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As an arithmetic geometer, I have no choice but to use topological methods hand in hand with algebraic methods. Very likely necessity has been the mother of aesthetics here, but I find proofs of linear algebra facts using genericity arguments to be beautiful and insightful.

Qiaochu has shown how to answer the OP's question using these methods [he uses the "analytic" -- i.e., usual -- topology on $\mathbb{C}^n$, but close enough] assuming the Cayley-Hamilton theorem. Here I want to show that one can also prove the Cayley-Hamilton theorem quickly by these methods.

Step 1: To prove C-H as a polynomial identity, it is enough to prove that it holds for all $n \times n$ matrices over $\mathbb{C}$.

Proof: Indeed, to say C-H holds as a polynomial identity means that it holds for the generic matrix $A = \{a_{ij}\}_{1 \leq i,j \leq n}$ whose entries are independent indeterminates over the ring $R = \mathbb{Z}[a_{ij}]$. But this ring embeds into $\mathbb{C}$ -- indeed into any field of characteristic zero and infinite absolute transcendence degree -- and two polynomials with coefficients in a domain $R$ are equal iff they are equal in some extension domain $S$.

Step 2: C-H is easy to prove for complex matrices which have $n$ distinct eigenvalues.

Proof: Indeed, such matrices are diagonalizable. Since the characteristic polynomial is a conjugacy invariant, it is easy to reduce to the case of matrices which are diagonal A$with$n$distinct eigenvalues :$A = \operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Then the \lambda_1,\ldots,\lambda_n$.

Proof: The characteristic polynomial evalued evaluated at $A$ is $P(A) = \prod_{i=1}^n(\lambda_i I_n - A)$prod_{i=1}^n(A-\lambda_i I_n)$. Let$e_1,\ldots,e_n$be a basis of$\mathbb{C}^n$such that each$e_i$is an eigenvector for$A$with eigenvalue$\lambda_i$. Then -- using the fact that the matrices$A - \lambda_i I_n$all commute with each other -- we have that for all$e_i$,$P(A)e_i = \left(\prod_{j \neq i} (\lambda_j I_n - A A-\lambda_j I_n)\right) \right) (\lambda_i A-\lambda_i I_n- A) e_i = 0.$Since$P(A)$kills each basis element, it is in fact identically zero. Step 3: The set of complex matrices with$n$distinct eigenvalues is a Zariski-open subset of$\mathbb{C}^n$: indeed this is the locus of nonvanishing of the discriminant of the characteristic polynomial. Since we can write down diagonal matrices with distinct entries, it is certainly nonempty. Therefore it is Zariski dense, and any polynomial identity which holds on a Zariski dense subset of$\mathbb{C}^{n^2}$holds on all of$\mathbb{C}^{n^2}$. 1 As an arithmetic geometer, I have no choice but to use topological methods hand in hand with algebraic methods. Very likely necessity has been the mother of aesthetics here, but I find proofs of linear algebra facts using genericity arguments to be beautiful and insightful. Qiaochu has shown how to answer the OP's question using these methods [he uses the "analytic" -- i.e., usual -- topology on$\mathbb{C}^n$, but close enough] assuming the Cayley-Hamilton theorem. Here I want to show that one can also prove the Cayley-Hamilton theorem quickly by these methods. Step 1: To prove C-H as a polynomial identity, it is enough to prove that it holds for all$n \times n$matrices over$\mathbb{C}$. Proof: Indeed, to say C-H holds as a polynomial identity means that it holds for the generic matrix $A = \{a_{ij}\}_{1 \leq i,j \leq n}$ whose entries are independent indeterminates over the ring$R = \mathbb{Z}[a_{ij}]$. But this ring embeds into$\mathbb{C}$-- indeed into any field of characteristic zero and infinite absolute transcendence degree -- and two polynomials with coefficients in a domain$R$are equal iff they are equal in some extension domain$S$. Step 2: C-H is easy to prove for complex matrices which have$n$distinct eigenvalues. Proof: Indeed, such matrices are diagonalizable. Since the characteristic polynomial is a conjugacy invariant, it is easy to reduce to the case of matrices which are diagonal with$n$distinct eigenvalues:$A = \operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Then the characteristic polynomial evalued at$A$is$P(A) = \prod_{i=1}^n(\lambda_i I_n - A)$. Let$e_1,\ldots,e_n$be a basis of$\mathbb{C}^n$such that each$e_i$is an eigenvector for$A$with eigenvalue$\lambda_i$. Then -- using the fact that the matrices$A - \lambda_i I_n$all commute with each other -- we have that for all$e_i$,$P(A)e_i = \left(\prod_{j \neq i} (\lambda_j I_n - A )\right) (\lambda_i I_n - A) e_i = 0.$Since$P(A)$kills each basis element, it is in fact identically zero. Step 3: The set of complex matrices with$n$distinct eigenvalues is a Zariski-open subset of$\mathbb{C}^n$: indeed this is the locus of nonvanishing of the discriminant of the characteristic polynomial. Since we can write down diagonal matrices with distinct entries, it is certainly nonempty. Therefore it is Zariski dense, and any polynomial identity which holds on a Zariski dense subset of$\mathbb{C}^{n^2}$holds on all of$\mathbb{C}^{n^2}\$.