Faddeev–LeVerrier algorithm: Difference between revisions

Revision as of 14:44, 2 March 2016

Revision as of 16:21, 2 March 2016

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Thus,

:<math>

M_1= I ~, \quad c_{n-1} = - \mathrm{tr} A =-c_n \mathrm{tr} A ; \qquad </math>
:<math>M_2= A-I\mathrm{tr} A , \quad c_{n-2}=-\frac{1}{2}\Bigl(\mathrm{tr} A^2 -(\mathrm{tr} A)^2\Bigr) =

-\frac{1}{2} (c_n \mathrm{tr} A^2 +c_{n-1} \mathrm{tr} A) ;

</math>

:<math>M_3= A^2-A\mathrm{tr} A -\frac{1}{2}\Bigl(\mathrm{tr} A^2 -(\mathrm{tr} A)^2\Bigr) I, \qquad </math>
::<math>c_{n-3}=- \tfrac{1}{6}\Bigl( (\operatorname{tr}A)^3-3\operatorname{tr}(A^2)(\operatorname{tr}A)+2\operatorname{tr}(A^3)\Bigr)=-\frac{1}{3}(c_n \mathrm{tr} A^3+c_{n-1} \mathrm{tr} A^2 +c_{n-2}\mathrm{tr} A); </math>

etc.,<ref>Zadeh, Lotfi A. and Desoer, Charles A. (1963, 2008). ''Linear System Theory: The State Space Approach'' (Mc Graw-Hill; Dover Civil and Mechanical Engineering) ISBN 9780486466637 , pp 303&ndash;305</ref>

&nbsp; ...;

&nbsp; :<math>\cdots ; \qquad c_{n-m}= -\frac{1}{m}(c_n \mathrm{tr} A^m+c_{n-1} \mathrm{tr} A^{m-1}+...+c_{n-m+1}\mathrm{tr} A)= -\frac{1}{m}\sum_{k=1}^{m} c_{n-m+k} \mathrm{tr} A^k ~ ; \qquad \cdots ...</math>

Observe {{math|''A⁻¹ {{=}} − M_n /c₀'' {{=}} (−)^''n''−1''M_n''/det''A''}} terminates the recursion at {{mvar| λ}}. This could be used to obtain the inverse or the determinant of {{mvar|A}}.