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    Trifocal Tensor: Exploring Depth, Motion, and Structure in Computer Vision
    Trifocal Tensor: Exploring Depth, Motion, and Structure in Computer Vision
    Trifocal Tensor: Exploring Depth, Motion, and Structure in Computer Vision
    Ebook191 pages1 hour

    Trifocal Tensor: Exploring Depth, Motion, and Structure in Computer Vision

    By Fouad Sabry

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    About this ebook

    What is Trifocal Tensor


    Within the realm of computer vision, the trifocal tensor is a numerical array that boasts dimensions of 3×3×3 and encompasses all the geometric relationships that are projective among the three perspectives. The coordinates of matching points or lines in three different views are related to one another by this method, which is independent of the structure of the scene and relies solely on the relative motion between the three views as well as the intrinsic calibration parameters of each view. As a result, the trifocal tensor can be thought of as the generalization of the fundamental matrix in three different perspectives. In spite of the fact that the tensor is composed of 27 elements, it is important to highlight that only 18 of those elements are genuinely independent.


    How you will benefit


    (I) Insights, and validations about the following topics:


    Chapter 1: Trifocal_tensor


    Chapter 2: Rank_(linear_algebra)


    Chapter 3: Trace_(linear_algebra)


    Chapter 4: Principal_component_analysis


    Chapter 5: Translation_(geometry)


    Chapter 6: Kronecker_product


    Chapter 7: Eigenvalues_and_eigenvectors


    Chapter 8: Three-dimensional_space


    Chapter 9: Fundamental_matrix_(computer_vision)


    Chapter 10: Corner_detection


    (II) Answering the public top questions about trifocal tensor.


    (III) Real world examples for the usage of trifocal tensor in many fields.


    Who this book is for


    Professionals, undergraduate and graduate students, enthusiasts, hobbyists, and those who want to go beyond basic knowledge or information for any kind of Trifocal Tensor.

    LanguageEnglish
    PublisherOne Billion Knowledgeable
    Release dateMay 1, 2024
    Trifocal Tensor: Exploring Depth, Motion, and Structure in Computer Vision

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      Book preview

      Trifocal Tensor - Fouad Sabry

      Chapter 1: Trifocal tensor

      Regarding computer vision, the trifocal tensor (also tritensor) is a 3×3×3 array of numbers (i.e, a tensor) that represents the whole set of relations between three viewpoints in projective geometry.

      Three-dimensional coordinates of parallel lines and points are related, having nothing to do with the topology of the scene and everything to do with relative mobility (i.e, stance) between the three perspectives and their innate calibration factors.

      Hence, The trifocal tensor can be thought of as a three-dimensional version of the basic matrix.

      It is pointed out that although the tensor consists of 27 elements, Only 18 of them can truly be considered self-sufficient.

      There are 11 degrees of freedom, or independent elements, in the so-called calibrated trifocal tensor, which encodes the relative posture of the cameras up to global scale by relating the coordinates of points and lines in three views given their intrinsic properties. Fewer correspondences need to be fitted into the model due to the decreased degrees of freedom, but this comes at the expense of higher nonlinearity.

      The tensor can also be seen as a collection of three rank-two 3 x 3 matrices {\mathbf T}_1, \; {\mathbf T}_2, \; {\mathbf T}_3 known as its correlation slices.

      Assuming that the projection matrices of three views are {\mathbf P}=[ {\mathbf I} \; | \; {\mathbf 0} ] , {\displaystyle {\mathbf {P} }'=[{\mathbf {A} }\;|\;{\mathbf {a} }_{4}]} and {\displaystyle {\mathbf {P} ''}=[{\mathbf {B} }\;|\;{\mathbf {b} }_{4}]} , the correlation slices of the corresponding tensor can be expressed in closed form as

      {\mathbf T}_i={\mathbf a}_i {\mathbf b}_4^t - {\mathbf a}_4 {\mathbf b}_i^t, \; i=1 \ldots 3

      , where {\mathbf a}_i, \; {\mathbf b}_i are respectively the ith columns of the camera matrices.

      In practice, however, The tensor is computed by comparing each of the three perspectives for point and line matching.

      Linear relationships between lines and points in three images are one of the most useful results of the trifocal tensor.

      More specifically, for triplets of corresponding points {\displaystyle {\mathbf {x} }\;\leftrightarrow \;{\mathbf {x} }'\;\leftrightarrow \;{\mathbf {x} }''} and any corresponding lines {\displaystyle {\mathbf {l} }\;\leftrightarrow \;{\mathbf {l} }'\;\leftrightarrow \;{\mathbf {l} }''} through them, The three-linear restrictions listed below are valid:

      {\displaystyle ({\mathbf {l} }^{\prime t}\left[{\mathbf {T} }_{1},\;{\mathbf {T} }_{2},\;{\mathbf {T} }_{3}\right]{\mathbf {l} }'')[{\mathbf {l} }]_{\times }={\mathbf {0} }^{t}}{\displaystyle {\mathbf {l} }^{\prime t}\left(\sum _{i}x_{i}{\mathbf {T} }_{i}\right){\mathbf {l} }''=0}{\displaystyle {\mathbf {l} }^{\prime t}\left(\sum _{i}x_{i}{\mathbf {T} }_{i}\right)[{\mathbf {x} }'']_{\times }={\mathbf {0} }^{t}}{\displaystyle [{\mathbf {x} }']_{\times }\left(\sum _{i}x_{i}{\mathbf {T} }_{i}\right){\mathbf {l} }''={\mathbf {0} }}{\displaystyle [{\mathbf {x} }']_{\times }\left(\sum _{i}x_{i}{\mathbf {T} }_{i}\right)[{\mathbf {x} }'']_{\times }={\mathbf {0} }_{3\times 3}}

      where [\cdot]_{\times} denotes the skew-symmetric cross product matrix.

      The location of a point in a third view can be determined from a pair of matched points in two views using only the trifocal tensor of those views. Point transfer describes this phenomenon, which also applies to lines and conics. Transferring generic curves as conics is possible by first modeling them as osculating circles in a local differential curve. However, the issue of uncalibrated trifocal tensors is still up in the air.

      The standard case involves six-point correspondences with three solutions.

      Recent years have seen the solution to the problem of calculating the trifocal tensor using a mere nine line correspondences.

      The calibrated trifocal tensor estimation is said to be extremely challenging and calls for four point correspondences. The same method also proved to be minimum with degree 216 for the combined situation of three point correspondences and one line correspondence.

      {End Chapter 1}

      Chapter 2: Rank (linear algebra)

      The rank of a matrix A is the number of dimensions of the vector space formed (or spanned) by its columns in linear algebra. Therefore, the nondegenerateness of the linear equations and linear transformations stored by A can be quantified in terms of its rank. Rank can be understood in a variety of ways. The rank of a matrix is a very elementary property.

      Rank is typically represented by the symbols rank(A) or rk (A); The rank of a matrix is defined in this section. There is a wide range of potential meanings, some of which are explored under Alternative Meanings.

      If A is a set, then its column rank is the number of elements in its column space, and its row rank is the number of elements in its row space.

      In linear algebra, it is a first-order result that the rank of a column always equals the rank of a row.

      (Three proofs of this result are given in § Proofs that column rank = row rank, below.) Consider this sum (i.e, Counting the number of unique rows and columns in A yields its rank.

      If a matrix has the same number of rows and columns as the largest conceivable matrix of those dimensions, then it is said to have full rank. If a matrix does not have full rank, we say that it is rank-deficient. If a matrix has fewer rows than it does columns, we say that it has a rank deficiency.

      The rank of a linear map or operator \Phi is defined as the dimension of its image:

      {\displaystyle \operatorname {rank} (\Phi ):=\dim(\operatorname {img} (\Phi ))}

      where {\displaystyle \dim } is the dimension of a vector space, and {\displaystyle \operatorname {img} } is the image of a map.

      The matrix

      {\displaystyle {\begin{bmatrix}1&0&1\\-2&-3&1\\3&3&0\end{bmatrix}}}

      has rank 2; there are at least two linearly independent columns (the first and second), therefore the rank is at least 2, but the rank is not greater than 3 since the third column is a linear combination of the first two (the first minus the second).

      The matrix

      {\displaystyle A={\begin{bmatrix}1&1&0&2\\-1&-1&0&-2\end{bmatrix}}}

      has rank 1; there are some columns, hence the rank is nonzero; yet, any two columns are linearly dependant on one another. In a similar vein, the inversion

      {\displaystyle A^{\mathrm {T} }={\begin{bmatrix}1&-1\\1&-1\\0&0\\2&-2\end{bmatrix}}}

      at the first spot, A.

      Indeed, given that A's transposition has the same column vectors as A's row vectors, To say that a matrix has the same rank as its transposition is comparable to saying that its column rank is the same as its row rank, i.e, rank(A) = rank(AT).

      Finding a matrix's rank typically involves transforming it into a simpler form, called row echelon form, using only simple row operations. Due to their invertibility, row operations do not alter the row space (and thus the row rank) and instead translate the column space to an isomorphic space (hence do not change the column rank). Row echelon form makes the rank equal to the sum of the numbers of pivots (or basic columns) plus the number of non-zero rows.

      Consider the matrix A, which can be written as

      {\displaystyle A={\begin{bmatrix}1&2&1\\-2&-3&1\\3&5&0\end{bmatrix}}}

      can be written down in compact row-echelon form using the subsequent basic row operations::

      {\displaystyle {\begin{aligned}{\begin{bmatrix}1&2&1\\-2&-3&1\\3&5&0\end{bmatrix}}&\xrightarrow {2R_{1}+R_{2}\to R_{2}} {\begin{bmatrix}1&2&1\\0&1&3\\3&5&0\end{bmatrix}}\xrightarrow {-3R_{1}+R_{3}\to R_{3}} {\begin{bmatrix}1&2&1\\0&1&3\\0&-1&-3\end{bmatrix}}\\&\xrightarrow {R_{2}+R_{3}\to R_{3}} \,\,{\begin{bmatrix}1&2&1\\0&1&3\\0&0&0\end{bmatrix}}\xrightarrow {-2R_{2}+R_{1}\to R_{1}} {\begin{bmatrix}1&0&-5\\0&1&3\\0&0&0\end{bmatrix}}~.\end{aligned}}}

      As a result, the rank of matrix A is 2, as the final matrix (in row echelon form) has two rows that are not 0.

      Basic Gaussian elimination (LU decomposition) can be unstable when employed in floating point computations on computers; a rank-revealing decomposition is recommended instead. Singular value decomposition (SVD) is a powerful alternative, although there are cheaper alternatives that are even more numerically robust than Gaussian elimination, such as QR decomposition with pivoting (so-called rank-revealing QR factorization). A practical choice that depends on both the matrix and the application is needed for the numerical determination of rank, such as a criterion for selecting when a value, such as a singular value from the SVD, should be considered as zero.

      In linear algebra, the equality of the column and row ranks of a matrix is a fundamental property.

      Numerous examples have been provided.

      One of the most elementary ones has been sketched in § Rank from row echelon forms.

      An alternative proof is presented here:

      It's easy to demonstrate that a basic row operation has no effect on either the row rank or the column rank. Due to the nature of Gaussian elimination, which involves only simple row operations, a matrix's row rank and column rank are both preserved in its reduced row echelon version. The matrix can be transformed into an identity matrix by performing a few more elementary column operations, such as adding a row of zeros to each side. Once again, this has no effect on the order of the rows or columns. The number of matrices with nonzero entries is directly proportional to the rank of each row and column.

      Two more proofs of this result are provided. The first one is field-independent and employs just elementary features of linear combinations of vectors. Wardlaw is the foundation of the proof (2005).

      Let A be an m × n matrix.

      Take r to be A's column rank, and let c1, .., cr be any basis for the column space of A.

      Place these as the columns of an m × r matrix C.

      Each and every one of A's columns can be written as a linear combination of C's r columns.

      This means that there is an r × n matrix R such that A = CR.

      R is the matrix whose ith column is formed from the coefficients giving the ith column of A as a linear combination of the r columns of C.

      To rephrase, R is the matrix which contains the multiples for the bases of the column space of A (which is C), in order to complete the letter A.

      Now, Linearly combining the r rows of R yields each row of A.

      Therefore, The rows in R span both the row space A and B, according to the Steinitz exchange lemma, r is the maximum row rank that A can have.

      It's evidence that A has a lower row rank than it does a column rank.

      Any matrix can benefit from this conclusion, then, use the solution for A's transposition.

      When A is transposed, its row rank becomes its column rank, and its column rank becomes its row rank, This proves the inverse inequality and gives us A's row rank and column rank being equal.

      (For a related concept, see Rank factorization.)

      Let A be an m × n matrix with entries in the real numbers whose row rank is r.

      Therefore, A has r rows, hence r is the dimension of its row space.

      Let x1, x2, …, xr be a basis of the row space of A.

      We claim that the vectors Ax1, Ax2, …, Axr are linearly independent.

      To understand, consider a linear homogeneous relation involving these vectors with scalar coefficients c1, c2, …, cr:

      {\displaystyle 0=c_{1}A\mathbf {x} _{1}+c_{2}A\mathbf {x} _{2}+\cdots +c_{r}A\mathbf {x} _{r}=A(c_{1}\mathbf {x} _{1}+c_{2}\mathbf {x} _{2}+\cdots +c_{r}\mathbf {x} _{r})=A\mathbf {v} ,}

      where v = c1x1 + c2x2 + ⋯ + crxr.

      We note two things: A linear combination of vectors in the row space of A is denoted by v, thus v must exist in the A rows, Moreover (b), because Av = 0, Each row vector of A and the vector v are orthogonal to one another, hence, is perpendicular to every vector in A's row space.

      Inferring that v is orthogonal to itself from (a) and (b), that either v = 0 or, According

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