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# matrix Q P6 watch

1. 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...

2. 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
3. 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
4. how'd you find the other eigenvector? method?
5. determinant of the matrice (A - λI) = 0 where I is unit matrice
6. ooooooh, yeah, am I dumb or WOT?
7. 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...
8. that's right... any order is alright.
9. oh great... that's a load off my mind... hehe

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