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Linear algebra

  1. Jun 1, 2005 #1
    1. If I: W-->W is the identity linear operator on W defined by I(w) = w for w in W, prove that the matrix of I repect with to any ordered basis T for W is a nXn I matrix, where dim W= n

    2. Let L: W-->W be a linear operator defined by L(w) = bw, where b is a constant. Prove that the representation of L with respect to any ordered basis for W is a scalar matrix.

    3. Let X,Y, Z be sqaure matrices. Show that: (a) X is similar to Y. (b) If X is similar to Y then Y is similar to X. (c) If X is similar to Y and Y is similar to Z, then X is similar to Z.
     
  2. jcsd
  3. Jun 1, 2005 #2
    Let [itex]E_{ij}[/itex] be the elements of the matrix of the identity operator in some ordered basis of W, with basis vectors [itex]\vec{e}_1, \vec{e}_2, ... , \vec{e}_n[/itex]. If [itex]w_j[/itex] are the coordinates of any vector w in that basis, then

    [tex]w_i^\prime = \sum_j E_{ij} w_j[/tex]

    By definition, the identity operator transforms the vector w back into itself, so that [itex]w_i^\prime = w_i[/itex]. Then using the elements [itex]\delta_{ij}[/itex] (kronecker delta) of the identity matrix, we have

    [tex] w_i = \sum_j \delta_{ij} w_j = w_i^\prime = \sum_j E_{ij} w_j[/tex]

    or, after subtracting

    [tex]\sum_j (E_{ij} - \delta_{ij}) w_j = 0[/tex] for each i.

    Since the w_j's are arbitrary, we must have that [itex]E_{ij} = \delta_{ij}[/itex] for all i and j.

    edit: by the way, in the step where I set [itex]w_i^\prime = w_i[/itex] for all i, I have assumed that the coordinates of a given vector w in a particular basis are unique. This is easy to prove using the fact that the elements of the basis are linearly independent, by definition.
     
    Last edited: Jun 2, 2005
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