jordan normal Watch

furryvision
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#1
Report Thread starter 8 years ago
#1
my matrix is:
\left[ \begin{array}{cccc} 4 & 1 & -4 \\ 6 & 2 & -7 \\ 6 & 1 & -6 \end{array} \right]

i've found the characteristic polynomial to be x^3 - 3x + 2, which gives the eigenvalues 1 and -2.

the corresponding eigenvector for 1 is (1, 1, 1)
the corresponding eigenvector for -2 is (1, 2, 2)

in the answer sheet it states that the algebraic multiplicity of the eigenvalue is 2.
the geometric multiplicity of the eigenvalue is 1.
are they talking about eigenvalue 1 or -2? also, how do we find these algebraic and geometric values?

the jordan normal matrix is then
\left[ \begin{array}{cccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{array} \right]
i have no idea how this jordan matrix was found!

any help pleeeeease? i have an exam coming up and i'm completely stuck! thank you!
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Zhen Lin
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Report 8 years ago
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I already answered this question last time, didn't I? It's the eigenvalue 1 that has algebraic multiplicity 2 and geometric multiplicity 1.
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