Cayley Hamilton Theorem Intuition Why should, intuitive (not a formal proof, just motivations ) be true that the square matrix satisfy its own characteristic equation? 
 A: Some may consider this "intuition" sacrilegious, but many people find the Cayley-Hamilton theorem intuitive because 
$$
\chi_A(A)=\det(A\cdot I-A)=\det(A-A)=\det(O_{n\times n})=0
$$
Of course, I say this may viewed as sacrilegious because this is a notorious bogus proof of the theorem.
A: I know I'm late to the scene, but I think my proof makes the theorem intuitive.

Intuition For any endomorphism $\Phi$, we have a factorization of the determinant $\text{det}(\Phi)$ into the adjugate and the matrix itself:
$$ \text{det}(\Phi) = \text{adj}(\Phi) \Phi$$
This is related to the factorization of the characteristic polynomial $p(t)$ of $\phi$ into $f(t)(t-\phi)$:
$$p(t) = f(t)(t - \phi)$$
These two factorizations are analogous, and in fact, if we get the formality right, we can view these as corresponding factorizations in isomorphic rings.

This correspondence also tells you how to prove the theorem, since we already have the factorization of determinant into the adjugate and the matrix itself. The part that remains to be explained is, "which rings do we choose to make this work"?

Theorem: Let $V$ be a finitely generated $k$-module. If $\phi : V \rightarrow V$ is a $k$-linear map, then the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$. In particular, $\text{End}_k (V)$ is integral over $k$.

Proof: Viewing $t - \phi$ as a $k[t]$-endomorphism of $V \otimes_k k[t]$, we have $\text{det} (t - \phi) 1_{V \otimes_k k[t]} = \text{adj}(t - \phi) (t - \phi)$ as elements of $\text{End}_{k[t]} (V \otimes_k k[t])$, where $\text{adj}(t - \phi)$ is the adjugate matrix. The isomorphism $\text{End}_{k[t]} (V \otimes_k k[t]) \rightarrow \text{End}_{k} (V)[t]$ sends $\text{det} (t - \phi) 1_{V \otimes_k k[t]}$ to $\text{char}(\phi)$, and $t - \phi$ to $t - \phi$. Since $t - \phi$ divides $\text{det} (t - \phi) 1_{V \otimes_k k[t]}$ in $\text{End}_{k}(V)[t]$, $t - \phi$ divides $\text{char}(\phi)$ in $\text{End}_{k}(V)$. So $\text{char}(\phi)$ has $\phi$ as a root in $\text{End}_k(V)$, so that the evaluation homomorphism $\text{ev}_{\phi} : k[t] \rightarrow \text{End}_k (V)$ sends the characteristic polynomial $\text{char}(\phi)$ to $0$.
A: This is an expansion of the idea in Arturo's comment.
Assume that we know the spectrum of $A$. Then the characteristic polynomial is
$$\eqalign{
p(\lambda) &= \prod_{k=1}^n \big(\lambda-\lambda_k\big) \cr
}$$
Evaluate the polynomial for $A$, and multiply by any eigenvector of $A$.
$$\eqalign{
p(A)\,v_j
 &= \prod_{k=1}^n \big(A-\lambda_kI\big)v_j \cr
 &= \prod_{k=1}^n \big(\lambda_j-\lambda_k\big)v_j =0 \cr
}$$
This is not a complete proof, but it strongly suggests that 
$\,\,p(A)=0$
