Question

Let V be the subspace of "vectors" in Hamilton's sense, that is, quat ernions with zero real part. Given a nonzero quaternion q, show that the mapping T V V defined by T(v) is an orthogonal mapping. This means that T(v). T(w) = u·w for all vectors u, w E V (again, V = the purely imaginary quat ernions) What is the mapping when q is an imaginary unit? Give its matrix for the basis i,j,k. For any nonzero quaternion q, the map has an eigenvalue 1. What is the associated eigenvector? Explain how you could use this information to start identifying T as a rotaticon The distributive property and associativity for scalar multiplicains actully all that's required here. That is just another way of saying multiplication is bilinear. We can define a bilinear map on a basis, and this is what's being done. But then we should prove it's actually associative for all products, not just scalar times quaternion Spoiler alert: it is 4. The common definition is: x w is the vector in R meeting these conditi ons perp endi cular to v and w, . having length equal to llell . Il 'll . I sin(θ), and det(, w, v xu) >0 The last requirement nails down the direction of u × w along the line perpendicular to the plan e spanned by e and w. It says that v,w, x w form a posivly-oriented ordered basis of R3. This is one obeying the "right-hand rule" if you curl the fingers of your right hand from v to w, the direction your thumb points is the direction of × w. So to prove that our definition agrees with this, you just have to show that u. (v × w) = 0 = w. (v × w) and the orientation condition. Do so.

Answer 1

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O-1 co non 군 en eigen valne 1

2 (1-021)