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Cuzkatzimhut (talk | contribs) |
Cuzkatzimhut (talk | contribs) |
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Line 51:
:<math>\sum_{k=1}^{n} \lambda^{k} \Big ( M_{1+n-k} - AM_{n-k} +c_k I\Big )= 0~,</math>
which thus dictates the recursion
:<math>\therefore \qquad M_{m} =A M_{m-1} +c_{n-m+1} I ~,</math>
for {{mvar|m}}=1,...,{{mvar|n}}. Note that ascending index amounts to descending in powers of {{mvar|λ}}, but the polynomial coefficients {{mvar|c}} are yet to be determined in terms of the {{mvar|M}}s and {{mvar|A}}.
Line 65:
:<math>\sum_{m=1}^{n-1} \lambda^{n-m} \Big ( m c_{n-m} + \operatorname{tr}A M_{m}\Big )= 0~,</math>
and finally
:<math>\therefore \qquad c_{n-m} = -\frac{1}{m} \operatorname{tr}A M_{m} ~.</math>
This completes the recursion of the previous section, unfolding in descending powers of {{mvar|λ}}.
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