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Diagonalize a Matrix A - Normalize eigenvectors

  1. Jan 16, 2005 #1
    I have a homework problem here I am a little at a loss on due to not very good examples in class and the part of the book that explains them is 4 chapters ahead and loaded with words I just do not understand yet. :bugeye: If someone could give a definition or two and get me started on this bad boy, I'd appreciate it.

    The problem itself is:

    Diagonalize the matrix A below. Normalize the eigenvectors so that they are unit vectors.

    A = \left( {\begin{array}{*{20}c}
    3 & {\sqrt 5 } \\
    {\sqrt 5 } & { - 1} \\
    \end{array}} \right)

    Code above is in work to look right.... until then.... remove the :

    A = 3 : sgrt(5)
    ::sqrt(5) : -1

    Now that the problem is stated, I will show my thoughts and what I am lacking in....

    Diagonalizing the matric is basically taking A and getting A', where a'11, a'22, and a'33 (the diagonal) are the eigenvalues.

    Side Note: The only definition I have of an eigenvalue is "Matricies that are true with Hermetian Conjugate have all real eigenvalues". But how do you define a word using the same word in the definition? Google search brings up a ton of pages that confuse me. So I stopped looking there!! I do know that they mean different things depending on their application... for instance in molecular vibrations they would be the frequency or in classical L=IW they would be the moments of inertia.... but that gets away from my main issue.

    Now the equations I have of diagonalizing a matrix is, knowing CC^-1 is a unit matrix, IC = CI'. I beleive this is using the similarity transform.

    Thats about where I stand. Do I simply create the unit matrix for C and multiply that by A and that will give me A' with my eigenvalues in the diagonal? If so, what would I do to normalize them? Or I guess the better question would be "What is normalization?"

    Appreciate any insight. :biggrin:
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 16, 2005 #2


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    Your text should have a theorem on diagonalizing a matrix, and examples on it. Try looking for the equation [itex]A = PDP^{-1}[/itex]. (The choice of letters used seems fairly universal)

    You are also probably looking at the wrong spot for the definition of eigenvector and eigenvalue -- the definition is that [itex]\lambda[/itex] is an eigenvalue of A, and [itex]\vec{v}[/itex] an eigenvector (associated with that eigenvalue) iff [itex]A\vec{v} = \lambda\vec{v}[/itex]. You may have seen this equation in your text for finding eigenvalues: [itex]\det (A - \lambda I) = 0[/itex].
  4. Jan 16, 2005 #3
    I think I got the same equation in my notes but it is in the form....

    I' = (C^-1)(I)(C) where C is some matrix such that CC^-1 is a unit matrix.

    Anyway, I found a good resource on the internet and also plugged through my notes quite a bit and I got somewhere. I just do not know if I answer the question or if there is still more to be done. Of course, I could have the answers right here but do not know their place (meaning).

    I got the eigenvalues for the problem to be u=4 and u=-2 from using the matrix:

    3-u : sqrt(5)
    sqrt(5) : -1-u

    Note: Is there some convention that shows how you combine the matrix with u to get the above matrix? I've seen it done a few times without the explanation so I just know to subtract u from a11 and a22 but dont know WHY I do that.....

    So, plugging those eigenvalues into the equations generated by the matrix above, I end up with the following eigenvectors:

    sqrt(5/6) : 1/sqrt(6)

    -1sqrt(6) : sqrt(5/6)

    So, does this mean I am done? Did I diagonalize the matrix? Unfortunantly, I have not been instructed on what that actually means.... what numbers go in a matrix with all 0's except where i=k? The eigenvalues are the diagonal or are the eigenvectors the diagonal?

    I think I have all the math part worked out pretty good except knowing how to get the u in the matrix as I had said above. Now I just need to figure out what values go in the diagonalized matrix... and the definitions.....

    PS: Does the Latex Generator really take that long or did I convert to the wrong version of Latex from MathType?
  5. Jan 16, 2005 #4


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    Well, algebraically, you've diagonalized a matrix A iff you've found an invertible matrix P and a diagonal matrix D such that [itex]A = PDP^{-1}[/itex].

    This is related to similarity transformations -- geometrically we picture it as choosing a basis in which A is a diagonal matrix. In particular, that means the basis is formed entirely from eigenvectors. (A fact that lets you write down the matrix P)

    The usual way to find all the eigenvectors is to simply solve the equation [itex]A\vec{v} = \lambda\vec{v}[/itex] for [itex]\vec{v}[/itex], for each eigenvalue.

    (Note that's the same as solving [itex](A - \lambda I)\vec{v} = 0[/itex]. Do you see why?)
    Last edited: Jan 16, 2005
  6. Jan 16, 2005 #5
    I did not want to try and type all this out in text so I will give a link to a .jpg that is a scan of the work I did. This will allow you to see all my work (I did not skip many steps on purpose) and see what I am looking at.


    Edit: Also, how would you find P and D matrix if your only given A? I understand their relationship... just not sure how to get them.
  7. Jan 16, 2005 #6


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    The results look right.

    Do you know how to apply a change of basis? Try using those two eigenvectors as your basis vectors... (they are independent, right?)
  8. Jan 16, 2005 #7
    I am not sure on what you mean....

    I found that the professor added the diagonalization contect to our current work in Chapter 3 (where we started the course Monday) from Chapter 10! So, after finding that in Chapter 10 and reading it, I am very clear on what the eigenvalues, eigenvectors, and the diagonal of the Matrix is. I am also clear now on what they label as [itex]C^-1MC = D[/itex] or as you were calling it with the [itex]PDP^-1[/itex].

    The 'diagonalized' matrix actually turns out to be nothing more than having the elements of the matrix with i=k (a11, a22, a33, ect) are the eigenvalues.

    I also found out how / why you take the original matrix and have the u appear in the equations. It is simply subtracting the right hand matrix to the left hand side of the equation to make it all = 0.

    Yeah, once I found the place in the book that talks about this stuff it is all very clear now. Too bad I didnt know that from the beginning. :)

    If you have a chance to pick up an extra book, I highly recommend the one we have. Very clear and understandable! Mathematical Methods in the Physical Sciences - 2e - Mary L Boas.
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