    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

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    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

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    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

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    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

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    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

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    More succinctly, if det(A)= 0, then det(xA), for any scalar x, is what?
     
    Last edited by a moderator: 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

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    Yes, that's correct.
     
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