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I am confused with how to show that an orthogonal matrix with determinant 1 must always be a rotation matrix.

My approach to proving this was to take a general matrix $\begin{bmatrix}a&b \\c&d\end{bmatrix}$ and using the definition of a matrix being orthogonal, work out some restrictions on $a,b,c,d$ such that the matrix must be a rotation matrix.

Doing this; I end up with $\begin{bmatrix}a&b \\c&d\end{bmatrix}^T\begin{bmatrix}a&b \\c&d\end{bmatrix}=\begin{bmatrix}a^2+c^2&ab+cd \\ab+cd&b^2+d^2\end{bmatrix}=\begin{bmatrix}1&0 \\0&1\end{bmatrix}$

We also know $ad-bc=1$ from the determinant restriction.

I thought we could say since $a^2+c^2=1$ then we could say $a=\cos(\theta)$ and $c=\sin(\theta)$. From here we could chose the second column of the matrix to a vector such that $b=-\cos(\pi/2-\theta)$ and $d=\sin(\pi/2-\theta)$ which gives $b=-\sin(\theta)$ and d=$\cos(\theta)$.

This would give $\begin{bmatrix}\cos(\theta)&-\sin(\theta) \\\sin(\theta)&\cos(\theta)\end{bmatrix}$ which is the rotation matrix I need to show.

Is this correct and is there a better approach to help me prove that any orthogonal matrix with determinant 1 must be a rotation matrix?

EDIT:

We can say $a=\cos(\theta)$ and $c=\sin(\theta)$ as we know $a^2+c^2=1$ and so $a$ and $c$ must lie on the unit circle, hence we can parameterise the variables in the matrix as such.

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    $\begingroup$ Notice that the last matrix equality generates 3 distinct equations: $a^2 + c^2 = 1$, $ab+cd =0$, $b^2+d^2=1$. Furthermore, the determinant gives another: $ad-bc =1$. So you have $4$ equations in $4$ unknowns. See what you can do to eliminate parameters and reduce to everything as a function of $1$ variable. Note that these equations are non-linear.... $\endgroup$
    – erfink
    Apr 1, 2017 at 21:37
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    $\begingroup$ Could we say that the points $a$ and $c$ lie on a unit circle and therefore can be represented via their parametric form? $\endgroup$
    – user258521
    Apr 1, 2017 at 21:40
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    $\begingroup$ But knowing $a^2 + c^2 = 1 \implies |a| \leq 1$ so that $a$ really is the cosine of some angle. It could be more spelled out, but there's nothing wrong there (IMO). $\endgroup$
    – pjs36
    Apr 1, 2017 at 21:41
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    $\begingroup$ @mathreadler agreed---fallacy is probably a bit too strong of a word. My point is that this is really the crux of the entire problem and needs a sentence of exposition about why it is true. $\endgroup$
    – erfink
    Apr 1, 2017 at 21:44
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    $\begingroup$ You should specify if your question is for $2\times 2$ case only. $\endgroup$
    – A.Γ.
    Apr 1, 2017 at 22:05

1 Answer 1

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The sentence "From here we could chose the second column of the matrix..." is ambiguous. You cannot choose $b$ and $d$, they are given. You need to prove that they are $\sin$ and $\cos$.

For example, since $$ a^2+c^2=1,\\ d^2+b^2=1, $$ the points $(a,c)$ and $(d,b)$ are on the unit circle and can be parameterized as $$ (a,c)=(\cos\theta,\sin\theta),\\ (d,b)=(\cos\phi,\sin\phi). $$ Now another two equations that you know are $$ ab+cd=0 \quad\Leftrightarrow\quad\cos\theta\sin\phi+\sin\theta\cos\phi=\sin(\theta+\phi)=0,\\ ad-bc=1\quad\Leftrightarrow\quad\cos\theta\cos\phi-\sin\theta\sin\phi=\cos(\theta+\phi)=1. $$ It gives $\theta+\phi=2\pi k$, which makes $$ b=\sin(2\pi k-\theta)=-\sin\theta,\\ d=\cos(2\pi k-\theta)=\cos\theta. $$

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