Is there a way to prove the Cayley-Hamilton Theorem without the use of cofactors, adjoints, etc?
Like is there another way to natural prove general matrix will satisfy its own characteristic polynomial?
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1This is an interesting question. – hamam_Abdallah Nov 15 '20 at 23:28
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Axler's book does this. https://linear.axler.net/ – Will Jagy Nov 15 '20 at 23:34
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Howard Straubing, A combinatorial proof of the Cayley-Hamilton theorem. – darij grinberg Nov 15 '20 at 23:43
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1You can treat the theorem as a universal polynomial identity, which allows you to reduce the proof the case of diagonalizable matrices over $\mathbf C$, where the calculation is very simple. See Theorem 3.4 in https://kconrad.math.uconn.edu/blurbs/linmultialg/univid.pdf. – KCd Nov 15 '20 at 23:44
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Yes. The characteristic polynomial of $A$ maps every eigenvector to $0$, so you get $p(A)=0$ directly for diagonalizable matrices because eigenvectors form a basis. In general, approximate $A$ by diagonalizable matrices and note that their polynomials converge to its in the limit. – Conifold Nov 16 '20 at 00:52
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If done in the right way, one can simply substitute $A$ for $x$ in $\rm{det}(xI-A)=0$. Look at my question On the Cayley-Hamilton theorem