[proofplan] The defining identity for an [orthogonal matrix](/page/Orthogonal%20Matrix) is $Q^\top Q = I_n$. Taking determinants of both sides and using the multiplicativity of the determinant, the determinant of a transpose, and the determinant of the identity matrix gives $(\det Q)^2 = 1$. Since $\det Q$ is a real number, this forces $\det Q = 1$ or $\det Q = -1$. [/proofplan] [step:Take determinants of the orthogonality identity] Since $Q \in O(n)$, by definition of the real [orthogonal group](/page/Orthogonal%20Group) we have \begin{align*}Q^\top Q = I_n.\end{align*} Applying the determinant map $\det: \mathbb{R}^{n \times n} \to \mathbb{R}$ to both sides gives \begin{align*}\det(Q^\top Q) = \det(I_n).\end{align*} Using the determinant identity $\det(AB)=\det(A)\det(B)$ for square real matrices $A,B \in \mathbb{R}^{n \times n}$, the determinant identity $\det(Q^\top)=\det(Q)$, and $\det(I_n)=1$, we obtain \begin{align*}\det(Q^\top Q)=\det(Q^\top)\det(Q).\end{align*} Therefore \begin{align*}\det(Q^\top Q)=(\det Q)^2.\end{align*} Combining this with $\det(Q^\top Q)=\det(I_n)$ gives \begin{align*}(\det Q)^2 = 1.\end{align*} [guided] The goal is to convert the matrix identity defining orthogonality into a scalar identity about $\det Q$. Since $Q \in O(n)$, the defining condition for the real orthogonal group is \begin{align*}Q^\top Q = I_n.\end{align*} Now apply the determinant map $\det: \mathbb{R}^{n \times n} \to \mathbb{R}$ to this equality. Because equal matrices have equal determinants, we get \begin{align*}\det(Q^\top Q)=\det(I_n).\end{align*} We now evaluate both sides. The product $Q^\top Q$ is a product of two matrices in $\mathbb{R}^{n \times n}$, so the multiplicativity of the determinant applies: \begin{align*}\det(Q^\top Q)=\det(Q^\top)\det(Q).\end{align*} The determinant is invariant under transpose, so $\det(Q^\top)=\det(Q)$. Substituting this into the preceding identity yields \begin{align*}\det(Q^\top Q)=(\det Q)(\det Q).\end{align*} Thus \begin{align*}\det(Q^\top Q)=(\det Q)^2.\end{align*} Finally, the determinant of the identity matrix is $1$, so $\det(I_n)=1$. Combining this with $\det(Q^\top Q)=\det(I_n)$ gives the scalar equation \begin{align*}(\det Q)^2=1.\end{align*} This is exactly the determinant constraint imposed by orthogonality. [/guided] [/step] [step:Solve the resulting scalar equation] Define $d \in \mathbb{R}$ by \begin{align*}d := \det Q.\end{align*} The preceding step gives $d^2=1$, hence \begin{align*}d^2 - 1 = 0.\end{align*} Factoring the polynomial $t^2-1$ over $\mathbb{R}$ gives \begin{align*}(d-1)(d+1)=0.\end{align*} Since $\mathbb{R}$ has no zero divisors, $d-1=0$ or $d+1=0$. Therefore $d=1$ or $d=-1$, so \begin{align*}\det Q \in \{-1,1\}.\end{align*} This proves the theorem. [/step]