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Homework Help: A is similar to B is equivalent to A^k is similar to B^k?

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Given two matrices, [tex] A [/tex] and [tex] B.[/tex] Is the following statement true?

    [tex] A [/tex] is similar to [tex] B [/tex] [tex] \Longleftrightarrow [/tex] [tex] A^k [/tex] is similar to [tex] B^k. [/tex]

    2. Relevant equations
    By definition, A and B are similar if there exists an invertible matrix P such that B = P-1AP.

    3. The attempt at a solution

    Clearly, the [tex] \Rightarrow [/tex] portion of the statement holds. For example, B2 = (P-1AP)(P-1AP) = P-1A(PP-1)AP = P-1A2P.

    However, I am not certain about the statement in the reverse direction. I haven't spent a terrible amount of time on it, but I can't think of any counterexamples straight off the top of my head.

    Any hints or suggestions? Thanks for your time.
  2. jcsd
  3. Jul 10, 2010 #2
    I'm hesitant in replying to my own question, but I figured it out. My intuition was right: the two are not equivalent statements.

    Let [tex]
    A =
    \left[ {\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array} } \right]
    [/tex]. Then A^2 = 0. Let B=0. Then A^2 is similar to B^2. But A is not similar to B.
  4. Jul 10, 2010 #3
    Uhm, I don't think that logic is right.
    I'm not sure whether or not similarity can be applied to the 0 matrix, but assuming it can;

    As you said if A is similar to B, then [tex]B=P^{-1}AP[/tex] but remember that A is a diagonal matrix containing the eigenvalues of B.
    Now if we consider this, we can see that [tex]A^{2}[/tex] is in fact similar to [tex]B^{2}[/tex] because both eigenvalues of [tex]B^{2}[/tex] will be 0 and [tex]A^{2}[/tex] is the 0 matrix, therefore technically it is a 'diagonal' matrix with eigenvalues of B along its diagonal.
    Using this same logic, we can see that A IS in fact similar to B because both eigenvalues of A are also 0.

    Now I'm hoping someone else will see this and confirm it but then again similarity might not apply to the 0 matrix in the same way that the 0 vector is not really an eigenvector. The main reason I question this is that if we use the 0 matrix, we get all eigenvalues being 0 and hence all eigenvectors being 0 (which are what make up the columns of P) and I would think that means that there isn't an invertable matrix P that exists and hence 0 is not similar to any matrix (using this same logic I'd think your matrix A isn't similar to anything either).
    Last edited: Jul 10, 2010
  5. Jul 10, 2010 #4
    The matrix A is diagonalizable if it is similar to a diagonal matrix, i.e., if [tex]P^{-1}AP[/tex] is a diagonal matrix. However, the condition A is similar to B by [tex]B=P^{-1}AP[/tex] in and of itself does not imply B is a diagonal matrix (or contains the eigenvalues of A).

    A is diagonalizable [tex] \Longleftrightarrow [/tex] A has n linearly independent eigenvectors.
    However, that is not the case for the counterexample I provided. For [tex]

    A =
    \left[ {\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array} } \right]

    [/tex] and B = 0, the zero vector is the only eigenvalue of A (it appears twice) and also clearly the only eigenvalue of B. Consequently, the sole eigenvector of A is v = <1,1>. Thus, A is not diagonalizable, i.e. it is not similar to a diagonal matrix.

    I don't see why the zero matrix would not be subject to the same rules of similarity as other matrices. It's only special in the sense that the only matrix similar to the zero matrix is the zero matrix itself itself (hence my reason for choosing it for my counterexample).
  6. Jul 10, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You are correct, the implication doesn't go both ways
  7. Jul 10, 2010 #6
    Ah yep that makes sense.
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