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    I have the matrix [2 3;3 2] and wish to diagnoalize it.

    First I find the eigenvalues of : -1 and 5 and put them into a matrix as:

    P=[-1 0;0 5]

    Then find the eigenvectors for each value and put them into a matrix
    D= [1 1; -1 1]

    Then find the inverse of p

    -1/5 [-1 0; 0 5]

    and then what?
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    (Original post by Carlo08)
    I have the matrix [2 3;3 2] and wish to diagnoalize it.

    First I find the eigenvalues of : -1 and 5 and put them into a matrix as:

    P=[-1 0;0 5]

    Then find the eigenvectors for each value and put them into a matrix
    D= [1 1; -1 1]

    Then find the inverse of p

    -1/5 [-1 0; 0 5]

    and then what?
    You've diagonalised it. It may be better to call your diagonal matrix of eigenvalues D and your invertible matrix of eigenvectors P

    Then, calling your matrix A, you have A=PDP^{-1}

    You may want to check that AP=PD and that P is invertible (e.g. has determinant 0 non-zero determinant).

    By the way, it isn't the inverse of the diagonal matrix of eigenvalues you require, instead the invertible matrix of eigenvectors.
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    (Original post by TheEd)
    You've diagonalised it. It may be better to call your diagonal matrix of eigenvalues D and your invertible matrix of eigenvectors P

    Then, calling your matrix A, you have A=PDP^{-1}

    You may want to check that AP=PD and that P is invertible (e.g. has determinant 0).

    By the way, it isn't the inverse of the diagonal matrix of eigenvalues you require, instead the invertible matrix of eigenvectors.
    * non zero
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    (Original post by DeanK22)
    * non zero
    Correct. P should have non-zero determinant so that it is invertible
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    Just to elaborate on what TheEd said (because understanding the concept means it's easier to remember):

    The idea of a diagonalised matrix is that it takes a two particular vectors (the eigenvectors) and scales them.

    Now, as it is, the matrix \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} tells you that it sends \begin{pmatrix}1 \\ 0 \end{pmatrix} to \begin{pmatrix} 2 \\ 3 \end{pmatrix} and \begin{pmatrix}0 \\ 1 \end{pmatrix} to \begin{pmatrix} 3 \\ 2 \end{pmatrix}. However, you also know that it sends \begin{pmatrix}1 \\ 1 \end{pmatrix} to 5 \times \begin{pmatrix} 1 \\ 1 \end{pmatrix} and \begin{pmatrix}-1 \\ 1 \end{pmatrix} to -1 \times \begin{pmatrix} -1 \\ 1 \end{pmatrix}, so what you want to get is a matrix which:

    1. Sends \begin{pmatrix}1 \\ 0 \end{pmatrix} to \begin{pmatrix} 1 \\ 1 \end{pmatrix} and \begin{pmatrix}0 \\ 1 \end{pmatrix} to \begin{pmatrix} -1 \\ 1 \end{pmatrix} (i.e. it sends the basis vectors to the eigenvectors)
    2. Scales these two vectors appropriately
    3. Takes you back to where you started

    Since these both do the same thing to both the basis vectors, they must be the same thing, so they're equal.

    This is why when you diagonalise you get something in the form A=P^{-1}DP, where D is diagonal: the matrix P sends your basis vectors to the eigenvectors of the matrix, then D scales them, and finally P^{-1} takes you back to where you were.
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    (Original post by nuodai)
    Just to elaborate on what TheEd said (because understanding the concept means it's easier to remember):

    The idea of a diagonalised matrix is that it takes a two particular vectors (the eigenvectors) and scales them.

    Now, as it is, the matrix \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix} tells you that it sends \begin{pmatrix}1 \\ 0 \end{pmatrix} to \begin{pmatrix} 2 \\ 3 \end{pmatrix} and \begin{pmatrix}0 \\ 1 \end{pmatrix} to \begin{pmatrix} 3 \\ 2 \end{pmatrix}. However, you also know that it sends \begin{pmatrix}1 \\ 1 \end{pmatrix} to 5 \times \begin{pmatrix} 1 \\ 1 \end{pmatrix} and \begin{pmatrix}-1 \\ 1 \end{pmatrix} to -1 \times \begin{pmatrix} -1 \\ 1 \end{pmatrix}, so what you want to get is a matrix which:

    1. Sends \begin{pmatrix}1 \\ 0 \end{pmatrix} to \begin{pmatrix} 1 \\ 1 \end{pmatrix} and \begin{pmatrix}0 \\ 1 \end{pmatrix} to \begin{pmatrix} -1 \\ 1 \end{pmatrix} (i.e. it sends the basis vectors to the eigenvectors)
    2. Scales these two vectors appropriately
    3. Takes you back to where you started

    Since these both do the same thing to both the basis vectors, they must be the same thing, so they're equal.

    This is why when you diagonalise you get something in the form A=P^{-1}DP, where D is diagonal: the matrix P sends your basis vectors to the eigenvectors of the matrix, then D scales them, and finally P^{-1} takes you back to where you were.

    Thank you both. So D is actually the diagonalized matrix?

    The question I am given asks for the modal matrix is this the P? And then it asks me to find the product : PAP-1 so is this then giving me D?
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    (Original post by Carlo08)
    Thank you both. So D is actually the diagonalized matrix?

    The question I am given asks for the modal matrix is this the P? And then it asks me to find the product : PAP-1 so is this then giving me D?
    Yup, P is the modal matrix, i.e. the matrix whose columns are eigenvectors. It's sometimes called the "change of basis matrix" for other reasons.

    Because A=P^{-1}DP we must have PAP^{-1}=D, which is diagonal (assuming A can be diagonalised of course!)
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    (Original post by Carlo08)
    Thank you both. So D is actually the diagonalized matrix?

    The question I am given asks for the modal matrix is this the P? And then it asks me to find the product : PAP-1 so is this then giving me D?
    A is the diagonalised matrix, D is the diagonal matrix of eigenvalues, P is the modal matrix.

    So if the diagonalised form of A=P^{-1}DP then D=PAP^{-1} and the question is asking you to check this.
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    My lecture notes give me D = P^-1 A P

    So does that mean that A = PDP^-1 which is backwards of what is written above
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    (Original post by Carlo08)
    My lecture notes give me D = P^-1 A P

    So does that mean that A = PDP^-1 which is backwards of what is written above
    This is also correct (actually the way I learnt it and how I said it in my first post)
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    (Original post by TheEd)
    This is also correct (actually the way I learnt it and how I said it in my first post)
    Sorry for being dumb here but how does A=PDP-1 and = P-1 DP when matrix multiplication is dependent on the order.
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    (Original post by Carlo08)
    Sorry for being dumb here but how does A=PDP-1 and = P-1 DP when matrix multiplication is dependent on the order.
    Notice the difference between D=P^{-1}AP and A=PDP^{-1} (I got the Ps and P^{-1}s mixed up in my previous posts... oops).
 
 
 
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