# Why degree of freedom of orthographic projection is 5?

This problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 171.

With the camera center at infinity, projecting a point in 3d projective space into it's image point in a 2d projective plane can be modeled as an affine camera,especially in an orthographic projection case, the camera model is $$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix} \begin{bmatrix}R & \mathbf t \\ {\mathbf0}^T & 1\end{bmatrix}_{4 \times 4}$$ Here $R$ is a rotation matrix of $3 \times 3$, denote $R$ as $\begin{bmatrix}r_1^T \\ r_2^T \\r_3^T \end{bmatrix}$. So the camera matrix become $$\begin{bmatrix}r_1^T & t_1 \\r_2^T & t_2 \\ \mathbf 0^T & 1 \end{bmatrix}_{3 \times 4}$$. My question is how to figure out that degree of freedom of above transformation is 5? Particularly, I cannot understand why matrix block $\begin{bmatrix}r_1^T \\ r_2^T \end{bmatrix}$ contributes 3 degrees of freedom.

• Basically, because matrix $R$ has 9 entries and system of equations $R^{T} R = I$, which defines an orthogonal matrix, has 6 equations (you might say nine, but because of transposing some of these equations are repeated). If you very carefully apply implicit function theorem you can locally describe the solution space of this equation as some 3-dimensional object, hence it has 3 degrees of freedom. Dec 5, 2017 at 8:12
– MvG
Dec 5, 2017 at 16:37
• Yeah, I could, but then as a fair person I should put a reference to a text where three-dimensionality is proven or even put a proof in my answer. And I can't find this reference from a get go or don't want to prove it myself :) Dec 5, 2017 at 18:58
• @Evgeny: I think even without a proof it's valuable. I think it is very clear on an intuitive base. And if you happen to find that reference, you can always edit your post. Until then, you can add a comment that you consider your answer incomplete. Personally I'd consider it more complete than what I wrote, but of course you might well argue that I posted a rather incomplete answer and shouldn't have done so. I think any real answer adds value, and votes will help determine that value. I don't envision downvotes for lack of a reference in this case.
– MvG
Dec 5, 2017 at 20:03

Removing one of the rows of the matrix doesn't really change the information contained therein: any row of an orthogonal matrix can be computed using the cross product of the other two. So you still need all three degrees of freedom for the $2\times3$ submatrix you described. When you go to a single row, then you loose one degree of freedom since a unit length vector essentially has one degree of freedom less than it has elements.
• To my understanding, a rotation matrix $\begin{bmatrix}r_1^T \\ r_2^T \\r_3^T\end{bmatrix}$ can be factorized into three Givens rotations, three degrees of freedom.for rotation matrix come from three rotation angles. But my question in this post is why the matrix block $\begin{bmatrix}r_1^T \\ r_2^T \end{bmatrix}$ still contributes three dof Dec 7, 2017 at 5:59