# A proof that an orthogonal matrix with a determinant 1 is a rotation matrix

Reading proof(starting on page 5) for item 1 of "Rotation Matrix Theorem" in this doc i'm stuck at understanding its last step. Matrix A being an orthogonal Matrix, at this step the conclusion that A is a rotation matrix is reached based on these facts:

• $u_1$ is a unit vector such that $A u_1 = u_1$
• $u_2$ is a unit vector perpendicular to $u_1$
• $u_3 = u_1 \times u_2$
• $A u_2 = \cos(\theta)u_2 + \sin(\theta)u_3$
• $A u_3 = -\sin(\theta)u_2 + \cos(\theta)u_3$

I can't realize how that conclusion is reached, although I clearly understand a matrix of the form: \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \\ \end{bmatrix} is a rotation matrix and applying A to the unit vectors has the effect of rotating them about the axis through $u_1$ by the angle $\theta$.

• Since matrix $A$ under the new basis $u_1,u_2,u_3$ is in this form, it is a rotation matrix that rotate any vector about the $u_1$ axis by angle $\theta$. – KittyL Aug 5 '15 at 9:21
• well the problem is "to prove A is a rotation" means to prove A is a rotation under the current basis not the new one(new one being $u_1$,$u_2$ and $u_3$) – Pooria Aug 5 '15 at 9:34
• What is the definition of a rotation matrix? If it rotate vectors about any axis, it should be a rotation matrix. It doesn't have to rotate vectors about $x,y,z$. – KittyL Aug 5 '15 at 9:37
• let me be more precise, I think we should prove that there's a rotation such that for an orthonormal basis, applying the rotation to vectors of that basis gives $A_1$, $A_2$ and $A_3$(each is a vector). – Pooria Aug 5 '15 at 9:47
• Think about it in a 3D space. No matter where the rotation is, (again it could be about any axis), as long as it rotate objects, it is a rotation matrix. It has nothing to do with basis. So it is to prove that there exists an orthonormal basis, such that $A$ can be written in this standard form under this basis. – KittyL Aug 5 '15 at 9:51

I understand your statement now. I will summarize my answers here.

As I mentioned in my comments, $A$ is a rotation matrix since it can be written in the "standard" form under the new basis. In other words, it rotate any vector about the axis that is in the direction of $u_1$.

Notice that this last statement "it rotate any vector about the axis that is in the direction of $u_1$" is independent of the basis.

Your statement "to prove that there's a rotation such that for an orthonormal basis, applying the rotation to vectors of that basis gives A1, A2 and A3" is also correct. Because it also says the columns of $A$ is an orthonormal basis, which is true. But I think this is kind of a confusing way to prove a matrix is a rotation. Also you'll have to construct another rotation to prove it. And in what form would you construct another rotation?

A more straightforward way, as above, is to prove that the matrix $A$ rotates vectors.

Edit:

This is with regard to the matrix $A$ represented in terms of basis $u_1, u_2, u_3$. We can write a linear transformation $T$ as a matrix in terms of any basis using the following way. Given basis $e_1, e_2, e_3$, the matrix in terms of this basis is calculated by $$[Te_1\quad Te_2 \quad Te_3]$$

where $Te_i$ is the column coordinate vector after you apply $T$ to $e_i$ in terms of this basis.

Looking at the question in this way, we see that $Au_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $Au_2=\begin{pmatrix}0\\ \cos\theta\\ \sin\theta\end{pmatrix}$, $Au_3=\begin{pmatrix}0\\ -\sin\theta\\ \cos\theta\end{pmatrix}$.

Then $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \\ \end{bmatrix}$$

is exactly the matrix $A$ in terms of the basis $u_1, u_2, u_3$.

• +1 :) Well I think those facts I listed(in question body) mean rotating i, j and k about the axis through $u_1$ by the angle $\theta$ gives $A_1$,$A_2$ and $A_3$ respectively and it's that meaning that proves A is a rotation matrix – Pooria Aug 5 '15 at 18:29
• @Pooria: I am not sure whether you understand the linear transformation and change of basis. It becomes clearer if you understand that well. I'll edit. – KittyL Aug 5 '15 at 18:40
• Well I am well aware of those, I can't understand why you thought I'm not! and by the way I talk about i,j k because it seems nice to do so, I know it can be any of infinite other orthonormal bases. – Pooria Aug 5 '15 at 19:00
• @Pooria: Great! I hope you solved your problem then? – KittyL Aug 5 '15 at 19:15
• Well not much of a problem as I only need a proof, once I get it I'll post it here too :) – Pooria Aug 6 '15 at 10:36