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anonymous
 3 years ago
Find all the eigenvalues for the given matrices.
if it is a 3x3. and a polynomial. how do i know what to break it down into?
anonymous
 3 years ago
Find all the eigenvalues for the given matrices. if it is a 3x3. and a polynomial. how do i know what to break it down into?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0[ 4 0 1 1 6 2 5 0 0 ] \[\lambda ^3+2 lambda ^2  29 lambda 30]/

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2\[\textbf Av=\lambda v\] \[\left(\textbf A\lambda \textbf I\right)v =0\] \[\left\textbf A\lambda \textbf I\rightv =0\] \[\left\left(\begin{array}\ 4&0&1\\1&6&2\\5&0&0\end{array}\right)\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right=0\] \[\left\left(\begin{array}\ 4\lambda&0&1\\1&6\lambda&2\\5&0&0\lambda\end{array}\right)\right=0\] \[(4\lambda)[2\times0(6\lambda)(0\lambda)]0[1\times(0\lambda)]+1[1\times0(2\times5)]=0\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2*\[(4\lambda)[2\times0(6\lambda)(0\lambda)]\]\[\qquad0[1\times(0\lambda)(5\times2)]\]\[\qquad\qquad+1[1\times0(6\lambda)]=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, i know how to do all the calculations but when you get the the last step: when its all combined how do you know what lamda itself is equal to? ( like, you look at the constant and see its factors or something and do something..idk)_)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2well at the last spet you'll have a bunch of factors =0 so to get zero at least one of the factors is zero

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2something like this only i have made mistakes \[(\lambda4)[(6+\lambda)\lambda]+[6+\lambda]=0\]\[(6\lambda+\lambda^2244\lambda)\lambda+6+\lambda=0\]\[\lambda^3+2\lambda^224\lambda+\lambda+6=0\]\[\lambda^3+3\lambda^224\lambda+6=0\]\[(\lambdaa)(\lambdab)(\lambdac)=0\] lambda would be a,b,c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but how do you know to break it into a b c?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2how to factorize? you could try polynomial long division

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lambda^3 +2\lambda^229\lambda30\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how would you go about doing this? possibilities: +/ : 1,2,3,5,6,10,30,15

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you plug in 1, you get =0.then what?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2your expression does factorizes nicely, try dividing by \((\lambda+1)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont know how... i knwo that sounds dumb but how do u do that?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1341885937030:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1341885983310:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2so\[\lambda^3 +2\lambda^229\lambda30=(\lambda+1)(\lambda^2+\lambda30)\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2now just factorize the quadratic factor...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks. also: do you know about diagnolization?
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