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Completing the basis of a matrix, Jordan form related 
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#19
Dec312, 10:24 PM

PF Gold
P: 3,188

It's over 1 am for me so I will analyze your last post tomorrow.
You made me remind of the way to get the Jordan form I've been taught. 1)First get all the eigenvalues of A. 2)Take an eigenvalue that you have not taken yet and do ##(A\lambda I)^m## for m=1, 2, etc. and calculate each time the dimension of the kernel of ##(A\lambda I)^m## until that number remains constant for some ##m_0 \leq n##. So you get ##dim (\ker (A\lambda I)^m)=dim (\ker (A\lambda I)^{m_0})## for all ##m \geq m_0##. For m=1, 2, ..., ##m_0##, calculate ##d_m=2 dim (ker [(A\lambda I)^m]) dim (ker [(A\lambda I)^{m1}]) dim (ker [(A\lambda I)^{m+1}])##. So that ##d_m >0## is equal to the number of times that the Jordan block ##J_m (\lambda )## of dimension mxm with lambda on its diagonal appear in the Jordan form. The geometric multiplicity of lambda is equal to the sum of the ##d_m##'s. The algebraic multiplicity is equal to the sum of the products ##md_m##. 3)Go to 2) until you've covered all lambda's. 


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