The question is motivated from the following problem:

Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is
$$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$

Since $A=A^{-1}$, $A^2=I$. If the converse of Cayley-Hamilton Theorem is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.

Here are my questions:

  1. Is $C$ the answer to the quoted problem?
  2. Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?

No, the converse of Cayley-Hamilton is not true for $n\times n$ matrices with $n\gt 1$; in particular, it fails for $2\times 2$ matrices.

For a simple counterexample, notice that if $p(A)=0$, then for every multiple $q(x)$ of $p(x)$ you also have $q(A)=0$; so you would want to amend the converse to say "if $p(A)=0$, then $p(a)$ is a multiple of the characteristic polynomial of $A$". But even that amended version is false

However, the only failure in the $2\times 2$ matrix case are the scalar multiples of the identity. If $A=\lambda I$, then $p(x)=x-\lambda$ satisfies $p(A)=0$, but the characteristic polynomial is $(x-\lambda)^2$, not $p(x)$.

For bigger matrices, there are other situations where even this weakened converse fails.

The concept that captures the "converse" of Cayley-Hamilton is the minimal polynommial of the matrix, which is the monic polynomial $p(x)$ of smallest degree such that $p(A)=0$. It is then easy to show (using the division algorithm) that if $q(x)$ is any polynomial for which $q(A)=0$, then $p(x)|q(x)$. (Be careful to justify that if $m(x)=r(x)s(x)$, then $m(A)=r(A)s(A)$; this is not immediate because matrix multiplication is not in general commutative!) So we have:

Theorem. Let $A$ be an $n\times n$ matrix over $\mathbf{F}$, and let $\mu(x)$ be the minimal polynomial of $A$. If $p(x)\in \mathbf{F}[x]$ is any polynomial such that $p(A)=0$, then $\mu(x)$ divides $p(x)$.

The Cayley-Hamilton Theorem shows that the characteristic polynomial is always a multiple of the minimal polynomial. In fact, one can prove that every irreducible factor of the characteristic polynomial must divide the minimal polynomial. Thus, for a $2\times 2$ matrix, if the characteristic polynomial splits and has distinct roots, then the characteristic and minimal polynomial are equal. If the characteristic polynomial is irreducible quadratic and we are working over $\mathbb{R}$, then again the minimal and characteristic polynomials are equal. But if the characteristic polynomial is of the form $(x-a)^2$, then the minimal polynomial is either $(x-a)$ (when the matrix equals $aI$), or $(x-a)^2$ (when the matrix is not diagonalizable).

As for solving this problem: if $\lambda$ is an eigenvalue of $A$, and $A$ is invertible, then $\lambda\neq 0$, and $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$: for if $\mathbf{x}\neq\mathbf{0}$ is such that $A\mathbf{x}=\lambda\mathbf{x}$, then multiplying both sides by $A^{-1}$ we get $\mathbf{x} = A^{-1}(\lambda \mathbf{x}) = \lambda A^{-1}\mathbf{x}$. Dividing through by $\lambda$ shows $\mathbf{x}$ is an eigenvector of $A^{-1}$ corresponding to $\frac{1}{\lambda}$.

Since $A=A^{-1}$, that means that if $\lambda_1,\lambda_2$ are the eigenvalues of $A$, then $\lambda_1 = \frac{1}{\lambda_1}$ and $\lambda_2=\frac{1}{\lambda_1}$; thus, each eigenvalue is either $1$ or $-1$.

If the matrix is diagonalizable, then we cannot have both equal to $1$ (since then $A=I$), and they cannot both be equal to $-1$ (since $A\neq -I$), so one eigenvalue is $1$ and the other is $-1$. Since the trace of a square matrix equals the sum of its eigenvalues, the sum of the eigenvalues is $0$.

Why is $A$ diagonalizable? If it has two distinct eigenvalues, $1$ and $-1$, then there is nothing to do; we know it is diagonalizable. If it has a repeated eigenvalue, say $1$, but $A-I$ is not the zero matrix, pick $\mathbf{x}\in \mathbb{R}^2$ such that $A\mathbf{x}\neq \mathbf{x}$; then $$\mathbf{0}=(A-I)^2\mathbf{x} = (A^2-2A + I)\mathbf{x} = (2I-2A)\mathbf{x}$$ by the Cayley Hamilton Theorem. But that means that $2(A-I)\mathbf{x}=\mathbf{0}$, contradicting our choice of $\mathbf{x}$. Thus, $A-I=0$, so $A=I$ and $A$ is diagonalizable. A similar argument shows that if $-1$ is the only eigenvalue, then $A+I=0$. . (Hiding behind that paragraph is the fact that if the minimal polynomial is squarefree and splits, then the matrix is diagonalizable; since $p(x)=x^2-1=(x-1)(x+1)$ is a multiple of the minimal polynomial, the matrix must be diagonalizable).

So this completes the proof that the trace must be $0$, given that $A\neq I$ and $A\neq -I$.

  • $\begingroup$ Why couldn't it be that $\lambda_1 = \frac{1}{\lambda_2}$ and $\lambda_2 = \frac{1}{\lambda_1}$ ? $\endgroup$ – Aritra Das Apr 29 '16 at 6:42
  1. If $A^2 = 1$ then the eigenvalues of $A$ satisfy $\lambda^2 = 1$, so they are either $+1$ or $-1$. As they cannot both be $+1$ or $-1$, we must have one each, and their sum (the trace) is $0$.

  2. If $p(A) = 0$ then $p(A)$ is divisible by the minimal polynomial of $A$. As an extreme example, take $A=0$. Then $p(A) = 0$ for lots of polynomials, but the characteristic polynomial is $x \mapsto x^{\dim A}$.


The multiple choice way to answer this is to note that $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ is an example with trace $0$, so if the question is valid then the answer must be C.

If $p$ is a polynomial such that $p(A)=0$, then it is true that every eigenvalue of $A$ is a zero of $p$. For if $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $0=p(A)v=p(\lambda)v$, which implies $p(\lambda)=0$. From this, you know as Yuval already pointed out that the possible eigenvalues are $1$ and $-1$.

The possible characteristic polynomials are thus $x^2-1$, $(x-1)^2$, and $(x+1)^2$. To rule out the last two cases, you can consider the triangular forms of $A$. For example, having characteristic polynomial $(x-1)^2$ implies that $A$ is similar to a matrix of the form $\begin{bmatrix}1&a\\0&1\end{bmatrix}$. But then the only way for $A=A^{-1}$ to be true would be if $a=0$, contradicting the hypothesis that $A\neq I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.