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Below are two supposed proofs for the Cayley-Hamilton theorem. The first is a bogus proof; I would really appreciate comments as to whether my explanation as to why it is a bogus proof is correct.

The second proof is more of a standard Cayley-Hamilton proof. Again, comments regarding any mistakes would be appreciated. Thanks.

Theorem: Let $A$ be a square matrix over a commutative ring, then $A$ satisfies its own characteristic polynomial.

Bogus Proof

$p_A(\lambda)=\det(\lambda I-A)$ and substituting $A$ for $\lambda$, $p_A(A)=\det(AI-A)=\det(A-A)=\det(0)=0$.

Any proof that substitutes $A$ for $\lambda$ in $p_A(\lambda)$ is incorrect. $\lambda I-A$ is a polynomial matrix with entries of polynomials in the variable $\lambda$. $\lambda I-A$ therefore takes its entries from a polynomial ring $\mathbb F[\lambda]$, where $\mathbb F$ is the field of coefficients and $\lambda$ is the fixed symbol in polynomials in $\mathbb F[\lambda]$. We therefore cannot substitute $A$ for $\lambda$.

Proof:

Take the identity $\det(\lambda I_n-A)I_n=(\lambda I_n-A)=\operatorname{adj}(\lambda I_n-A)$

By definition $p_A(\lambda):=(\lambda I_n-A)$, therefore $p_A(\lambda)I_n=(\lambda I_n-A)\operatorname{adj}(\lambda I_n-A)\;$

The LHS of $*$ can be written as a linear combination of constant matrices. By definition, $p_A(\lambda)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c\lambda+c_0$, therefore $p_A(\lambda)I_n=\lambda^nI_n+c_{n-1}\lambda^{n-1}I_n+\cdots+c\lambda I_n+c_0I_n$

The RHS of $*$ can also be written as a linear combination of constant matrices. $\operatorname{adj}(\lambda I_n-A)$ is a polynomial matrix and can therefore be expressed as a linear combination of constant matrices. As the entries of $\operatorname{adj}(\lambda I_n-A)$ are the minors of the matrix $\lambda I_n-A$, the entries of $\operatorname{adj}(\lambda I_n-A)$ are polynomials of degree $n-1$ or less. Therefore:

$\operatorname{adj}(\lambda I_n -A)=\lambda^{n-1} B_{n-1}+\lambda^{n-2} B_{n-2}+\cdots+\lambda^1 B_1+\lambda^0 B_0=\displaystyle \sum_{i=0}^{n-1} \lambda^i B_i$

Using this to expand the RHS of $*$:

$(\lambda I_n -A)\displaystyle \sum_{i=0}^{n-1} \lambda^i B_i=\displaystyle \sum_{i=0}^{n-1} \lambda I_n \lambda^i B_i-\displaystyle \sum_{i=0}^{n-1} A\lambda^i B_i=\displaystyle \sum_{i=0}^{n-1} \lambda^{i+1} B_i-\displaystyle \sum_{i=0}^{n-1} \lambda^i AB_i=\displaystyle \sum_{i=0}^{n-1} (\lambda^{i+1} B_i-\lambda^i AB_i)$

$=\lambda^n B_{n-1}+\displaystyle \sum_{i=1}^{n-1} \lambda^i (B_{i-1}-AB_i)-AB_0$

Now for $*$ both sides are polynomials (linear combinations of constant matrices with $\lambda^i$ as variables). When two polynomials are equal their coefficients are equal; equating coefficients of $\lambda^i$:

\begin{aligned}\lambda^n:&\;&I_n&=B_{n-1}\\\\\lambda^{n-1}:&\;&\;c_{n-1}I_n&=B_{n-2}-AB_{n-1}\\\\\vdots&\;&\;&\;\;\vdots\\\\ \lambda^1:&\;&c_1I_n&=B_0-AB_1\\\\ \lambda^0:&\;&\;\;c_0I_n&=-AB_0 \end{aligned}

As the coefficients are equal, the sum of the LHS of coefficients is equal to the sum of the RHS:

$I_n+c_{n-1}I_n+\ldots+c_1I_n +c_0I_n=B_{n-1}+B_{n-2}-AB_{n-1}+\ldots+B_0-AB_1-AB_0$

Multiplying both sides by $A^i$ the equality holds:

$A^n+c_{n-1}A^{n-1}+\ldots+c_1A+c_0I_n=A^nB_{n-1}+A^{n-1}B_{n-2}-A^nB_{n-1}+\ldots+AB_0-A^2B_1-AB_0$

The LHS of $**$ is the characteristic polynomial of $A$, $p_A(A)$, and the RHS of $**$ is a telescoping sum that equals the zero matrix. Therefore, $A$ satisfies its own characteristic polynomial.

Q.E.D.

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  • $\begingroup$ What is the difference to your earlier question? math.stackexchange.com/questions/3571379/… $\endgroup$ – daw Mar 18 '20 at 17:47
  • $\begingroup$ What I wrote before was incorrect, hopefully the above is correct but I'm not sure. $\endgroup$ – VN7 Mar 18 '20 at 21:42
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    $\begingroup$ Another way to expose the bogus proof is to note $p_A(A)$ has to be a matrix, so can't be the scalar $0$, or the determinant of any matrix. $\endgroup$ – J.G. Mar 18 '20 at 23:08
  • $\begingroup$ Thanks J.G. that's correct for this bogus proof. There are other bogus proofs where $p_A(A)$ does equal the zero matrix where we substitute $A$ for $\lambda$. For example if we take the identity $p_A(\lambda)I_n=(\lambda I_n-A)$det$(\lambda I_n-A)$ and substitute $A$ for $\lambda$, we get $p_A(A)$ equalling the zero matrix, and not a scalar, however this is still not correct. $\endgroup$ – VN7 Mar 19 '20 at 13:46
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I'll just address the first question, about the bogus proof.

Arguing why a wrong proof is wrong is always very tricky, especially if the wrong proof is a wrong proof of a correct statement. So, obviously, arguing why an argument that another proof of a correct statement is wrong is itself wrong is doubly tricky, especially when the claim that the wrong proof of the correct statement is wrong is indeed correct. Still I'm gonna give it a shot...

First: I don't think your argument that the bogus proof is wrong is itself wrong, just that it is perhaps incomplete. I hope that I can explain why.

The core of you argument is that 'you can't substitute $A$ for $\lambda$'. The problem is: how do you counter the argument by someone who (falsely) believes the bogus proof is correct and answers 'What do you mean I can't, I just did'.

What I mean by 'just did' is that the notion of substituting $A$ for $\lambda$ is an essential part in the statement of the Cayley-Hamilton theorem. You have this polynomial $p(\lambda)$, you substitute $A$ for $\lambda$ and the claim is that you get $0$.

So somewhere you should explain how this 'allowed' form of substituting $A$ for $\lambda$ (I have a polynomial, i.e. element of $F[\lambda]$, I substitute $A$ for $\lambda$ and get a matrix with coefficients in $F$) is different from the 'not allowed' substitution that gives let's you substitute $A$ into $\det(A - \lambda I)$ and get $\det(A - AI)$.

You do give some explanation for this, but I think it could be more elaborate. As I understand it, your argument is: 'suppose I want to make the substitution [of $A$ for $\lambda$] first in the expression $A - \lambda I$ and then, after I did this I take the $\det$ of the result. Then I find that I already cannot make this first step, because $A - \lambda I$ is a matrix whose coefficients are polynomials in $\lambda$ and so substituting $A$ for $\lambda$ yields a matrix whose coefficients are matrices, which is a very weird notion that doesn't make sense'.

Now someone who believes in the bogus proof might object to this in two ways.

Case 1. Someone objects to the first sentence. He might say: 'well you may want to make the substitution first in $A - \lambda I$ and then compute the determinant of the result, but I first compute this determinant (which is a polynomial in $\lambda$) and then substitute $A$.' In a sense this person is right: that is what the Cayley-Hamilton theorem is talking about. But in the end you are more right, so you should try and convince this person that substituting into $A - \lambda I$ first and then taking the determinant is not your personal preference but really what the bogus proof is doing. I would recommend to make this part more explicit.

Case 2: the person might already understand this but have a different, more subtle objection. I think countering this is what really is missing from your (otherwise correct) refutation of the bogus proof. The person might say: 'well you think about about $A - \lambda I$ as a (single) matrix whose coefficients are (many) polynomials. But I think about the same thing as a (single) polynomial, whose (many) coefficients are matrices, i.e. an element of the ring $R[\lambda]$ where $R$ is the ring of all matrices of the correct size. Now both perspectives are equally valid in the sense that the ring of matrices whose entries are polynomials and the ring of polynomials whose coefficients are matrices are isomorphic, but from my perspective there is nothing wrong with substituting $A$ or any other matrix for $\lambda$: whenever you have a a polynomial in $\lambda$ over some ring $R$ you can substitute elements of $R$ into it. After all that's what we learn in high school that a polynomial is: a recipe of how to multiply a given element of $R$ with itself and other, given elements, and add up the outcomes to end up with a new element of $R$.'

I find this sounding quite convincing. But of course something is wrong: if the perspectives are indeed equivalent we can't have that substitution is possible in one perspective and not in the other. So I think your explanation of what is wrong with the bogus proof is stronger if you can add some argument that the substitution cannot be made regardless of whether you think about $A - \lambda I$ as a matrix (whose entries are polynomials) or as a polynomial (whose coefficients are matrices).

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  • $\begingroup$ Thanks Vincent for taking the time to answer, that's great. $\endgroup$ – VN7 Mar 19 '20 at 13:40

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