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Let T : $\mathbb{R^3} \rightarrow \mathbb{R^3}$ be an orthogonal transformation such that det $T = 1$ and T is not the identity linear transformation. Let S $ \mathbb{R^3}$ be the unit sphere, i.e., S = {$(x; y; z):x^2 + y^2 + z^2 = 1$}: Show that T fixes exactly two points on S.

can anyone help me to solve this problem.thanks.

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up vote 1 down vote accepted

Intuitively, these should be the points for which the axis or the rotation intersects the sphere. To find these for an explicit given $T$ we need to solve $T(v)=v$ since this geometrically characterizes the axis. I suppose you need to think about how to argue that $\lambda = 1$ is in fact the only real eigenvalue of $T$. Hint: the information that $T$ is not the identity is important to this goal. Also, for an orthogonal transformation the matrix $[T]=A$ will have $A^TA=I$. Take the determinant and recall the determinant is equal to the product of the eigenvalues (possibly complex). Think about this and you can construct the answer easily.

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I assume you know that $det(AB)=det(A)det(B)$. – James S. Cook Sep 23 '12 at 2:10

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