Formal power series
... Applying linear algebra to number-sequences and countingproblems. Any questions about the homework or the material? Fibonacci numbers, understood via matrices Domino tilings of 3-by-n rectangles For all non-negative n, let a_n = number of domino tilings of a 3-by-2n rectangle (a_0 = 1) and let b_n = ...
... Applying linear algebra to number-sequences and countingproblems. Any questions about the homework or the material? Fibonacci numbers, understood via matrices Domino tilings of 3-by-n rectangles For all non-negative n, let a_n = number of domino tilings of a 3-by-2n rectangle (a_0 = 1) and let b_n = ...
download
... space not passing through either of the two camera centres. The ray through the first camera centre corresponding to the point x meets the plane π in a point X. This point X is then projected to a point x0 in the second image. This procedure is known as transfer via the plane π. Since X lies on the ...
... space not passing through either of the two camera centres. The ray through the first camera centre corresponding to the point x meets the plane π in a point X. This point X is then projected to a point x0 in the second image. This procedure is known as transfer via the plane π. Since X lies on the ...
Linear Span and Bases 1 Linear span
... Point 1 implies in particular, that every subspace of a finite-dimensional vector space is finite-dimensional. Points 2 and 3 show that if the dimension of a vector space is known to be n, then to check that a list of n vectors is a basis it is enough to check whether it spans V (resp. is linearly i ...
... Point 1 implies in particular, that every subspace of a finite-dimensional vector space is finite-dimensional. Points 2 and 3 show that if the dimension of a vector space is known to be n, then to check that a list of n vectors is a basis it is enough to check whether it spans V (resp. is linearly i ...
Tensor principal component analysis via sum-of
... where “=” denotes equality in the ring R[x]/(kxk2 − 1) and where s1 , . . . , sk have bounded degree, when such certificates exist. (The polynomials {si } and {t j } certify that h(x) 6 c. Otherwise c − h(x) would be negative, but this is impossible by the nonnegativity of squared polynomials.) Our ...
... where “=” denotes equality in the ring R[x]/(kxk2 − 1) and where s1 , . . . , sk have bounded degree, when such certificates exist. (The polynomials {si } and {t j } certify that h(x) 6 c. Otherwise c − h(x) would be negative, but this is impossible by the nonnegativity of squared polynomials.) Our ...
maximal compact normal subgroups and pro-lie groups
... We note that any open normal subgroup of G contains an isomorphic image of a subgroup Nq of FxF such that [A, A-1] G N0 for each A G F. To see this we note that, if A;G Y\Ki has zth coordinate 1 and all other coordinates 0, and / e JZ í¿ has zth coordinate [A,I], and all other coordinates [1,1], the ...
... We note that any open normal subgroup of G contains an isomorphic image of a subgroup Nq of FxF such that [A, A-1] G N0 for each A G F. To see this we note that, if A;G Y\Ki has zth coordinate 1 and all other coordinates 0, and / e JZ í¿ has zth coordinate [A,I], and all other coordinates [1,1], the ...
1 SUBSPACE TEST Strategy: We want to see if H is a
... Looking at the last two entries we see that a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 . However, abc = (u1 + v 1 )(u 2 + v 2 )(u 3 + v 3 ) ≠ u1 * u 2 * u 3 + v 1 * v 2 * v 3 , in general. ...
... Looking at the last two entries we see that a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 . However, abc = (u1 + v 1 )(u 2 + v 2 )(u 3 + v 3 ) ≠ u1 * u 2 * u 3 + v 1 * v 2 * v 3 , in general. ...
Laplace-Beltrami Eigenfunctions for Deformation Invariant Shape
... embedding [EK03, JZ07]. One considers the matrix of pairwise geodesic distances between points on the surface. Spectral embedding, for example Multidimensional Scaling, is used to “flatten” this structure – to get an embedding of these points into the Euclidean space such that Euclidean distances di ...
... embedding [EK03, JZ07]. One considers the matrix of pairwise geodesic distances between points on the surface. Spectral embedding, for example Multidimensional Scaling, is used to “flatten” this structure – to get an embedding of these points into the Euclidean space such that Euclidean distances di ...
Removal Lemmas for Matrices
... Here a set of pairwise-disjoint A-copies in M is a set of s × t submatrices of M , all equal to A, such that any entry of M is contained in at most one of the submatrices. Theorem 1.2 is an analogue for binary matrices of the non-induced graph removal lemma. However, in the graph removal lemma, δ −1 ...
... Here a set of pairwise-disjoint A-copies in M is a set of s × t submatrices of M , all equal to A, such that any entry of M is contained in at most one of the submatrices. Theorem 1.2 is an analogue for binary matrices of the non-induced graph removal lemma. However, in the graph removal lemma, δ −1 ...
Simple exponential estimate for the number of real zeros of
... invariant linear subspace for all monodromy operators. Conversely, if the monodromy group is reducible, and /i,..., /^ 0 < k < n, span the corresponding invariant subspace, then by the classical Riemann theorem one can construct an operator L^ G S) of order A;, satisfied by /i,..., fjc. But then one ...
... invariant linear subspace for all monodromy operators. Conversely, if the monodromy group is reducible, and /i,..., /^ 0 < k < n, span the corresponding invariant subspace, then by the classical Riemann theorem one can construct an operator L^ G S) of order A;, satisfied by /i,..., fjc. But then one ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.