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Why do we need to convert to a diagonal matrix?

  1. Dec 15, 2012 #1
    Apart from simplifying matrix powers, why do we want to diagonalize a matrix? Do they have any appealing application which can be used to motivate to study diagonal matrices.
    Thanks for any answers.
     
  2. jcsd
  3. Dec 15, 2012 #2
    Diagonal matrices are nice because ANY matrix calculation is simpler with diagonal matrices. For example, any two diagonal matrices compute. Also, it is trivial to find the inverse of a diagonal matrix. Apart from these facts and the fact that numerical computations are easier and much more stable, there are theoretical reasons to want to deal with diagonal matrices. Proving things about diagonal matrices is quite a bit easier than proving things about general matrices.
     
  4. Dec 17, 2012 #3
    One example in which diagonalization is important is the study of quadratic forms.
    http://en.wikipedia.org/wiki/Quadratic_form

    A quadratic form can be written as
    [itex] Q(x) = x^TAx [/itex]
    where x is a column vector and A is a symmetric matrix. It is a theorem that you can always diagonalize A by a rotation of coordinates. For example, in 2D, if you have an equation such as: ax^2+2bxy+cy^2 = D, then by rotating your coordinate axes you can rewrite the equation as
    [itex] A\bar{x}^2+B\bar{y}^2 = D[/itex]
    in your new coordinates. Therefore, the original equation represents an ellipse or a hyperbola (or possibly a pair of parallel lines if one of the eigenvalues A or B is zero.)

    Quadratic forms are important, for example, because a general function f(x,y,z) has a local Taylor polynomial approximation
    [itex] f = f(P) + df + Q_f + higher order terms [/itex]
    The second order term is a quadratic form which is determined by the Hessian matrix. So, for example, at a critical point (where the differential df =0), the first nonzero term in [itex] \Delta f [/itex] is the quadratic form determined by the Hessian. Since all quadratic forms can be diagonalized by a rotation of coordinates, that means that by a rotation of coordinates,
    [itex] \Delta f = A\bar{x}^2+B\bar{y}^2+C\bar{z}^2 + higher order terms[/itex]
    A, B, C are the eigenvalues of the Hessian. One thing you can do with this knowledge is determine whether a critical point is a maximum. To do that you check the eigenvalues of the Hessian matrix. If they are all negative, then you have a relative maximum.
     
  5. Dec 19, 2012 #4
    A good application is in the study of 2nd order differential equations.
    This can be seen in Structural dynamics

    Equations can arise in the form of

    My'' + y' + Ky = F

    where M,C,K are (nxn) matrices and y'',y',y, and F are (nx1) vectors

    The n equations are coupled with each other. If we can diagonalize M,C,K then we uncouple them and we can then solve n independent equations.

    for example, lets assume there is a matrix [itex]\Phi[/itex] such that

    [itex]\Phi^{T}[/itex] M [itex]\Phi[/itex] = [itex]M[/itex] (Diagonal)
    [itex]\Phi^{T}[/itex] C [itex]\Phi[/itex] = [itex]C[/itex] (Diagonal)
    [itex]\Phi^{T}[/itex] K [itex]\Phi[/itex] = [itex]K[/itex] (Diagonal)

    Then if we let y = [itex]\Phi[/itex]u then

    M[itex]\Phi[/itex]u'' + C[itex]\Phi[/itex]u' + K[itex]\Phi[/itex]u = F

    Multiply by transpose [itex]\Phi^{T}[/itex] to get

    [itex]\Phi^{T}[/itex]M[itex]\Phi[/itex]u'' + [itex]\Phi^{T}[/itex]C[itex]\Phi[/itex]u' + [itex]\Phi^{T}[/itex]K[itex]\Phi[/itex]u = [itex]\Phi^{T}[/itex]F

    which simplifies to

    [itex]M[/itex]u'' + [itex]C[/itex]u' + [itex]K[/itex]u = [itex]\Phi^{T}[/itex]F
    which is just n independent equations which can be solved separately to find each component in the vector u.

    Once that is done, the vector y can be found by y = [itex]\Phi[/itex]u

    I hope this helps and I hope this was readable.
     
  6. Dec 22, 2012 #5
    Diagonalizing matrices can help computer run-times as well.
     
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