Camera Center from Projection Matrix

mediumPROCamera modelCamera ModelComputer Vision 1

Given a $3 \times 4$ projection matrix $P$ , compute the camera center $C$ in world coordinates.

The camera center is the point in 3D space where all projection rays converge. It satisfies:

$P \cdot \tilde{C} = \mathbf{0}$

where $\tilde{C} = [C_x, C_y, C_z, 1]^T$ is the camera center in homogeneous coordinates. This is the null space of $P$ .

Algorithm using SVD:

Compute the SVD of $P$ : $P = U \Sigma V^T$
The camera center in homogeneous coordinates is the last column of $V$ (or last row of $V^T$ )
Convert from homogeneous to Euclidean by dividing by the last element: $C = [V_{0,3}/V_{3,3}, V_{1,3}/V_{3,3}, V_{2,3}/V_{3,3}]$

Round each coordinate to 4 decimal places.

Example

Input

P = [[1, 0, 0, -5],
     [0, 1, 0, -3],
     [0, 0, 1, -2]]

Output

[5.0, 3.0, 2.0]

Reasoning

Input: P is a $3 \times 4$ camera matrix acting on homogeneous world points $(X, Y, Z, W)$ . Here its left $3 \times 3$ block is the identity and its last column is $(-5, -3, -2)$ .
Math: the center is the 4-vector that $P$ sends to zero. Row by row, $X - 5W = 0$ , $Y - 3W = 0$ , $Z - 2W = 0$ . Three equations in four unknowns leave one free direction: every multiple of $(5, 3, 2, 1)$ .
Normalised to unit length (sign arbitrary), that direction is $(5, 3, 2, 1)/\sqrt{39} \approx (0.8006, 0.4804, 0.3203, 0.1601)$ .
Back to 3D: divide the first three entries by the fourth, using exact values: $\frac{5/\sqrt{39}}{1/\sqrt{39}} = 5$ , and likewise $3$ and $2$ . The $\sqrt{39}$ and any sign flip cancel.
Output: [5.0, 3.0, 2.0], rounded to 4 decimal places. Notice that for $P = [I \mid -C]$ the last column is just $-C$ . Scaling $P$ by 2 leaves the center unchanged, because the null direction stays the same.

Constraints

Input: A 3x4 projection matrix P as a list of lists
Use numpy for SVD computation
Return: A list [X, Y, Z] representing the camera center
Round to 4 decimal places
The last element of the null space vector is guaranteed to be non-zero

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Linear AlgebraFoundations Camera Models & Calibration3D Computer Vision

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Background Knowledge

The problem involves understanding the concept of a camera model in computer vision, specifically the projection matrix $P$ . The projection matrix is a $3 \times 4$ matrix that maps 3D points in world coordinates to 2D points in image coordinates. It is a fundamental concept in computer vision, as it allows us to relate the 3D world to the 2D images captured by a camera. The camera center, also known as the optical center, is the point in 3D space where all projection rays converge.

The equation $P \cdot \tilde{C} = \mathbf{0}$ indicates that the camera center lies in the null space of the projection matrix $P$ . The null space of a matrix is the set of vectors that, when multiplied by the matrix, result in the zero vector. In this case, the null space represents the direction in which the camera is pointing, and the camera center is the point in 3D space that lies on this direction. To find the null space, we can use techniques such as Singular Value Decomposition (SVD), which is a factorization of a matrix into the product of three matrices: $U$ , $\Sigma$ , and $V^T$ .

SVD is a powerful tool in linear algebra, and it has many applications in computer vision and machine learning. In the context of this problem, SVD allows us to decompose the projection matrix $P$ into its constituent parts, which can then be used to find the camera center. The SVD decomposition of a matrix $P$ is given by $P = U \Sigma V^T$ , where $U$ and $V$ are orthogonal matrices, and $\Sigma$ is a diagonal matrix containing the singular values of $P$ . The last column of $V$ (or last row of $V^T$ ) corresponds to the null space of $P$ , which in this case represents the camera center in homogeneous coordinates.

Algorithm/Approach

The general approach to solving this problem involves using SVD to find the null space of the projection matrix $P$ . This can be achieved by:

Computing the SVD of $P$
Extracting the last column of $V$ (or last row of $V^T$ ), which represents the camera center in homogeneous coordinates
Converting the homogeneous coordinates to Euclidean coordinates by dividing by the last element

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