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I am reading "Linear Algebra" by Ichiro Satake.
The author proved that $B_0,B_1,\dots,B_{n-1}$ commute with $A$.
But I think we don't need to prove that $B_0,B_1,\dots,B_{n-1}$ commute with $A$ to prove Cayley-Hamilton theorem.
I think if $C$ is any matrix which commutes with $A$, we can substitute $C$ for the variable $x$.
So, I think if $C$ is any matrix which commutes with $A$, $$f_A(C)=(C-A)(C^{n-1}B_0+C^{n-2}B_1+\dots+B_{n-1})$$ and $$f_A(C)=(B_0C^{n-1}+B_1C^{n-2}+\dots+B_{n-1})(C-A)$$ hold.
In particular, $A$ commutes with $A$ itself, so $$f_A(A)=(A-A)(A^{n-1}B_0+A^{n-2}B_1+\dots+B_{n-1})=O$$ and $$f_A(A)=(B_0A^{n-1}+B_1A^{n-2}+\dots+B_{n-1})(A-A)=O$$ hold.

Am I wrong?

Cayley-Hamilton Theorem

tchappy ha
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  • If $C$ is commute with $A$, does $C^nM_0+C^{n-1}M_{n-1}+\dots+M_0=(C-A)(C^{n-1}B_0+C^{n-2}B_1+\dots+B_{n-1})$ hold? I want to know that. – tchappy ha Dec 28 '21 at 14:39
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    Wiki gives a good explanation about polynomials with matrix coefficients and why commuting is important. – A.Γ. Dec 28 '21 at 14:39
  • @A.Γ. Thank you very much for your link. – tchappy ha Dec 28 '21 at 14:47
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    @tchappyha This depends on whether "substituting" $C$ into a polynomial of the form $g(x)=x^kM$ means $C^kM$ or $MC^k$. If $g(C)$ means $C^kM$, then yes, it suffices to require $C$ to commute with $A$ to guarantee that $f_A(C)=0$. However, if $g(C)$ means $MC^k$, the case is different. I guess the author wants to prove Cayley-Hamilton theorem in a way such that which convention to use is unimportant. – user1551 Dec 28 '21 at 14:53
  • @user1551 Thank you very much for your comment. – tchappy ha Dec 29 '21 at 03:40

1 Answers1

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To summarize: yes, one can get away with proving that $A$ and $B_k$ commute to the price of explaining how to understand a polynomial with matrix coefficients, e.g. all powers of $x$ to the left, prove that, in this case, for the first line in (7) to hold it is enough that $x$ and $A$ commute and, at the end, remark on that $p(A)$ could be understood as an ordinary polynomial (i.e. no need to collect all powers of $A$ to the left any more) as $A$ and $c_kI$ also commute. It is quite close to what the first (direct algebraic) proof in Wiki does implicitly.

It feels that to prove commutation of $A$ and $B_k$ (simply by comparing the coefficients in the first and the second lines in (7)) and forget the trouble is somewhat easier.

A.Γ.
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