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Subspace criteria applied to square matrices, proof help

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Let n [tex]\geq[/tex]2. Which of the conditions defining a subspace are satisfied for the following subsets of the vector space Mnxn(R) of real (nxn)-matrices? (Proofs or counterexamples required.

    There are three subsets, i will start with the one where The subset V is that of square matrices with determinant = 0.

    So in this case i know that these matrices aren't invertible but it isnt a criteria of subspaces for there to be a multiplicative inverse. The first axiom is that there is a zero vector 0v such that x of V time 0v = 0. But how would i write a proof for this?

    Then the other axioms i need to show are closure under addition and scalar multiplication, an additive identity element (0), and an additive inverse. I dont need to show multiplicative inverse, no? I mean in any case to show the set is a subspace?

    Any help is appreciated, thanks
    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2
    I'm not sure where you're getting that first axiom. To be a subspace you need to show the three things you listed in your last paragraph: closure under addition, closure under scalar multiplication, and that the space contains the 0 vector. In this example, the crucial part is closure under addition. Can you prove or give a counterexample to the statement that if det(A) = 0 = det(B) then det(A+B) = 0?
  4. Mar 2, 2008 #3
    Oh, that is great thinking. How about the two matrices,
    1 1
    0 0


    1 0
    1 0
    adding these two you get

    2 1
    1 0

    and the inverse is ad - bc = 0 - 1 = -1 so it isnt of the set V.

    Thats great , thanks. To show that it has a zero vector:

    0v =

    Then A of the set V,

    0v.A = 0
    0v.a1 a2 = 0
    a3 a4
    = 0 0
    0 0 = 0

    Does that satisfy as a proof for the existence of the zero vector?
  5. Mar 2, 2008 #4
    The 0 vector is characterized by A+0 = A = 0+A For all A in the vector space. It doesn't have to do with multiplication since, in general, you don't have a notion of multiplying vectors (you just happen to in this case).
  6. Mar 2, 2008 #5
    Sorry my mistake, so it should be easy to prove that one.

    Okay so that is the first subset dealt with. The two are where the square matrices have:

    rank(A) [tex]\leq[/tex]1

    and trace(A) = 0

    Where trace(A) denotes the sum aij of the diagonal element of the matrix A =(aij)

    Im unsure about rank right now so will read on it, but i have no idea where to start with the trace = 0.
  7. Mar 2, 2008 #6
    Oh, i can use that trace(A+B) = trA +trB
    and Tr(xA) = x(TrA)
  8. Mar 2, 2008 #7
    To show that scalar multiplication is closed for the case when det(A)=0, how do i show a general determinant for any sized square matrix? Because i need to show the determinant is still zero after sclr multiplication.
  9. Mar 2, 2008 #8
    It is a well known fact that det(AB) = det(A)det(B). Further, if c is a real number and M is a matrix then cM = (cI)M where I is the identity matrix. Using these facts can you deduce what det(cM) is if you know det(M)?
  10. Mar 2, 2008 #9
    It has to be zero then, so by multiplying the scalar by the identity allows me to calculate the identity for cM using the rules for dterminants of matrices multiplied together. Thats so great, thanks guys.

    The only one im having trouble with now is the rank matrices, where rank is less than or equal to one. What does this mean and where do i start?
  11. Mar 2, 2008 #10
    If rank is less than or equal to one, doesn this mean that the set is made up of 2x2 matrices with just one element in them that is a number other than zero?
  12. Mar 3, 2008 #11


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    Science Advisor

    There is no longer any point in worrying about the zero vector. You have shown, by counterexample, that the set of matrices with determinant 0 is not closed under addition. You don't need to look at the other conditions- if it doesn't satisfy one, it can't satisfy all and so is not a subspace.

    By the way, you never have to show, separately, that the 0 vector is in the subspace. After you have shown the set is closed under scalar multiplication, it follows from the fact that [itex]0\vec{v}= \vec{0}[/itex].
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