## A property of 3×3 orthogonal matrices with determinant -1

Posted: May 22, 2018 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem. Let $A \in M_3(\mathbb{R})$ be orthogonal and suppose that $\det(A)=-1.$ Find $\det(A-I).$

Solution. Since $A$ is orthogonal, its eigenvalues have absolute value $1$ and it can be be diagonalized. Let $D$ be a diagonal matrix such that $PDP^{-1}=A$ for some invertible matrix $P \in M_3(\mathbb{C}).$ Then

$\det(D)=\det(A)=-1, \ \ \det(D-I)=\det(A-I).$

We claim that the eigenvalues of $A$ are $\{-1,e^{i\theta},e^{-i\theta}\}$ for some $\theta.$ Well, the characteristic polynomial of $A$ has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since $\det(A)=-1,$ the eigenvalues of $A$ are either all $-1,$ which is the case $\theta=\pi,$ or two of them are $1$ and one is $-1,$ which is the case $\theta = 0,$ or one is $-1$ and the other two are in the form $\{e^{i\theta}, e^{-i\theta}\}$ for some $\theta.$ So

\displaystyle \begin{aligned} \det(D-I)=\det(A-I)=(-2)(e^{i\theta}-1)(e^{-i\theta}-1)=-4(1-\cos \theta). \ \Box \end{aligned}

Note that given $\theta,$ the matrix

$\displaystyle A=\begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}$

is orthogonal, $\det(A)=-1$ and $\det(A-I)=-4(1-\cos \theta). \ \Box$