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Vector space proof

  1. Jan 20, 2007 #1
    Please help me proove the following:

    Let V be a vector space over all n-by-n square matrices. Let W be a non-trivial subspace of V satisfying the following condition: if A is an element of W and B is an element of V then AB, BA are both elements of W.

    Proove that W = V.

    And here is what I am thinking about it...

    1. If W contains the identity matrix then this equivalence is quite obvious. Whatever matrix B from V you give me, I multiply it by the identity matrix I and the B I = B is also element of W.

    2. When W contains a regular matrix it is quite similar - because then W must contain the identity matrix. Let A be a regular matrix from W, then if you give me an inverse matrix, when I multiply them then what I will get is the identity matrix and due to the aforesaid condition the identity matrix is an element of W.

    But somehow I do not know how to continue.... Please help
     
  2. jcsd
  3. Jan 20, 2007 #2

    radou

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    I don't know if this will be helpful, but try to prove that W is a subset of V (which is trivial), and V is a subset of W in order to prove the equality.

    Edit: corrected.
     
    Last edited: Jan 20, 2007
  4. Jan 20, 2007 #3

    matt grime

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    is this some deliberate attempt to misspell the word 'prove'?

    W is non trivial. Pick w=/=0 in W. Now just show you can generate all of V. Hint: pick some obvious basis of V, show every element of this basis can be obtained by multiplying w by some appropriate choices.
     
  5. Jan 20, 2007 #4

    HallsofIvy

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    Matt, your methods seems unnecessarily complicated. The method rudo outlined, together with radou's hint works nicely. At first I puzzled over this because I misread the problem- I thought that W and V were subspaces of some other space! Since W is a subspace of V and V is the space of all 4 by 4 matrices, If w is any invertible matrix in W, then w-1 is contained in V and so ww-1= I is contained in W. from that, if v is any matrix in V then Iv= v is in W.

    Hmm, suppose there are no invertible matrices in W? Is it possible a set of non-invertible matrices to be a subspace?
     
  6. Jan 20, 2007 #5
    Of course. For example, span{M}, where M is invertible, is an example.

    We must show that a subspace consisting only of non-invertible matrices (which must be of the form span{M1, M2,...Mk}, where M1, M2,...Mk are each non-invertible) cannot satisfy the property "if A is an element of W and B is an element of V then AB, BA are both elements of W." Unfortunately a product of a non-invertible matrix with any matrix is still non-invertible, making the contradiction non-immediate. However, a sum of non-invertible matrices needs not be non-invertible.

    The case of a 1-dimensional such subspace is ruled out since wv cannot be a multiple of w for all v. I believe a subspace, of dimension greater than 1, consisting only of non-invertible matrices cannot exist. Right?

    I'm beginning to think Matt Grime's method is perhaps easier. Take w and multiply it by elementary matrices from V to get the matrix with 1 in the ijth entry and 0's everywhere else. Obtaining a basis for V that also exists in W. Hence W=V. The property that "AB, BA are BOTH elements of W." is needed to get the appropriate row and column operations to achieve the matrix transformations.
     
    Last edited: Jan 20, 2007
  7. Jan 21, 2007 #6

    matt grime

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    Unnecessarily complicated? No, sorry, I have to disagree there. My method takes about 3 lines if you try hard to make it long winded.

    1. W contains a non-zero element
    2. W contains an elementary matrix
    3. W contains all elementary matrices.
     
  8. Jan 21, 2007 #7

    matt grime

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    Yes. Trivially, and very important ones they are too: the set of strictly upper triangular matrices.

    Of course, none that are two sided ideals.

    If I wanted to be fancy I would have asked you to prove that the only automorphisms of M_n are inner, and that M_n is a simple algebra.
     
    Last edited: Jan 21, 2007
  9. Jan 21, 2007 #8
    Thanks all of you for your help...
     
  10. Jan 22, 2007 #9

    HallsofIvy

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    Having realized that the subspace does not necessarily contain an invertible matrix, I now agree with you!
     
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