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anonymous

  • 5 years ago

example finding the basis and dimension of the subspace of X of a 2 x 2 matrix?

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  1. anonymous
    • 5 years ago
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    What exactly do you need? Do you have a specific question?

  2. anonymous
    • 5 years ago
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    If you post your question and I'm not around, I'll get an e-mail about it and I'll help you out as soon as I can.

  3. anonymous
    • 5 years ago
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    The question is --- finding the basis and dimension of the subspace of X of a 2 x 2 matrix?

  4. anonymous
    • 5 years ago
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    The basis will be dependent upon the form of the subspace. I mean, if we're taking a normal 2 x 2 matrix, M = |a b| |c d| the standard basis is the set: B = { |1 0|, |0 1|, |0 0|, |0 0|} |0 0| |0 0| |1 0| |0 1| and the dimension is, by definition, the cardinality of the basis (i.e. the number of elements in the basis). So \[\dim(M_{2 \times 2})=4\]

  5. anonymous
    • 5 years ago
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    You can get basis sets that have a different number of elements (and therefore, dimension) depending on special conditions. For example, if the 2 x 2 matrix is to be symmetric, then you'd have, S = |a b| = a|1 0| + b|0 1| + c|0 0| |b c| |0 0| |1 0| |0 1|. Therefore the set { |1 0| , |0 1| , |0 0| } |0 0| |1 0| |0 1| spans the subset of all symmetric 2 x 2 matrices, and can be shown to be linearly independent, and so forms a basis for this subset. The dimension in this case would be 3.

  6. anonymous
    • 5 years ago
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    Incidentally, it's easy to show that these basis elements are linearly independent, since by the definition of linear independence, a set of vectors in linearly independent if the equation \[c_1v_1+c_2v_2+...+c_nv_n=0\]has only the trivial solution,\[c_1=c_2=...=c_n=0\]

  7. anonymous
    • 5 years ago
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    For the first basis, letting each of the v_i be one of the matrices, \[c_1v_1+c_2v_2+c_3v_3+c_4v_4\] = |c_1 c_2| |c_3 c_4| = |0 0| |0 0| if and only if\[c_1=c_2=c_3=c_4=0\]the trivial solution.

  8. anonymous
    • 5 years ago
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    You can do similar with the second basis.

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