# Thread: Verifying that a matrix is a rotation matrix

1. ## 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)

2. Just check it is orthogonal and that each row has its norm equal to 1. That won't be very troublesome

3. Thank you!

Regarding checking that it is orthogonal, the best way is to show that $PP^{T}=P^{T}P=I$?

4. Yes in that case it is easy to verify that $PP^T=I$.

5. 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 $PP^T=I$.

6. 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.