Can You Solve This Linear Transformation Equation?

[katie_3011]

1. R\circF\circR^-1=S
where F denotes the reflection in the x-axis
where S is the reflection in the line y=x
where R = R_\pi/4 : R² \rightarrow R²

3. An attempt

I have found that the standard matrix for R = [cos\theta sin\theta]
[sin\theta cos\theta]
So therefore, the inverse of R would be the same matrix.

The standard matrix for F = [1 0]
[0 -1]

When I multiplied the matrices together, I got a matrix [1 -1]
[1 1],
which does not equal S, which should be [0 1]
[1 0].

I have tried multiplying out the matrices a few times, and I'm pretty sure this is where my mistake is, but I'm not entirely sure how to multiply cos\theta and sin\theta with actual numbers.

Thanks in advance for your help

 

[Mark44]

1. R\circF\circR^-1=S
where F denotes the reflection in the x-axis
where S is the reflection in the line y=x
where R = R_\pi/4 : R² \rightarrow R²

3. An attempt

I have found that the standard matrix for R = [cos\theta sin\theta]
[sin\theta cos\theta]

No, this is not the matrix. To rotate a vector counterclockwise by an angle of theta, the entry in row 1, column 2 should be -sin(theta).

So therefore, the inverse of R would be the same matrix.

Nope, that's not true, either.

The standard matrix for F = [1 0]
[0 -1]

When I multiplied the matrices together, I got a matrix [1 -1]
[1 1],
which does not equal S, which should be [0 1]
[1 0].

I have tried multiplying out the matrices a few times, and I'm pretty sure this is where my mistake is, but I'm not entirely sure how to multiply cos\theta and sin\theta with actual numbers.

Thanks in advance for your help

 

[HallsofIvy]

1. R\circF\circR^-1=S
where F denotes the reflection in the x-axis
where S is the reflection in the line y=x
where R = R_\pi/4 : R² \rightarrow R²

3. An attempt

I have found that the standard matrix for R = [cos\theta sin\theta]
[sin\theta cos\theta]
So therefore, the inverse of R would be the same matrix.

As Mark44 said, a rotation matrix is anti-symmetric, not symmetric. If the angle is, as here, \pi/4 so that cos(\pi/4)= sin(\pi/4)= \sqrt{2}/2 and the matrix is
\begin{bmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}

Further, the inverse of "rotation by angle \theta" is "rotation through angle -\theta". cos(-\theta)= cos(\theta), sin(-\theta)= -sin(\theta) so changing from \theta to -\theta changes the sign on the "sin" (off-diagonal) but not on the "cos" (diagonal) terms. The matrix rotating by angle -\theta is
\begin{bmatrix}\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{bmatrix}

The standard matrix for F = [1 0]
[0 -1]

When I multiplied the matrices together, I got a matrix [1 -1]
[1 1],
which does not equal S, which should be [0 1]
[1 0].

I have tried multiplying out the matrices a few times, and I'm pretty sure this is where my mistake is, but I'm not entirely sure how to multiply cos\theta and sin\theta with actual numbers.

Thanks in advance for your help

 

[katie_3011]

I'm pretty sure that the first matrix (the one for R) is correct. These are my assumptions:

If the line for e1 is at an angle theta from the x-axis (assuming that theta is less than pi/4), then the line for e2 would still be in the first quadrant, therefore all of the values would still be positive.

If this is wrong, can you explain to me why?

 

[HallsofIvy]

I'm pretty sure that the first matrix (the one for R) is correct. These are my assumptions:

If the line for e1 is at an angle theta from the x-axis (assuming that theta is less than pi/4), then the line for e2 would still be in the first quadrant, therefore all of the values would still be positive.

If this is wrong, can you explain to me why?

There was no "e1" or "e2" in what you wrote before so I have no idea what a "line for e2" or "line for e2" would be.

 

[Mark44]

Both HallsofIvy and I are telling you that your rotation matrix is not correct.

 

[katie_3011]

Thank you, I understand now