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What property to look for to see if a matrix is diagonalizable?

  1. Aug 14, 2005 #1

    Just say you are given a 3x3 matrix with a value k in it. What would be the best way to find out if it is diagonalizable or not for certain values of k? The matrix in question is

    -1 0 0
    k 1 0
    -2 1 -1
  2. jcsd
  3. Aug 14, 2005 #2


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    My text gives a Test for Diagonalization:

    1. The characteristic polynomial of the matrix A splits
    2. For each eigenvalue [itex]\lambda[/itex] of A, the multiplicity of [itex]\lambda[/itex] equals n - rank(T - [itex]\lambda[/itex]I), where n is the dimension of the vector space on which A is an operator.

    An operator on an n-dimensional vector space is represented by an nxn matrix, so if you have a 3x3 matrix, then n = 3. We say a polynomial splits if it can be written in the form (x - [itex]\lambda _1[/itex])...(x - [itex]\lambda _k[/itex]). The polynomial x² + 1 does not split over the real numbers, but over the complex numbers, it can be written (x - i)(x + i). The multiplicity of an eigenvalue [itex]\lambda[/itex] is the exponent of the factor (x - [itex]\lambda[/itex]) in the characteristic polynomial. So if your characteristic polynomial is (x - 2)²(x - 0)(x - 1)³, then 2 has multiplicty 2, 0 has multiplicty 1, and 1 has multiplicity 3. So everything you need to know is now there to check if a matrix is diagonalizable.

    You will find the char. poly. to be (1 + x)²(1 - x), so it clearly splits. Eigenvalues are 1 (with multiplicity 1) and -1 (with multiplicity 2). First, we want to see if:

    multiplicity-of-1 = n - rank(A - 1I)
    1 = 3 - rank(A - I)
    rank(A - I) = 2

    and it's easy to see that this is indeed the case. Next, we want to see if:

    multiplicity-of-(-1) = n - rank(A - (-1)I)
    2 = 3 - rank(A + I)
    rank(A + I) = 1

    The matrix A + I is:

    0 0 0
    k 2 0
    -2 1 0

    this has rank 1 if and only if it has 1 linearly independent row, so the bottom two rows must be scalar multiples of each other. Since 2 is double 1, then k must be double -2, so k = -4. I believe this is the number you're looking for. Hopefully giving the answer away like this but taking you through solving the problem will allow you to do similar problems on your own. However, I did this in a bit of a rush so I might have made a mindless error somewhere, so check over it yourself.
  4. Aug 14, 2005 #3


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    An n by n matrix is diagonalizable if and only if there are n independent eigenvectors.
    It's easy to see that that is true if all eigenvalues are distinct (but not "only if").
    It is also true that an n by n real matrix has n independent eigenvectors if the matrix is symmetric. ( complex matrix: Hermitian)
    Last edited: Aug 15, 2005
  5. Aug 14, 2005 #4


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    I like
    1. the minimal polynomial splits
    2. the minimal polynomial has no repeated roots
  6. Aug 15, 2005 #5


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    halls tests are the easiest to use in practice:
    i.e. either check that

    1) the amtrix is symmetric (in which case it is also ortogonally diagonalizable)

    or 2) that the characteristic polynomial of degree n has n distinct real roots.

    either of these is sufficient, but not necessary for diagonalizability.

    the nec and suff condition is that the minimal polynomial splits into distinct linear factors, but it is not so easy to find the minimal polynomial.

    It is also useful to realize that a sufficient condition for a non zero matrix NOT to be diagonalizable is that the matrix is nilpotent, i.e. some power is zero. also if it is a sum of a nilpotent and a diagonalizable matrix, I think it is not diagonalizable.

    here is a simple property, essentially contained above for NON diagonalization:

    if there is a root c of the characteristic polynomial such that (T-c)^2 kills some vector not killed by T-c, then T is not diagonalizable.
    Last edited: Aug 15, 2005
  7. Aug 15, 2005 #6


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    A handy criterion for unitary diagonalization is:

    A square matrix is unitarily diagonalizable if and only if it is a normal matrix.

    (A square matrix is normal if it commutes with its hermitian conjugate).
  8. Aug 15, 2005 #7
    Thanks a bunch guys. Had my exam today!
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