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As most are surely aware, there are lots of proofs of the famous Cayley Hamilton Theorem.

I was told by a friend of a proof which is claimed to be rather direct and short. Its strategy goes as follows.

Let $V$ be a vector space over a field $F$ with dimension $n$. As in most proofs of $C-H$, we first notice that the theorem holds for diagonalizable matrices. Then we adjoin $n^2$ indeterminates to our field and consider its algebraic closure. Then, the $n\times n$ matrix with entries these indeterminates is diagonalizable. It follows that Cayley-Hamilton holds as a polynomial identity over our original field.

I am having some difficulty following these steps. In particular, do such indeterminates always exist so that we can adjoin them, and if so, what allows us to claim that there exist $n^2$ of these? If we take the field $\mathbb{C}$ for example, we know that it is algebraicaly closed by itself, i.e., it is its own closure.

Also, why is the resulting matrix diagonalizable-do we construct a basis of $F$ by using these intederminants to prove it? And finally what does it mean to hold as a polynomial identity over the original field?

If one could reference me to a source of the proof or explain some of its steps so that I could recreate it, I would be thankful!

Jean Marie
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MathematicianByMistake
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    How do you prove "easily", or whatever, that CH holds for diagonalizable matrices? I can't see it... – DonAntonio May 10 '16 at 19:38
  • @Joanpemo Neither can I.. Nor the other steps are really obvious to me I fear.. – MathematicianByMistake May 10 '16 at 19:44
  • The "proof" in yellow in your question is wrong: to work over algebraically closed field doesn't make all the matrices diagonalizable. That's just not true. – DonAntonio May 10 '16 at 19:52
  • The easiest proof I know of the complete CH theorem is for example here: http://algebra.math.ust.hk/eigen/06_jordan/lecture1.shtml .This is the same proof that in the book Linear Algebra by Lipschutz of the Schaum series. It uses the adjugate (or classical adjoint) matrix. – DonAntonio May 10 '16 at 20:09
  • @Joanpemo Thank you for the reference. I will look into it! – MathematicianByMistake May 10 '16 at 20:13
  • @Joanpemo I think I found a related result here-one that shows that indeed they are diagonilazible. Check this- http://math.stackexchange.com/questions/107945/diagonalizable-matrices-with-complex-values-are-dense-in-set-of-n-times-n-comp – MathematicianByMistake May 10 '16 at 20:21
  • Thank you. I can't see what you see in that link. For example, the matrix $;\begin{pmatrix}1&1\0&1\end{pmatrix};$ isn't diagonalizable over the complex numbers (alg. closed), and there are infinite examples of non diagonalizable matrices: all the ones for which their minimal polynomial has multiple roots, say. – DonAntonio May 10 '16 at 20:29

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