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An orthogonal matrix is a square matrix Q such that its transpose is equal to its inverse: \(Q^T = Q^{-1}\). In other words, the product of an orthogonal matrix and its transpose is the identity matrix: \(QQ^T = I\).

Given that λ is an eigenvalue of Q, and x is its eigenvector, the eigenvalue equation is \(Qx = \lambda x\), where x is a nonzero vector.

First, we take the transpose of both sides of the eigenvalue equation: \((Qx)^T=(\lambda x)^T\) Since taking the transpose of a scalar multiple is equivalent to multiplying the transpose by the scalar, we can rewrite this as: \((Q^T)(x^T)=\lambda (x^T)\) Now, we know that the transpose of an orthogonal matrix is the inverse, so we can substitute the inverse in place of the transpose: \((Q^{-1})(x^T)=\lambda (x^T)\) Multiplying both sides by Q on the left: \(Q(Q^{-1}x^T)=Q(\lambda x^T)\) This simplifies to \(x^T=\lambda Qx^T\). Remember the original eigenvalue equation: \(Qx = \lambda x\). We can multiply both sides of this equation by \(x^T\) on the left: \((x^T)Qx=(x^T)(\lambda x)\) Using the fact that the product of Q and its transpose equals to identity matrix, we get: \(x^T x = \lambda (x^T x)\) Since x is a nonzero vector, \(x^T x\) is also nonzero. Thus, we can cancel out \(x^T x\) on both sides: \(\lambda = 1\) The magnitude of the eigenvalue λ must be equal to 1.

Recall the definition of orthogonal matrices: \(QQ^T = I\). Now, taking the determinant of both sides, we have: \(\det(QQ^T)=\det(I)\) Using the property that the determinant of a product is the product of determinants: \(\det(Q)\det(Q^T)=1\) As Q is orthogonal, we know that \(Q^T = Q^{-1}\). Then, the determinant of Q transpose equals the reciprocal of the determinant of Q: \(\det(Q)\frac{1}{\det(Q)}=1\) Multiplying both sides by \(\det(Q)\), we get: \(\det(Q)^2=1\) Taking the square root of both sides, we find that the magnitude of the determinant of Q is equal to 1: \(|\det(Q)|=1\) This completes the proof.