Let A = [-3,2,6 0,-1,6 0,0,-3] Find an invertible matrix P and a diagonal matrix D such that D = P^{-1}AP. find P and D I worked out the polynomial -7x^3-7x^2-15x-9 to give x= -1,-3,-3 My answer for P= [1,0,0 1,-3,-3 0,1,1] and D= -1,0,0 0,-3,0 0,0,-3 something's wrong!

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Let A = [-3,2,6 0,-1,6 0,0,-3] Find an invertible matrix P and a diagonal matrix D such that D = P^{-1}AP. find P and D I worked out the polynomial -7x^3-7x^2-15x-9 to give x= -1,-3,-3 My answer for P= [1,0,0 1,-3,-3 0,1,1] and D= -1,0,0 0,-3,0 0,0,-3 something's wrong!

Linear Algebra
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what type of decomposition are you doing here ?
I think you get something wrong at P, since eigenvalue -3 has 2 dimensions and they are not yours! recheck
you should get D=dia(-3,-1,-3)

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yes, surely, because they are eigenvalues of A, what I need is checking the order this guy arrange in P to have corresponding of numbers, like (-3,-1,-3) or (-1,-3,-3)
ok I will try!
We cannot guess, right, kid? if we mess the order up, although we know it is the Diagonal matrix of A but the P is in different order of eigenvectors, We will be crazy when checking what wrong is, I had that experience. It made me crazy
tr(A)=-3-1-3=-7 A11+A22+A33=3+9+3=15 |A|=-3(3)=-9 so, the characteristic polynomial would be\[\Delta(t)=t^3+7t^2+15t+9\]
well, as long as the eigen vector columns are arranged in proper order, you should be good.
still no luck...
no
eigen vectors=? for t=-1: \[-2x+2y+6z=0\\6z=0\\-3z=0\implies v_1=(1,1,0)^T\]
yup
and v2 and v3 I put at (0,-3,1)
for t= -3: \[ 2y+6z=0\\ 2y+6z=0\\ v_{2,3}=(\alpha,-3\beta,\beta)^T \]
@thdrbird you cannot have them two identical
oh,ok..
you can use v2 when a=0,b=1 v3 when a=1,b=1
or v3 when a=1,b=0 ANY combination of alpha and beta that do not give the same vectors..
can you give me an example?
got it!
I forgot to set alpha!
thanks @electrokid, great help!
@Hoa as well! Thanks!
I did nothing guy!!!

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