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Diagonalizable matrix

  1. Oct 27, 2012 #1

    What steps do we follow to calculate the diagonalize matrix of 3x3

    (1) If we get a matrix of 3x3 do we first calculate the eigenvalues of lambda1,l2 and l3?

    (2) Then form a matrix with the eigen vectors?

    (3) Then apply P inv.AP?

    Is that the way to calculate?

    -- Shounak
  2. jcsd
  3. Oct 27, 2012 #2


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    Hey shounakbhatta.

    The process you have described is correct but your diagonalization will be PDP^(-1) = A where A is your original matrix.
  4. Oct 27, 2012 #3
    Find the eigenvalues and put that eigenvalues as diagonal elements (preferably in increasing order). And you are done. It worked for the almost every problem that i worked. please inform whether there is any exception to it...
  5. Oct 28, 2012 #4


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    If you already found the eigenvalues, then it is unnecessary to calculate (P^(-1))AP, for the result will be the diagonal matrix D with the eigenvalues on the main diagonal, which can just be written down without calculations.

    With "diagonalization" it is ususally meant that A should be written in terms of D, that is: A=PD(P^(-1)). Note that this is equivalent to D=(P^(-1))AP, but it is the other form that is usually required. Still, it could be useful to calculate D=(P^(-1))AP to check the answer, so that no mistake was made in earlier calculations.
  6. Oct 28, 2012 #5
    Hello Sreerajit,

    Thanks for the wonderful help. Yes, it worked out.

    Thanks all for the help.
  7. Oct 28, 2012 #6

    In that case where do we need to calculate the eigenvectors or do we don't need at all?
  8. Oct 28, 2012 #7


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    It depends. If you only need the diagonal matrix, you don't need the eigenvectors. But in most cases, one also wants a basis of eigenvectors to obtain the connection between A and D: A=PD(P(^1)), where the columns in P are eigenvectors. One cannot use diagonalization to very much if one does not know both D and P.
  9. Oct 28, 2012 #8
    So, to summarize if I have separate eigen values I can just put them diagonally and get the diagonal matrix.

    If any lambda's value is a multiplicity then I need to find the corresponding eigen vectors and see whether the eigen vectors are different, if they are different, then it is diagonalizable, otherwsie not?

    Am I right?
  10. Oct 28, 2012 #9
    If I do have a eignvalue which has multiplicity, then how to know that the matrix is diagonalizable or not?
  11. Oct 28, 2012 #10


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    For an eigenvalue with multiplicity > 1 (as a root of the characteristic equation), you must find the eigenvectors to this eigenvalue and check if the subspace they consitute (the eigenspace) has the same dimension as the multiplicity. If it is so for all (multiple) eigenvalues, then the matrix is diagonalizable, otherwise not. (For a simple eigevalue, with multiplicity 1, the eigenspace will always have dimension 1.)

    Sorry I was a little imprecise in my previous post. I did not consider multiple eigenvalues there.
  12. Oct 28, 2012 #11
    Dimension of eigenspace means?
  13. Oct 28, 2012 #12


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    Let l be an eigenvalue of an n x n-matrix A. The set of all eigenvectors belonging to l, together with the zero vector, is a subspace of R^n (or C^n, if we allow complex elements in vectors and matrices). This subspace is called the eigenspace belonging to l, and, as a subspace of R^n, it has a dimension.

    What you do is that you solve the homogenous system (A-lI)x=0. The (nontrivial) solutions to this are the eigenvectors belonging to l. The eigenspace is the null space of A-lI, and if you use Gauss-Jordan elimination to find the solution, this gives a basis for this nullspace, i.e. a maximal linearly independent set of eigenvectors, and the number of vectors in the basis is the dimension of the eigenspace.
  14. Oct 30, 2012 #13
    The exception is if your characteristic polynomial is inseparable that is it has repeated roots. In this case you can not diagonalise the matrix, but you can get it into a canonical block diagonal form known as the Jordan form using what are known as generalized eigenvectors.
  15. Oct 30, 2012 #14
    Thank you Jim. Thanks for the help.
  16. Oct 30, 2012 #15


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    The matrix can be diagonalizable even if the characteristic equation has repeated roots. This is the case if the dimension of each eigenspace is the same as the multiplicity of the corresponding root.
    But it might not be so, and then the Jordan block matrix is not a diagonal matrix.
  17. Oct 30, 2012 #16
    Hello Erland,

    It would be very kind of you if you please post an example to illustrate.


    -- Shounak
  18. Oct 30, 2012 #17


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    The problem is that I don't know how to write matrices in TeX-notation here...

    But one can always borrow from others. Look up the examples here:

  19. Oct 30, 2012 #18
    The second question in that link gives us 3 eigenvalues in which 2 of them are equal. But the diagonalized matrix has no change.????
  20. Oct 31, 2012 #19
    I'll do a simple example when you can't diagonalise. This will be a simple example to illustrate generalized eigenvectors and the Jordan canonical form.

    Consider the matrix [tex]A=\left(\begin{array}{cc} 4 & 1 \\
    -1 & 2 \end{array} \right)[/tex] it's easy to check that it has only one eigenvalue namely

    [tex]\lambda = 3[/tex]

    A generalized eigenvector, v, of rank k of A associated with eigenvalue [tex]\lambda[/tex] is one such that:

    [tex](\mathbf{A}-\lambda\mathbf{I})^k\mathbf{v}=0[/tex] and [tex](\mathbf{A}-\lambda\mathbf{I})^{k-1}\mathbf{v}\neq 0[/tex] . Note that when k equals one this is just the usual definition of an eigenvector. Let v be a generalized eigenvector of rank k associated with eigenvalue [tex]\lambda[/tex]

    Define [tex]\mathbf{v_k}=\mathbf{v}[/tex] and [tex]\mathbf{v_{k-1}}=(\mathbf{A}-\lambda \mathbf{I})\mathbf{v_k}[/tex]

    [tex]\mathbf{v_{k-2}}=(\mathbf{A}-\lambda \mathbf{I})^2\mathbf{v_{k-1}}[/tex] and so on.
    This way you can get a string of generalized eigenvectors. Let's see how it works in our simple example. We had [tex]\lambda=3[/tex] in our example. Plugging 3 into [tex](\mathbf{A}-\lambda\mathbf{I})[/tex] we get:

    [tex]\left(\begin{array}{cc} -1 & -1 \\
    1 & 1 \end{array} \right)[/tex]

    Note that [tex]\left(\begin{array}{cc} -1 & -1 \\
    1 & 1 \end{array} \right)^2=0[/tex] so any vector will suffice as a generalized eigenvector. Let's just pick [tex]\mathbf{v_2}=\left(\begin{array}{cc} 1 & 0\end{array} \right)^T[/tex]

    so [tex]\mathbf{v_1}=(\mathbf{A}-3\mathbf{I})\mathbf{v_2}=\left(\begin{array}{cc} -1\\ 1 \end{array} \right)[/tex]

    So we have [tex]\mathbf{P}=(\mathbf{v_1}|\mathbf{v_2})[/tex]

    So the Jordan form is [tex]\mathbf{J}=\mathbf{P^{-1}}\mathbf{A}\mathbf{P}=\left(\begin{array}{cc} 3 & -1\\ 0 & 3 \end{array} \right)[/tex] The matrix A cannot be diagonalised this is the best you can acheive. Note how the eigenvalue 3 appears in it, and although it is not diagonal it is block diagonal. Some times it will work out that matrix will be diagonal even with degeneracy.
  21. Oct 31, 2012 #20


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