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Question about determinants and characteristic polynomials.

  1. Dec 17, 2011 #1
    I know this may be a very stupid question, but I would really like to know. Is the determinant and the characteristic polynomial of an equation unique? I did several textbook questions and when I look at the solutions, they end up with completely different answers. Sometimes I am wrong and see my mistake, and at other times I am positive I did the cofactor expansion for a matrix correctly. Thanks in advance.
  2. jcsd
  3. Dec 17, 2011 #2


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    for a given matrix, yes, the determinant is unique. because the determinant is what they call "an alternating tensor", there's lots of plus and minus signs floating about in the various terms that go into it. this makes it very easy to make one small mistake and that ruins everything. so anything larger than a 4x4 matrix is extremely tedious to do by hand (unless the matrix has lots of 0's or some other "nice" form).
  4. Dec 17, 2011 #3
    Thanks Deveno.

    But, couldn't the determinate of a matrix larger than 4x4 be computed easily by first row reducing and have zeros in the first column and then expanding along it?
  5. Dec 17, 2011 #4


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    Yes, and that (and similar methods) is the most efficient way to calculate determinants, but the amount of arithmetic for an NxN matrix is still proportional to the cube of N. Working direct from the defintion of a determinant, the amount of arithmetic is proportional to N factorial which grows much quicker than N cubed.

    It's no big deal for a computer program to find the determinants of a "general" matrix of size 1000 x 1000 (in a few seconds, on a modern PC) and for matrices with special properties it's quite possible to work with sizes of 1,000,000 x 1,000,000 or bigger.

    As you have found out already, there isn't much to be learned from doing bigger matrices by hand, except that it's very easy to make mistakes.
  6. Dec 18, 2011 #5


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    unique up to sign, unless you require initial coefficient +1.
  7. Dec 19, 2011 #6


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    The determinant is a polynomial on R^n^2. For instance on R^4 it is xw-yz

    The characteristic polynomial is a determinant.
  8. Dec 19, 2011 #7


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    the determinant of a matrix is unique, period (unless you want to claim the choice of "even" and odd" permutations of indices is arbitrary).

    formally, we have, for an nxn matrix A = (aij):

    [tex]\det(a_{ij}) = \sum_{\sigma \in S_n} (\text{sgn}(\sigma)\prod_{i=1}^n a_{i\sigma(i)})[/tex]

    the sign of the "reordering of the column index" which determines the sign of every term, is unambiguous. for a 2x2 matrix, the term a11a22 is ALWAYS positive.

    the characteristic polynomial, however, is only unique up to sign, since both det(xI - A) and det(A - xI) have the same roots (and thus determine the same eigenvalues), so either one can be regarded as "the characteristic polynomial".

    if one views "a" determinant as being the volume of an oriented n-cube, there is, of course, the question of choosing the orientation (in 3 dimensions, this is called "the right-hand rule"). if one views a matrix as modelling a system of linear equations, this boils down to "which order we list the equations in". as geometric objects, there is no clear way to determine "which" direction of a vector (which way on a coordinate axis) ought to be "the positive one".
  9. Dec 19, 2011 #8


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    Geometrically, a non-zero determinant of an nxn real matrix can be seen as the oriented n-volume of a collection of n n-vectors:

    If the determinant is 0, then the figure determined by the vectors is degenerate, i.e., a square with collinear sides.

    If the determinant is non-zero, you get the (oriented) n-volume.

    I think this case of ℝn can be generalized to other fields (tho maybe more problematic for rings ).

    So, at least for the real case, if you accept that the m-volume determined by a collection of vectors is well-defined, then the determinant/m-volume is unique.
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