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I have some problems understanding the proof of the Caley-Hamilton theorem (saying that a matrix the root of ith characteristic polynomial), namely:

Why $A \cdot A^D = A^D \cdot A = \det A \cdot I$ ?

($A^D $ is $A$'s adjecency matrix and $I$ - identity matrix)

$A^D = [a_{ij}], \ \ a_{ij} = (-1)^{i+j} \cdot \det A_{ji}$, $A_{ji}$ - cofactor of $A$

Could you explain that to me?

Thank you.

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You mean the adjugate matrix. "Adjacency matrix" means something else ( – Qiaochu Yuan Apr 17 '13 at 20:27
Already edited. Thanks. – Andrew Apr 17 '13 at 20:57
up vote 1 down vote accepted

Note the fact $\det A=\sum_j \pm a_{ij} \det A_{ij}$ by expanding $\det A$ in row $i$, and that $0=\sum_j \pm a_{ij} \det A_{kj}$ for $k\neq i$, since row $i$ and row $k$ are the same.

Now you can prove the formula by expanding everything.

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Yes, I know that. But I still don't see how to prove the equality. Could you write a bit more? – Andrew Apr 17 '13 at 19:56
Collect all terms involving $a_{ij}$. Make sure you get the sign right. – hardmath Apr 17 '13 at 20:05
@Andrew I added another formula will be used in the proof. – Ma Ming Apr 17 '13 at 20:07
Thanks for the second formula, that was my main problem, because if we denote $AA^D = B$, then $b_{11} = \sum _{k} a_{1k} (-1)^{k+1} det A_{1k}$ and everything's fine, but $b_{12} = \sum _{k} a_{1k} (-1)^{k+2} det A_{2k}$ but now I see that is $0$. Thanks again. – Andrew Apr 17 '13 at 20:16
By the way, isn't there a typo in your answer. Shouldn't there be row i? – Andrew Apr 17 '13 at 20:19

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