Explain in detail please! Let SO(3) denote the set of 3 times 3 matrices A such

Question

Explain in detail please!

Let SO(3) denote the set of 3 times 3 matrices A such that A has real entries, Det(A) = 1, and AT = A-1. Show that the matrix. is in SO(3). This matrix can be described as "a rotation of R3 with axis v= [1,0,0]T and angle of rotation. Let v be any unit length vector v. Let p = [vv'v"] be the transition matrix from F to the standard basis, Show that for the matrix A sbove, if we define B = PAP-1, then B is in SO?(3) and Bv = v. If you look more carefully you will see that you can describe B as a rotation with axis v. The rest of this exercise is devoted to showing how the composition of two rotations about different axes by different angles result in a new rotatyion about a new axis with a new. Show that if A and B are in SO(3), then their product AB is also in SO(3). Show that if A is a 3 times 3 matrix then Det(-A) = - Det(A). Show that if A is in SO(3) then there is a vector v in R3 such that Av = v. To do this, you need to show that A-I is singular. Consider the determine of A-I. Use the fact AT = AT = A-1 to show that Det(A-I) = Det((A-I)A-1) = Det(I-AT)=Det(I-A) =Det(-(A-I)). From this you should be able to concude that Det(A-I) = 0. Knowing that Av = v, one then can choose an orthonormal basis for R3 of the form F = and show that. Restricted to the subspace eith basis the transformation is a rotation by some angle. All of this can be applied to any element of SO(3) to determine it as a rotation about an axis by some angle.

Explanation / Answer

(a)
By expanding by the first column det(A) = 1 * det(M)
Where M =
cos(a) -sin(a)
sin(a) cos(a)

det(M) = cos(a)^2+sin(a)^2=1

Hence det(A)=1

A is orthogonal since each column are orthogonal (inner product = 0) and the norm of each vector = 1.
So A^TA = I => A^T = A^-(1)

========>So A is in SO(3)


det(B)=det(PAP^(-1))=det(P)det(A)det(P^-1)=det(A)det(P)*1/det(P)=det(A)=1

Since P is the matrix of an orthonormal basis of R^3 then P is orthonormal and PP^T = I and P^(-1)=P^T
Then BB^T = PAP^T*PA^TP^T = PA(P^TP)AP^T=P(AA^T)P^T=PP^T=I, then B^T = B^(-1)

========>So B is in So(3)

Note that if E={e1,e2,e3} is the standard basis, we have Pe1 = v ( and then P^Tv=P^TPe1=e1) (e1=(1,0,0)^T)

So Bv = PAP^Tv = PAe1

But Ae1 = (1 0 0)^T = e1 by computation, then PAe1 = Pe1 = v

=====> So Bv=v


(b)
Suppose A,B is SO(3), then A^T=A^(-1), B^T=B^(-1) and det(A)=det(B)=1
det(AB)=det(A)det(B)=1*1=1
(AB)^T=B^TA^T=B^(-1)A^(-1)=(AB)^(-1)
=====> So AB is in SO(3)

(c)
For a matrix A = (c1 c2 c3) (3 columns)
det(-A)=det(-c1 -c2 -c3) = -det(c1 -c2 -c3) = +det(c1 c2 -c3) = -det(c1 c2 c3)=-det(A)

(d)
Since det(A)=1, then det(A^(-1))=1 and we can write :
det(A-I) = det(A-I)*det(A^(-1))=det((A-I)A^(-1))=det(I-A^(-1))=det(I-A^T)=det(I-A)=det(-(A-I))=-det(A-I)
So det(A-I)=0

So there exist a vector v such that (A-I)v = 0
=====> Av=v

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