We already showed that there's a part of a matrix that transforms like a scalar: the...
We already showed that there's a part of a matrix that transforms like a scalar: the trace, which is a contraction of the two indices in a type-(1,1) tensor (in index notation, tr M = M'). For (3 x 3)- matrices, it turns out that the antisymmetric part of M transforms like a dual vector! That is, given 0 a 6 an antisymmetric matrix M = (-a ö ), the object QM = (M}, M3, M4) = (c, –b, a) behaves like other dual vectors we have seen. In this next part you will demonstrate this by looking at a rotation. Let R be an active transformation by about the 2-axis, R= (-sino cose ©). -b-co cos sin 0 0 0 0 1 (d) What is the matrix representation of M', the matrix M after undergoing an active rotation by R? What is the matrix representation of ami? Show that this is exactly what we'd expect from performing an active rotation by R on the object am: