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Questions on Similar matrices [ All of same type ]

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    1. Let [tex]\theta[/tex] be a real number. Prove that the following two matrices are similar over the field of complex numbers:

    [tex]\left [\stackrel{cos \theta}{sin \theta} \stackrel{-sin \theta}{cos \theta} \right][/tex] and [tex] \left [\stackrel{ e^{i \theta} }{0} \stackrel{0}{e^ {-i \theta}} \right][/tex]

    2. Let W be the space of all nx1 column matrices over R. If A is an nxn matrix over R, then A defines a linear operator La on W through left multiplication : La (X) = AX. Prove that every linear operator on W is left multiplication by some matrix A.
    Now, if T,S be operators such that Tn = Sn = 0 but Tn-1 [tex]\neq[/tex] 0, Sn-1 [tex]\neq[/tex] 0 . Prove that T and S both have the same matrix A for some basis B for T and B' for S.
    Similarly show that if M and N are nxn matrices such that Mn = Nn = 0 but Mn-1 = Nn-1 [tex]\neq[/tex] 0, then M and N are similar.

    3. The attempt at a solution
    Sum number 1: I'm not sure how to start this.

    Sum number 2: The first part is okay. I can always find/make some matrix A such that the column space of A is the range of La.
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2


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    For (1), what's the definition of similar matrices? It involves finding a third matrix, called a change of basis matrix, right? Try writing down the definition and plugging in what you have. For 2x2 matrices you should be able to solve this by brute force.
  4. Oct 5, 2009 #3


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    Note that the second and third parts of question 2 are essentially asking the same thing. Two matrices are similar precisely when they represent the same linear map with respect to two choices of basis.

    How to prove question 2? It depends on what linear algebra you have covered so far. Do you know about eigenvalues? Do you know about any canonical forms, particularly the Jordan normal form?
  5. Oct 5, 2009 #4
    I solved for 1 by myself. Instead of doing what you said, I instead just used the fact that if both the matrices are similar, I can find some operator T such that if the first matrix is the matrix of T in some basis (which I am free to choose), then the second matrix will be the matrix of T in some other basis {p1, p2}. Then I just equated the two. Point is, if they are similar, the equation will yield such a basis, and it did. Piece of cake. No brute force. :)
  6. Oct 5, 2009 #5
  7. Oct 5, 2009 #6
    Actually, part 2.1 of my question asks to prove that there exists an invertible OPERATOR U such that T = USU-1.
    Part 2.2 asks me to show that there exists an invertible MATRIX U such that M = UNU-1

    Isn't there a difference? Part 1 shows that two totally different operators can have the same matrix representing them in two different bases.

    Part 2 asks to show that 1 operator can have 2 different matrices in different bases.
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