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If two matrices have the same determinant, are they similar?

  1. Jul 10, 2013 #1
    If two matrices are similar, it can be proved that their determinants are equal. What about the converse? I don't think it is true, but could someone help me cook up a counterexample? How does one prove that two matrices are not similar?


  2. jcsd
  3. Jul 10, 2013 #2


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    Take any non-zero matrix with vanishing determinant. This will obviously not be similar to the zero matrix.
  4. Jul 10, 2013 #3
    Like wbn showed, determinant is invariant under similarity, but it is not a complete invariant. This means that two nonsimilar matrices share the same determinant.
    It is actually not so easy to come up with a complete invariant of similarity. A possible complete invariant is given by the Jordan normal form.
  5. Jul 23, 2013 #4
    Like Micromass and WBN have explained, similar matrices have the same determinant, so if two matrices have different determinants they cannot be similar. There are a lot of other quick checks you can do: rank, determinant, trace, eigenvalues, characteristic polynomial, minimal polynomial. Do the Matrices define the same linear map with respect to different bases?

    It is easy to come up with counterexamples of how two matrices having the same determinant is not strong enough to guarantee similarity.
  6. Jul 23, 2013 #5


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    Take the 2x2n matrix representing the Fibonnacci series (take the diagonalized version to simplify) and find another 2x2 matrix with the same determinant , but scaled by x and 1/x (x not 0 , of course) . Since the new matrix does not generate the Fibonnacci series, the two matrices cannot be similar. Or , use the fact that all rotations in R^n have determinant 1. Then rotations by different amounts cannot be similar to each other.
  7. Jul 24, 2013 #6


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    look up jordan form. this gives a canonical representative of the similarity class. i.e. no 2 different jordan forms are similar.

    looking at the appearance of a jordan matrix it should be obvious you have a large amount of freedom to change the similarity class, i.e. the jordan form, without changing the determinant.

    or maybe clearer, look at a "companion matrix" used in rational canonical form, another version of a canonical choice of a representative of a similarity class.

    Even for a matrix that is "cyclic" i.e. whose rational canonical form is a single companion matrix, note that the matrix contains the coefficients of the entire characteristic polynomial, whereas the determinant is given by only the constant term of that polynomial.
    Last edited: Jul 24, 2013
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