MATH 105: Finite Mathematics 2
... Identity Matrix For any positive integer n, the identity matrix, In , is an n × n square matrix with 1s on the top-left to bottom-right diagonal and ...
... Identity Matrix For any positive integer n, the identity matrix, In , is an n × n square matrix with 1s on the top-left to bottom-right diagonal and ...
On Approximate Robust Counterparts of Uncertain Semidefinite and
... smaller than the sizes m of these matrices, so that this transformation reduces quite significantly the design dimension of the problem. As about LMIs, the 2L “large” (of the size m × m) LMIs X` ±A` [x] of the original problem are now replaced with 2(L − M ) “small” (of the sizes k` × k` ) LMIs (1 ...
... smaller than the sizes m of these matrices, so that this transformation reduces quite significantly the design dimension of the problem. As about LMIs, the 2L “large” (of the size m × m) LMIs X` ±A` [x] of the original problem are now replaced with 2(L − M ) “small” (of the sizes k` × k` ) LMIs (1 ...
How Much Does a Matrix of Rank k Weigh?
... basis of the row space of A, each row of A is a linear combination of the rows of R. That means there is a unique m × k matrix C such that A = C R. Note that the rank of C must also be k. Using the factorization A = C R we can express the set of rank k matrices as the Cartesian product of the set of ...
... basis of the row space of A, each row of A is a linear combination of the rows of R. That means there is a unique m × k matrix C such that A = C R. Note that the rank of C must also be k. Using the factorization A = C R we can express the set of rank k matrices as the Cartesian product of the set of ...
10.2 Linear Transformations
... Note that the above statement describes how a transformation T interacts with the two operations of vectors, addition and scalar multiplication. It tells us that if we take two vectors in the domain and add them in the domain, then transform the result, we will get the same thing as if we transform ...
... Note that the above statement describes how a transformation T interacts with the two operations of vectors, addition and scalar multiplication. It tells us that if we take two vectors in the domain and add them in the domain, then transform the result, we will get the same thing as if we transform ...
Document
... are in V, it follows that u + v is also in V. Hence the sum T(u) + T(v) = T(u + v) is in the range of T. (vector addition) 3. Let T(u) be a vector in the range of T and let c be a scalar. Because u is in V, it follows that cu is also in V. Hence, cT(u) = T(cu) is in the range of T. (scalar multiplic ...
... are in V, it follows that u + v is also in V. Hence the sum T(u) + T(v) = T(u + v) is in the range of T. (vector addition) 3. Let T(u) be a vector in the range of T and let c be a scalar. Because u is in V, it follows that cu is also in V. Hence, cT(u) = T(cu) is in the range of T. (scalar multiplic ...
full version
... We complement the above algorithmic result by the following lower bound. Theorem 1.5 (Informal Version). There is τ : N → R with τ 6 O(n3/4 / log(n)1/4 ) so that when T is an instance of Problem 1.1 with signal-to-noise ratio τ, with probability 1 − O(n−10 ), there exists a solution to the degree-4 ...
... We complement the above algorithmic result by the following lower bound. Theorem 1.5 (Informal Version). There is τ : N → R with τ 6 O(n3/4 / log(n)1/4 ) so that when T is an instance of Problem 1.1 with signal-to-noise ratio τ, with probability 1 − O(n−10 ), there exists a solution to the degree-4 ...
8. Linear Maps
... dim(Im L). So our assumption gives dim(Im L) = dim(V ), ie. Im L is a subspace of V of the same dimension as V . A corollary of the Dimension Theorem says that V = Im L. (b) For L surjective ie. Im L = V , our assumption says that dim(U ) = dim(Im L). So the dimension relation gives dim(Ker L) = 0. ...
... dim(Im L). So our assumption gives dim(Im L) = dim(V ), ie. Im L is a subspace of V of the same dimension as V . A corollary of the Dimension Theorem says that V = Im L. (b) For L surjective ie. Im L = V , our assumption says that dim(U ) = dim(Im L). So the dimension relation gives dim(Ker L) = 0. ...
Instability of standing waves for non-linear Schrödinger
... it follows that if WL n WL ^ {0} x (-1, +1), then there is a non-zero Po with (P o , T0) e WL n W+ and r 0 ^ ± 1. (2.5) then has a solution which decays to 0 as x -* ±00. Since there are stable and unstable manifolds associated to eigenvalues that lie off the imaginary axis, the decay is exponential ...
... it follows that if WL n WL ^ {0} x (-1, +1), then there is a non-zero Po with (P o , T0) e WL n W+ and r 0 ^ ± 1. (2.5) then has a solution which decays to 0 as x -* ±00. Since there are stable and unstable manifolds associated to eigenvalues that lie off the imaginary axis, the decay is exponential ...
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.