If $A$ is $2\times 2$ skew orthogonal with $A^TA=-I$ and $\det A=1$ then $\operatorname{tr} A=0$

Question

$A$ is skew orthogonal if $A^TA=-I$ in $\mathbb{Z_p}$ for $p>2$. The general form of $2\times 2$ characteristic polynomial: $x²-(\operatorname{tr} A)x+\det A$. It is given: $\det A=\pm 1$

If $\det A=-1$ then $\operatorname{tr} A= 0,1,-1$. But if $\det A=1$, then $\operatorname{tr} A$ must be $0$ only. What is the reason behind this?

Answer 1

Edited 27/01: As the OP has warned me privately, there was a mistake in a sign in the case $\det(A)=-1$. This makes a new type of solution to appear.

This answer includes also the case with $\det(A)=-1$.

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Let $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$. Then $A^T=\begin{pmatrix}a & c \\ b & d\end{pmatrix}$ and $$AA^T=\begin{pmatrix}a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2\end{pmatrix}=-I,$$

1) If $\det(A)=1$ then we have the identities $$\begin{align*} a^2+b^2&=-1 & (1)\\ c^2+d^2&=-1 & (2)\\ ac+bd&=0 & (3)\\ ad-bc&=1 & (4) \end{align*}$$ in $\mathbb{Z}_p$.

Multipying (3) by $c$ and (4) by $d$ we find $ac^2+bcd=0$, $ad^2-bcd=d$, so $$ac^2+ad^2=d\implies a(c^2+d^2)=d\implies -a=d$$ by (2) and therefore trace$(A)=a+d=-d+d=0$.

Observe that this proof is valid over any commutative unital ring, we have not used that it is $\mathbb{Z}_p$ in any way (nor that it is a field!).

2) If $\det(A)=-1$ then we have similar identities, changing (4) with $ad-bc=-1 \ (4')$. Then similar computations give us $a=d$ and $b=-c$.

Therefore $$A=\begin{pmatrix} a & b \\ -b & a\end{pmatrix}$$ with $a,b$ such that $a^2+b^2=-1$. Observe that this implies $AA^T=-I$, $\det(A)=-1$. Then trace$(A)=2a$, but this does not imply that trace$(A)\in\{-1,0,1\}$!

For example, we can pick $b=0$, $a^2=-1$. By quadratic reciprocity, such an $a$ exists in $\mathbb{Z}_p$ with $p$ prime if and only if $p\equiv 1(\text{mod}4)$. We can pick for example $p=13$. Then $8^2\equiv -1(\text{mod}{13})$, so $$A=\begin{pmatrix} 8 & 0 \\ 0 & 8\end{pmatrix}$$ satisfifes the hypotheses in $\mathbb{Z}_{13}$ but trace$(A)\equiv 16\equiv 3\not\equiv 0,\pm1(\text{mod}13)$.

Similarly, in $\mathbb{C}$ we can pick $a=i$ to get trace$(A)=2i\neq 0,\pm 1$.

Answer 2

Just for the sake of it, I'm going to give a proof which is more matrix-oriented and more generalizable.

Since $AA^T=-I$, $-A^T$ is a right inverse of $A$, so it is the inverse of $A$, $A^{-1}=-A^T$.

Now, if $p(X)$ is the characteristic polynomial of $A$ invertible of size $n\times n$, then $$q(X):=\frac{X^n}{(-1)^n\det(A)}p(X^{-1})$$ is the characteristic polynomial of $A^{-1}$ (since it is monic, of degree $n$, and has $A^{-1}$ as root).

In our case, by hypothesis $\det(A)=1$, so $p(X)=X^2-$trace$(A)X+1$ and hence $$q(X)=X^2-\text{trace}(A)X+1,$$ what implies that trace$(A^{-1})=$trace$(A)$. But trace$(A^{-1})=$trace$(-A^T)=-$trace$(A)$, so $$\text{trace}(A)=-\text{trace}(A)$$ and trace$(A)=0$ if there is no $2$-torsion.

Answer 3

For $\det(A)=1$, when we multiply both sides of the characteristic equation of $A$ by $A^T$, we get

$A^T(A^2-\text{tr}(A)A+\det(A)I)=0$

$A^TAA-\text{tr}(A)A^TA +A^T=0$

$-A+\text{tr}(A)I +A^T=0$

$ \text{tr}(A)I =A-A^T$, but $\text{tr}(A-A^T)=0$ $\Rightarrow \text{tr}(A)=0$.