On the Coriolis Forcing vector and its Matrix

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The context for the question is in the attachments (pg1.png, pg2.png, pg3.png), so there is some reading involved. Although, it is a short and simple read if anything. The inquiry is in (inquiry.png).
My understanding of the situation is that Q(t) abides by the differential equation
Q'(t)Q(t)T + Q(t)Q'(t)T = 0 .
So by something akin to Picard's existence theorem on differential equations, we can resolve exact trajectories for Q(t) given initial conditions. The peculiar thing is that
Q(t0) z(t0) = z(t0)
can likely be assumed to be true for all z(t0) ∈ ℝn, in which case
Q(t0) = I
should also be true.
Now if I let R = ( i(t0) , j(t0) , k(t0) ), then
RTR = RRT = I ,
and the question insists that
RTQ(t)R = Q ⇒ Q(t)R = RQ(t).
Although, our choice of the initial condition basis vectors for R is arbitrary, so how can we expect that R necessarily commutes with Q(t)? Unless of course, the question is asking for the matrix expression of Q(t) in the { i(t0) , j(t0) , k(t0) } basis. I am just wondering if I am to expect the matrix expression of Q(t) to remain invariant given any orthonormal basis of the same orientation?
(Also, how do I insert latex?)
 

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  • #2
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Well, by the looks of it, the matrix expression for Q(t) would change depending on the basis { i(t0) , j(t0) , k(t0) } used. So I suppose the answer is simply in the context of the trajectory of one point.
 

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