# Verifying that a matrix is a rotation matrix

• Oct 27th 2009, 09:58 AM
Jodles
Verifying that a matrix is a rotation matrix
Hi!

How do you, properly, verify that a given matrix is a rotation matrix?

Consider the following matrix P:
$
P =
\left[
\begin{array}{c c c c c c c}
1 & 0 & \cdots & \cdots & \cdots & 0 & 0 \\
0 & 0 & \cdots & & & \cdots & \cdots \\
\cdots & \cdots & cos \theta & \cdots & sin \theta & \cdots & \cdots \\
\cdots & \cdots & & & & \cdots & \cdots \\
\cdots & \cdots & -sin \theta & \cdots & cos \theta & \cdots & \cdots \\
\cdots & \cdots & & & & \cdots & \cdots \\
0 & 0 & \cdots & \cdots & \cdots & 0 & 1
\end{array}
\right]
$

where $P_{rr} = P_{ss} = cos \theta, P_{rs} = -P_{sr} = sin \theta, P_{ii} = 1$ (for $i ~= r, s$) and else $0$.

It's essentially an identity matrix with that familiar 2x2 rotation matrix setup sort of in the middle there. But mathematically, what conditions must I show are true to establish that this is a rotation matrix? I would think orthogonality is one of them?

(sorry for the missing \cdots, I exceeded the limit of latex code:))
• Oct 27th 2009, 10:03 AM
clic-clac
Just check it is orthogonal and that each row has its norm equal to 1. That won't be very troublesome :)
• Oct 27th 2009, 10:38 AM
Jodles
Thank you!

Regarding checking that it is orthogonal, the best way is to show that $PP^{T}=P^{T}P=I$?
• Oct 27th 2009, 11:05 AM
clic-clac
Yes in that case it is easy to verify that $PP^T=I$.
• Oct 27th 2009, 12:02 PM
Jodles
I'm sorry to bother you with more of these simple questions, but when verifying $PP^T=I$, is it ok notation-wise to omit the larger part of $P$ and only show calculations on a smaller part containing the cos and sin elements? Then show that they either go to 1 or 0 and thus $PP^T=I$.
• Oct 27th 2009, 12:43 PM
clic-clac
Well that depends on the corrector... So it's hard to answer you!

But even the more detailled version of the answer is not that bad: let's say $PP^T=(a_{ij})_{0

If $i$ or $j\neq r,s$ then for any $1\leq k\leq n,\ P_{ik}P_{jk}=1\Leftrightarrow i=k=j,$ otherwise it is zero, hence $a_{ij}=\sum_{k=1}^nP_{ik}P_{jk}=\delta_{ij}$

Now $i,j\in\{r,s\}.\ a_{rr}=a_{ss}=\cos^2\theta +\sin^2\theta=1$ and $a_{rs}=a_{sr}=-\sin\theta\cos\theta + \cos\theta\sin\theta=0$ ... I guess it is evident with the form of theese lines/rows.