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## Cayley Hamilton Theorem alternative proof

If $p=\det(A-\lambda I)$ is the characteristic polynomial of a square matrix (linear operator) $A$, then $p(A)=0$.

Proof: If all the eigenvalues of $A$ are distinct, then there is a base in which $A$ has diagonal matrix, which means that there exists a diagonal matrix consisting of eigenvalues of $A$, $D=diag(\lambda_1,...,\lambda_n)$ and an invertible matrix $T$ such that $A=TDT^{-1}$. Then if $q$ is a polynomial we have $q(A)=T\cdot diag(q(\lambda_1),...,q(\lambda_n)) \cdot T^{-1}$. If $q=p$ then obviously $p(A)=0$.

If $A$ does not have distinct eigenvalues, then, using this property there exists a sequence $(A_r)_{r \geq 1}$ of matrices which converges to $A$ and each $A_r$ has distinct eigenvalues. Therefore $p(A_r) \to p(A)$, since a polynomial is a continuous function. Now take for every matrix $A_r$ its characteristic polynomial $p_r$. Then $p_r(A_r)=0$ and furthermore, since the coefficients of $p_r$ are continuous functions of elements of $A_r$, we have $p_r(X)-p(X) \to 0$ for every matrix $X$, when $r \to \infty$. This means that $p_r(A_r)-p(A_r)\to 0$ when $r \to \infty$, from which we conclude that $p(A_r) \to 0$ and $p(A_r) \to p(A)$ when $r \to \infty$. Therefore $p(A)=0$.