# Rotation matrix between two orthonormal frames

1. Jul 7, 2015

### santos2015

I am reading a paper and am stuck on the following snippet.

Given two orthonormal frames of vectors $(\bf n1,n2,n3)$ and $(\bf n'1,n'2,n'3)$ we can form two matrices $N= (\bf n1,n2,n3)$ and $N' =(\bf n'1,n'2,n'3)$. In the case of a rigid body, where the two frames are related via rotations and translations only, we can can calculate the rotation matrix between the two frames as:
$R = N' N^{T}$.

My linear algebra is quite rusty and I am having some trouble understanding why this is true. Thanks for looking.

2. Jul 7, 2015

### fzero

Let $\{\mathbf{e}_i\}$ be the standard basis $\mathbf{e}_1 = (1,0,0)^T,\ldots$. Then $N^T$ is the change of basis from $\{ \mathbf{n}_i \}$ to $\{\mathbf{e}_i\}$, since we can show that $\mathbf{e}_1 = N^T \mathbf{n}_1$, etc. Similarly, $N'$ is the change of basis from $\{\mathbf{e}_i\}$ to $\{ \mathbf{n}'_i \}$. Therefore the change of basis from $\{ \mathbf{n}_i \}$ to $\{ \mathbf{n}'_i \}$ is given by the product $N' N^T$.