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Abstract Vector Spaces

  1. Apr 30, 2007 #1
    M22 is the set of all m x n matrices with real entries
    P3 is the set of all polynomials of degree at most n, together with the zero polynomial.

    1) Find a basis of M22 consisting of matrices with the property that A^2 = A.
    I only found 2 of the vectors with a lot of hard work...
    [1 0
    0 1]
    [0 0
    0 1]
    I need 2 more...but I can think of any more...
    By the way, is there any SYSTEMATIC method to solve this problem?

    2) Is it possible to have a basis of P3 consisting of polynomials whose coefficients sum to 0? Support your answer.
    [I have no idea how to go about doing this...]

    Can someone please help me? Thanks!
  2. jcsd
  3. May 1, 2007 #2
    What is the dimension of M22' = {M22,A²=A} ?
  4. May 1, 2007 #3


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    I'm guessing that you mean that "Mmn" would be the set of all real m x n matrices and that M22 is the set of real 2x2 matrices.

    If you put A=

    a b
    c d (I'm too lazy to do the LaTeX stuff)

    the equation A^2=A will give you four equations describing relationships between a, b, c and d. And if you take the determinant of both sides of the equation A^2=A, you will get a fifth equation (det A=1). Maybe you can use these equations to find your basis.

    There should be four matrices in the basis, since M22 is four-dimensional.
  5. May 1, 2007 #4


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    What you need to do to solve the second problem is to show that the set of polynomials that satisfy that condition is a three-dimensional subspace of P3. (P3 is four-dimensional).
  6. May 1, 2007 #5


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    To get a set that spans a space do the following steps:
    1) find the "general vector" (written with variables) of the set.
    2) break up the general vector into the sum of vectors that have only one kind of variable in them
    3) the vectors that you get in step 2 span the space.
    To get a base check if the set that you got is linearly independent.
  7. May 2, 2007 #6
    Hi, how can this help me to find the basis of 4 vectors with that specific property that A^2=A?
  8. May 2, 2007 #7


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    EDIT: Never mind, I misread the question. Now it makes a lot more sense


    1 0
    0 0

    also satisfies your property that A2 = A

    This isn't linearly independent of your other two vectors though
    Last edited: May 2, 2007
  9. May 2, 2007 #8


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    I'm surprised that you ask this, because what I suggested is that you solve the system of equations defined by the matrix equation A^2=A. Of course that will help.

    The calculations are pretty tedious though. I did most of them myself and it's pretty boring stuff. I also noticed that I made a mistake in the previous post. The determinant of A doesn't have to be =1. It can also be =0.

    Here's the reason why the calculations get tricky and boring. You have to investigate all of the following cases separately:

    1. det A=0 and b=c=0
    2. det A=1 and b=c=0
    3. det A=0 and not b=0
    4. det A=0 and not c=0
    5. det A=1 and not b=0
    6. det A=1 and not c=0

    Some of these are dead ends, and some tell you what you already know, but you will also find new basis vectors if you do this. (I did).

    Edit: All the basis vectors I've found have determinant 0. I don't think there are any with determinant 1, but I haven't proved that conclusively. I suggest you focus on number 3 above. I got two new basis vectors from that. When you've done 3 you can probably guess what you'd get from 4. After that, you have more matrices than you need, so you just have to pick four that are linearly independent. Note that when you do 3 (after you've eliminated c and d), you can pick a value for b (I suggest b=1) and check if your matrix A really satisfies A^2=A. You will find that that it does, for two particular values of a, but not for any other.
    Last edited: May 2, 2007
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