|Mar13-10, 12:50 PM||#1|
Linear Algebra with linear operators and rotations
Definie linear operators S and T on the x-y plane as follows: S rotates each vector 90 degress counter clockwise, and T reflects each vector though the y axis. If ST = S o T and TS = T o S denote the composition of the linear operators, and I is the indentity map which of the following is true?
A) ST = I
B) ST = - I
C) TS = I
D) ST = TS
E) ST = -TS
I know for a fact that A and B are false.
B) For A then we can start with a vector on the positive x axis. Reflect it over the y axis and the vector will lie on the negative x axis. Then rotate it 90 degress under S and it will lie on the negative y axis. This does not equal -I
For B) if we let the vector lie on the positive x axis, then if we reflect it under T it is on the negative real axis and then rotate it 90 degress under S then it lies on the negative y axis This does not equal I.
C) If we take a vector on the positive x axis again, rotate it 90 degress under S, then it will on the positive y axis, Under T it will not be reflected since it lies on the y axis and so TS does not equal I
By the way I deliberately picked the vector lying on the x axis because for that vector it does not work. The statement is true if the vector is lying somewhere in the first quadrant not on the axis but it has to be true for every vector.
I am not sure about D or E. Can someone check my work and help with D and E please?
I figure it out my providing counter-examples for A) through D).. Does anyone know how to show that ST = -TS for every vector in the x-y plane? Is there a way to prove it?
|Mar13-10, 01:28 PM||#2|
You could also write down the matrices representing S and T and calculate their products to see if any of the equations holds.
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