If you have an orthogonal matrix with a determinant of -1, how do you determine the plane of reflection?
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[Math] How to find plane of reflection from transformation matrix
geometrylinear algebramatricesreflection
Related Solutions
If you can find the reflection through the plane perpendicular to the middle of the $AB$ segment, then it automatically sends $A$ to $B$ and $B$ to $A$.
How do we do this reflection? First, suppose that the midpoint of $A$ and $B$ is the origin: $A=-B$. Now, note the mathematical concept of a projection:
Here, $u,v,w,x$ are being multiplied by a projection matrix $P$: one that sends them to their closest point on the line $m$. It turns out that the projection of a vector Now suppose that $m$ is the line from $A$ to the origin.
If we subtract the projection of the line $x$ on to $OA$ from $x$ twice, then we get the reflection of $x$ in the plane through $O$ perpendicular to $OA$. Now, if $a$ is the vector of the point $a$, then the projection of the vector $x$ on to the line $a$ is given by $(x.a)x$ (do you know how to work out the dot product?) If you know properties of the dot product, you can work this out for yourself. So the transformation that reflects $x$ in this plane is given by:
$$ x\mapsto x-2(x.a)x $$
This is a linear transformation, so it has a matrix, if you want to work it out: to work out the matrix, substitute in $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ for $x$, and the vectors you get out will be the columns of the matrix.
What if $A\neq-B$? Then, as I said in my comment above, there is no matrix that will encode the reflection, as any matrix must send the origin to itself. So we have to use a trick called conjugation. First, find the midpoint $c=\frac12(a+b)$, and then let $a'=a-c$ and $b'=b-c$, and for any point $x$, let $x'=x-c$. So we are 'changing our world view' in some sense: we are taking the point $c$ and mapping it to the origin. Then $a'+b'=0$, so we can encode the reflection in the plane between them as a matrix $P'$.
Then, if $x$ is an arbitrary vector, the reflection of $x'$ in the plane between $a'$ and $c'$ is given by $P'x'=P'(x-c)$. If we now add $c$ to everything, we see that the reflection of $x$ in the plane between $a$ and $b$ is given by $P'(x-c)+c$, which we can rearrange to $P'x+d$, where $d=c-P'c$. So the transformation isn't given by a matrix, but it is given by a matrix multiplication, followed by adding a constant vector.
The process of carrying out an operation (subtracting $c$), then carrying out another operation (the reflection) and then doing the inverse of the first operation (adding $c$ back again) is known as conjugation, and is very important in mathematics. Think of it as a temporary change of 'world view': we pretend that $c$ is the origin so we can do our reflection, and then we map everything back again at the end.
Let $\|u\|=1$ and $Q=I-2uu^\top$.
Of course, the first condition above says that $u^\top u=1$. Secondly, the meaning of $v$ is perpendicular to $u$" is that $v^\top u= u^\top v=0$.
compute Qu and simplify as much as possible. Does this just mean move the equation around to get Qu?
No, $Qu$ is a matrix product. Starting with the above, you have $Qu=(I-2uu^\top)u=Iu-2uu^\top u$. What does this reduce to?
Suppose v is orthogonal to u. Compute Qv. Im not sure how this is done.
Again, it's just a matrix product. $Qv=(I-2uu^\top)v=Iv-2uu^\top v$. What does this reduce to?
Then I'm asked to explain in plain English which subspace $Q$ is reflecting across.
This is an interesting question which is not as mechanical as the rest of the problem. For one of the two parts above, you'll discover that $Qx=-x$. Geometrically, this means that $Q$ just reversed the direction of that vector.
One of the other computations is going to come out to $Qy=y$, meaning that $Q$ didn't alter the vector at all! But remember that the above had you assume that $x$ and $y$ are perpendicular to each other, so this means that one direction was reversed, and all perpendicular directions were left alone. If you fix a vector $x$, what does the collection of all perpendicular vectors look like?
Compute the reflection matrix $Q_1=I−2u_1u^\top_1$ where $u_1=(0,1)$. Compute $Q_1x_1$, where $x_1=(0,1)$ and sketch the vectors $u_1, x_1$ and $Q_1x_1$ in the plane.
This is one is easy to start because it begins with "do this computation." What part is holding you back? Putting the givens into the equations? The matrix addition and multiplication?
Best Answer
Any real eigenvalue is $\pm 1,$ and it is $-1$ as you are in dimension 3 with negative determinant. And there is a real eigenvalue because the dimension is odd, the degree pf the characteristic polynomial is odd.
So, there is a -1 eigenvector $v,$ if the matrix is called $M$ we have $Mv=-v.$
$M$ also preserves angles, easy enough to prove. So, the plane orthogonal to $v$ is setwise fixed. Note that $M$ may also cause a rotation within that plane.
For example, compare $$ M \; = \; \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$
versus
$$ M \; = \; \left( \begin{array}{rrr} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right) $$
versus
$$ M \; = \; \left( \begin{array}{rrr} \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 2} & 0 \\ \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\ 0 & 0 & -1 \end{array} \right) . $$
Thought question, or called a "Gesundheit" experiment in German, what if $M=-I?$ Possibly Gedanken experiment, or Weltanschauung. Could be Schadenfreude.