# Constructing a 2x2 matrix R which represents reflection in the x-y plane,

Construct a 2x2 matrix R which represents reflection in the x-y plane through the line $$(cos(\theta)x+(sin(\theta)y=0$$, where $\theta$ is any real number.

(Let's call this line "L".)

Write an explicit formula for each element of the matrix R in terms of $\theta$, and use trigonometric identities to express your results in terms of cos(2$\theta$) and sin(2$\theta$). Check your answer in the following ways:

1. Evaluate $R^2$.

2. Evaluate the determinant of R.

3. Check that R is an orthogonal matrix.

4. Does R have the eigenvalues and eigenvectors you would expect it to have based on the reflection operation that it represents?

My work, so far: I plugged in some values for $\theta$ and noticed that, for a given $\theta$ that we fix and then get some line that we are reflecting through, we have that the vector (cos$\theta$, sin$\theta$) is orthogonal to this line =L.

Then a unit vector along this line of reflection is (-sin($\theta$), cos($\theta$)). Call this unit vector on this line "u".

This vector u is an orthonormal basis for the line (a subspace) =:L

Then, supposedly, the projection matrix, which projects any vector in the x-y plane to the line L, is given by $uu^T$.

But if I compute this matrix, I am getting this:

$uu^T$ = (-sin($\theta$), cos($\theta$))*$(-sin(\theta), cos(\theta))^T$

=$(sin^2(\theta) +cos^2(\theta))$

=(1).

I get....the "1 matrix"? Something's gone wrong and I'm not sure where and how.

The projection matrix offered in the solution is:

$$P:= \begin{bmatrix} sin^2(\theta) & -sin(\theta)cos(\theta) \\ -sin(\theta)cos(\theta) & cos^2(\theta) \\ \end{bmatrix}$$

Where did this come from?

And then the reflection matrix is given as: $$R =2P - I= \begin{bmatrix} 2sin^2(\theta)-1 & -2sin(\theta)cos(\theta) \\ -2sin(\theta)cos(\theta) & 2cos^2(\theta)-1 \\ \end{bmatrix}$$

Why do we use 2P - I? I have a vague idea of why we use 2P: if we take twice the value of the image of the projection, then that moves the final point to the other side of the line, with this point having the same distance to the line of reflection as the corresponding point on the other side of the line (the pre-image point). Is this reasoning correct? Or is there a better / correct reason?

Finally, I don't know why we need to subtract the identity matrix from 2P. That doesn't seem to contribute anything at all to the reflection operation that we want to construct...

Thanks in advance,

## 1 Answer

On the part about $uu^T$, your error is that $u$ needs to be considered a column vector, not a row vector.

As for the part about $2P - I$, you seem to have generally the right idea, but you're not executing it properly. If you start with a vector $v$, its reflection $Rv$ is symmetric to $v$ with respect to the projection $Pv$. This means that the vector from $v$ to $Rv$, which is $Rv - v$, is twice the vector from $v$ to $Pv$, which is $Pv - v$. Can you finish the calculation from there?

• Yes @user196235, I follow the computations and can arrive at R = 2P - I now, but I'm a little confused about taking twice the vector v: are we taking twice the norm of v to Pv, or are we taking twice the vector from v to Pv (which is Pv-v)? Because twice the vector would mean...they're exactly the same vectors. But, drawing this out on paper, I feel like the vector from v to Pv has a different direction from the vector from Pv to Rv. – User001 Nov 29 '14 at 4:49
• No, the vector from $v$ to $Pv$ is the same as the vector from $Pv$ to $Rv$. – user196235 Nov 29 '14 at 4:51
• Ok, got it. Thanks so much, @user196235 – User001 Nov 29 '14 at 5:00