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Linear Algebra - Identity matrices

  1. Oct 1, 2008 #1
    I am having some difficulty with identity matrices in linear algebra at the moment. I am sure it is fairly simple to solve, but I just cannot follow the logic behind this particular problem.

    I need to come up with a matrix B (2x2), such that B =/= I but B2 = I

    Code (Text):
    I = (1 0)
        (0 1)
    If I am attempting this correctly, I have to find a number that isnt 1, but if you square it, it equals 1. So I came up with this.
    Code (Text):
    B = (-1 0)
        (0 -1)
  2. jcsd
  3. Oct 1, 2008 #2


    User Avatar
    Science Advisor

    Well? Have you tried it? What is
    [tex]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right][/tex]
  4. Oct 1, 2008 #3
    well, dang, i don't know how to type matirices in latex, so i'll just try to give some hints.

    let the matrix [tex] B=[B_1,B_2][/tex] where [tex] B_1,B_2[/tex] are the columns of B, so

    [tex] B_1=[a.... b]^T, and, B_2=[c ... d]^T[/tex] now multiply [tex] BB=[BB_1 .... BB_2][/tex] now we want the following to hold

    [tex]BB_1=e_1,BB_2=e_2[/tex] where [tex] I=[e_1 .... e_2][/tex], [tex]e_1=[1 .... 0]^T, e_2=[0 .... 1]^T[/tex]

    now you will end up with some eq. involving in it a, b, c ,d so you can figure out the conditions that a, b, c, and d have to satisfy in order for BB=I. THis way you can find more than one such matrix.

    I hope i was of any help
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