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Why should, intuitive (not a formal proof, just motivations ) be true that the square matrix satisfy its own characteristic equation?

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    $\begingroup$ follow the steps of the proof toward enlightenment and thou shalt believe! $\endgroup$ – Alvin Lepik Apr 6 '19 at 17:45
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    $\begingroup$ For diagonalizable operators, because you can find a basis of eigenvectors, and each eigenvector is annihilated by a (linear) factor of the characteristic polynomial. For non-diagonalizable ones but whose characteristic polynomial splits, for similar reasons relative to a basis of generalized eigenvectors. So you can guess that it will satisfy it if you think of the matrix as having coefficients in the algebraic closure of its field of definition, and hence will just do so in general. $\endgroup$ – Arturo Magidin Apr 6 '19 at 17:47
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    $\begingroup$ I think you will find this paper entitled "Down with determinants" very interesting and relevant. $\endgroup$ – Carl Christian Apr 6 '19 at 21:08
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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 $\,\,p(A)=0$

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    $\begingroup$ This is a great answer. Perfect level of detail. I understand the theorem much better now. $\endgroup$ – littleO Apr 6 '19 at 22:05
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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.

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I know I'm late to the scene, but I think this 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 $0$.

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