0
$\begingroup$

I have done two proofs of the Cayley-Hamilton theorem below. I would really appreciate it if anyone could let me know whether they are correct or not. Also if there are any errors what these are. Many thanks for your help.

Theorem: Let $A_{nxn}$ be a matrix over a commutative ring, then $p_A(A)$ equals the zero matrix; that is $A_{nxn}$ satisfies its own characteristic polynomial.

Proof 1. Taking the characteristic polynomial definition $p_A(\lambda)$$=$det$(\lambda I-A)$ and substituting $A$ for $\lambda$ supposedly gives $p_A(A)$$=$det$(AI-A)$$=$det$(A-A)$$=$det$(0)$$=$$0$, however zero here is a scalar whereas $p_A(A)$ is a matrix and therefore this is not proof. However, how about the proof below?

By another theorem, det$(\lambda I_n -A)I_n=(\lambda I_n -A)$ adj$(\lambda I_n -A)$. Therefore $p_A(\lambda)I_n=(\lambda I_n -A)$ adj$(\lambda I_n -A)$. Substituting $A$ for $\lambda$:

$p_A(A)I_n=(A I_n -A)$ adj$(A I_n -A)$$\space$$\space$*

LHS of *: $p_A(A)I_n=A^nI_n+c_{n-1}A^{n-1}I_n+\cdots+cAI_n+c_0I_nI_n=A^n+c_{n-1}A^{n-1}+\cdots+cA+c_0I_n=p_A(A)$

RHS of *: $(A I_n -A)$ adj$(A I_n -A)=(A-A)$ adj$(A-A)=(0)$ adj$(0)=(0)$, where $(0)$ is the zero matrix.

Therefore $p_A(A)$ equals the zero matrix and $A_{nxn}$ satisfies its own characteristic polynomial.

Q.E.D.


Proof 2. This is more of a standard Cayley Hamilton proof however I would appreciate comments on any mistakes or improvements.

By another theorem, det$(\lambda I_n -A)I_n=(\lambda I_n -A)$ adj$(\lambda I_n -A)$. Therefore $p_A(\lambda)I_n=(\lambda I_n -A)$ adj$(\lambda I_n -A)$. Substituting $A$ for $\lambda$:

$p_A(A)I_n=(A I_n -A)$ adj$(A I_n -A)$$\space$$\space$*

LHS of *: $p_A(A)I_n=A^nI_n+c_{n-1}A^{n-1}I_n+\cdots+cAI_n+c_0I_nI_n=A^n+c_{n-1}A^{n-1}+\cdots+cA+c_0I_n=p_A(A)$

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

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 * with $\lambda$:

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

Substituting $A$ for $\lambda$ this becomes a telescopic series:

$\displaystyle \sum_{i=0}^{n-1} (A^{i+1} B_i-A^i AB_i)=\displaystyle \sum_{i=0}^{n-1} (A^{i+1} B_i-A^{i+1} B_i)=(0)$, where $(0)$ is the zero matrix.

As a telescopic series the RHS of * equals the zero matrix. However we must also check that the * matrix equality is true. A matrix equality is true if each matrix has the same entries. Therefore for * each matrix corresponding to each $\lambda^i$ must be equal. By definition two polynomials are equal if their coefficients are equal. Therefore the coefficients of $\lambda^i$ are equal, therefore corresponding matrices are equal, therefore the matrix equality is true.

Q.E.D.

$\endgroup$

1 Answer 1

2
$\begingroup$

In my opinion, both proofs still have the same problem like the false proof $p_A(A)=\det(A\cdot I-A)=0$. It is more hidden.

The identity $$ p_A(t)I = (tI-A)adj(tI-A) $$ is fine: Here, all matrices are considered as matrices with entries in the polynomial ring $F[t]$, where $F$ is the underlying field. However, the matrix $A$ is not an element in this ring. So one cannot plug in $A$ for $t$ at any point in the proof.

The left-hand side of the above equation can be written as $p_A(t) = \sum_{k=0}^n p_k t^k I$. The right-hand side is as you wrote equal to $\sum_{k=0}^{n-1}(t^{k+1}B_i-t^k AB_k)$. This is an identity in $F[t]^{n,n}$. So we can compare entries and coefficients, and it follows that $p_kI = B_{k-1}-AB_k$ for $k=1\dots n-1$, $p_0I=-AB_0$, $p_nI=I=B_{k-1}$. These identities can now be multiplied with the appropriate powers of $A$ and summed to obtain the claim. This is essentially the computation in your second proof, but with some intermediate step.

$\endgroup$
1
  • $\begingroup$ Outstanding, thanks for explaining $\endgroup$
    – VN7
    Mar 6, 2020 at 10:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .