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    given that A =

    1 0 4
    0 5 4
    4 4 3

    How do I verify that 2i – 2j +k is an eigenvector of A and find the corresponding eigenvalue??

    given that the third eigenvector of A is 2i + j – 2k, what’s the matrix P and diagonal matrix D such that P^T AP = D

    ??

    And can you explain that?

    EEEK...

    :eek:
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    to verify that 2i - 2j + k is an eigenvector, say v, you know that:

    Av = λv, where λ is v's corresponding eigenvalue

    so if you multiply the matrix by the eigenvector you get a multiple of the eigenvector, where the multiple is its eigenvalue
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    P is the matrix whose columns are orthogonal normalised eigenvectors. e.g. if an eigenvector is 2i - 2j + k, the first column in P is 2/3, -2/3, 1/3. They are over 3 since they have been normalised to give a unit vector by dividing by the modulus.

    The next two columns use the other two normalised eigenvectors which you should be able to find. Then just times P^T, A and P together to find D.

    It's a pain to do matrices on the computer so I'll let you do the working

    Hope this helps
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    how'd you find the other eigenvector? method?
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    determinant of the matrice (A - λI) = 0 where I is unit matrice
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    ooooooh, yeah, am I dumb or WOT?
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    oh and by the way, when you take det (A + bI) = 0, does it matter wot order you put the values in - I mean, if you get -2, 5, 2, does it matter if you put them in as 5, 2, -2? Cos I don't see how you tell what order they're in...
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    that's right... any order is alright.
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    oh great... that's a load off my mind... hehe
 
 
 
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