How can I describe geometrically the transformation ACA-1 ?

[Natasha1]

If A represents the matrix of a rotation of 45 degrees anti-clockwise around (0,0), B a shear with x-axis invariant and shear factor of 2, and C a reflection in the x-axis.

How can I describe geometrically the transformation ACA-1 ?

My answer is a rotation of 90 degrees anti-clockwise around (0,0) but surely there is more than that?

 

[HallsofIvy]

If A represents the matrix of a rotation of 45 degrees anti-clockwise around (0,0), B a shear with x-axis invariant and shear factor of 2, and C a reflection in the x-axis.

How can I describe geometrically the transformation ACA-1 ?

My answer is a rotation of 90 degrees anti-clockwise around (0,0) but surely there is more than that?

There is no B in this problem? Think about what happens to the unit vectors in the x and y directions. Since A corresponds to a 45 degree rotation anti-clockwise, A^-1 corresponds to its opposite, a 45 degree rotation clockwise. The vector i is changed by A^-1 to the vector $\frac{sqrt{2}}{2}(i- j)$, then by C to [itex]\frac{sqrt{2}}{2}(i+ j)[itex], then by A to j. Okay, that's been rotated 90 anti-clockwise. The vector j is changed by A^-1 to [itex]\frac{\sqrt{2}}{2}(i+ j)[\itex], then by C to $\frac{\sqrt{2}}{2}(i- j)$ and finally by A to i. No, that's not a rotation by 90 degrees anti-clockwise! It is a reflection about what line?

 

[Natasha1]

There is no B in this problem? Think about what happens to the unit vectors in the x and y directions. Since A corresponds to a 45 degree rotation anti-clockwise, A^-1 corresponds to its opposite, a 45 degree rotation clockwise. The vector i is changed by A^-1 to the vector $\frac{sqrt{2}}{2}(i- j)$, then by C to [itex]\frac{sqrt{2}}{2}(i+ j)[itex], then by A to j. Okay, that's been rotated 90 anti-clockwise. The vector j is changed by A^-1 to [itex]\frac{\sqrt{2}}{2}(i+ j)[\itex], then by C to $\frac{\sqrt{2}}{2}(i- j)$ and finally by A to i. No, that's not a rotation by 90 degrees anti-clockwise! It is a reflection about what line?

line y = x

 

[HallsofIvy]

Way to go!

 

[Natasha1]

That done I need to show that if P' is the image of P under D=BCB-1. If P is not on the x-axis, then PP' is bisected by the x-axis and is at a constant angle to the x-axis, for any choice of P?

I can visually see what's happening and can see that the angle is 90 degrees but how can I show it? :uhh:

 

[Natasha1]

Could anyone help me with the last question? Please