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    Linear algebra: Dimensions

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    Find the dimensions of each of the following vector spaces.

    a) The vector space of all diagonal n X n matrices

    b) The vector space of all symmetric n X n matrices

    c) The vector space of all upper triangular n X n matrices

    © BrainMass Inc. brainmass.com December 24, 2021, 10:32 pm ad1c9bdddf
    https://brainmass.com/math/linear-algebra/linear-algebra-dimensions-486814

    SOLUTION This solution is FREE courtesy of BrainMass!

    The dimension of a vector space is the number of the basis vectors.
    For example, let's look at the general space (all the matrices).
    They are of the form:
    (1.1)
    Since we have four independent variables (the matrix entry) we have four degrees of freedom and therefore we have four basis vectors.
    In this case, the standard basis is:
    (1.2)
    It is easy to see that indeed since:
    (1.3)
    And the basis matrices are linearly independent since if

    We must have
    Of course we can come up with other basis that will span this space, but we know that all of the bases must have the same number of vectors.

    Therefore the dimension of is
    (1.4)
    We can generalize this and say that for the space we need basis matrices, with the standard basis:
    (1.5)
    That is, all the entries of the matrix are zero except the cell which is 1.
    Thus we can say that
    (1.6)
    And as before we can say that if and only if for any

    All the diagonal matrices are of the form:
    (1.7)
    We see that we have n independent variables so we have n vectors in the basis.
    We can write and diagonal matrix as:
    (1.8)
    Thus:
    (1.9)

    Symmetric matrices are in the form:
    (1.10)
    In other words:
    (1.11)
    How many independent variables do we have?
    There are n diagonal elements and elements in the upper half of the matrix.
    Once we set them, the lower half is set as well.
    We can write it in term of the standard basis:
    (1.12)
    The first term is represents the diagonal elements, are the basis matrices with element above the diagonal and are the basis matrices with non-zero elements below the diagonal.
    Therefore we can define a new basis vectors
    (1.13)
    And the new basis is
    (1.14)
    In the original basis we had matrices. Out of those, n were the diagonal matrices. Thus are the off-diagonal matrices, and since we create a new basis vector from two standard matrices , we have new basis matrices.
    Therefore the new basis is has:
    (1.15)
    Therefore:
    (1.16)
    For example, for the matrices we have:
    (1.17)
    And
    (1.18)
    For the matrices we have:
    (1.19)
    (1.20)
    The same hold for the upper triangular matrices space
    (1.21)
    Again we have the n independent variables in the diagonal and the off-diagonal elements. This leaves us again with independent variables, only now we can write the matrix as:
    (1.22)
    Thus:
    (1.23)

    To summarize:
    (1.24)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:32 pm ad1c9bdddf>
    https://brainmass.com/math/linear-algebra/linear-algebra-dimensions-486814

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