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.

-
You mean the adjugate matrix. "Adjacency matrix" means something else (en.wikipedia.org/wiki/Adjacency_matrix). – Qiaochu Yuan Apr 17 '13 at 20:27
Already edited. Thanks. – Andrew Apr 17 '13 at 20:57

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.
Collect all terms involving $a_{ij}$. Make sure you get the sign right. – hardmath Apr 17 '13 at 20:05
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