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Math Help - Dimension of upper-triangular matrices

  1. #1
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    Question Dimension of upper-triangular matrices

    Show that the subspace of the set of upper triangular n x n matrices of M(nn) has dimension (n^2 + n)/2 for n = 3

    How do I set this up? Do I use simplified span or independent test method, and how?

    Thanks.
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  2. #2
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    Re: Dimension of upper-triangular matrices

    Hi,

    The answer to your question is encoded in the definition of "dimension of the set of upper triangular  n\times n matrices". You should count the number of independent variables that there exist in such matrices.
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  3. #3
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    Re: Dimension of upper-triangular matrices

    prove that the matrices \{E_{ij}\} where E_{ij} is the matrix with 1 in the i,j-th position, and 0's elsewhere, form a basis for i ≤ j.

    these matrices are clearly linearly independent, since they are a subset of a basis for Mat(n,F). showing they span the set of upper-triangular matrices is likewise fairly trivial.

    so....how many of these matrices are there?

    well there's 1 for every diagonal entry position, so n of these,

    1 for every super-diagonal entry position, so n-1 more,

    1 for every super-super diagonal position, so n-2 more,

    ....and so on until we get the the "last" super-diagonal, the upper-right corner, with one position.

    add these all up: we have 1 + 2 +...+ n-1 + n = n(n+1)/2.

    to see what i mean, in the space of 3x3 matrices, i am talking about the following set:

    \left\{ \begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&1&0\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&1\\0&0&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&1&0\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&0&1\\0&0&0 \end{bmatrix},\ \begin{bmatrix}0&0&0\\0&0&0\\0&0&1 \end{bmatrix}\right\}

    which has 1+2+3 = 6 elements.
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