Vector linearly independent

The matrices F and M can be found from straightforward integration of (5.9) with the initial conditions being N linearly independent vectors. Then the quasienergy partition function equals... 

We have noted that if is the energy-momentum four vector of a photon (i.e., P = 0, k0 > 0) there exist only two other linearly independent vectors orthogonal to ku. We shall denote these as tft k) and ejf fc). They satisfy... 

It has been shown that the p columns of an nxp matrix X generate a pattern of p points in 5" which we call PP. The dimension of this pattern is called rank and is indicated by liPP). It is equal to the number of linearly independent vectors from which all p columns of X can be constructed as linear combinations. Hence, the rank of PP can be at most equal to p. Geometrically, the rank of P can be seen as the minimum number of dimensions that is required to represent the p points in the pattern together with the origin of space. Linear dependences among the p columns of X will cause coplanarity of some of the p vectors and hence reduce the minimum number of dimensions. 

The length (norm) of the column vector v is the positive root Adv. A vector is normalized if its length is 1. Two vectors rj and rj of an n-dimensional set are said to be linearly independent of each other if one is not a constant multiple of the other, i.e., it is impossible to find a scalar c such that ri = crj. In simple words, this means that r - and Tj are not parallel. In general, m vectors constitute a set of linearly independent vectors if and only if the equation... 

For a given p two linearly independent vectors e are possible. If the z-axis is taken to be directed along p, these two vectors can be defined in terms of the unit vectors Xi and Xrn along the x and y-axes respectively. 

Since B depends on the choice of the linearly independent vectors used to form d> , all possible combinations must be explored in order to determine if one of them satisfies (5.96) and (5.97). Any set of linearly independent columns of d> that yields a matrix B satisfying (5.96) and (5.97) will be referred to hereinafter as a mixture-fraction basis. 

In order to show that no mixture-fraction basis exists, it is necessary to check all possible reference vectors. For each choice of the reference vector, there are three possible sets of linearly independent vectors that can be used to compute B. Thus, we must check a total of 12 possible mixture-fraction bases. Starting with c(0) as the reference vector, the three possible values of B(0) are... 

The left nullspace E of the stoichiometric matrix N is defined by a set of linearly independent vectors ej that are arranged into a matrix E that fulfills [50, 96]... 

If r = n, then the linearly independent vectors span the entire... 

It is also true that for a general vector space any set of linearly independent vectors can be combined in analogous fashion to give a set of orthonormal vectors. In this case the scalar product is defined by... 

Linear operators in finite-dimensional spaces. It is supposed that an n-dimensional vector space Rn is equipped with an inner product (, ) and associated norm a = / x, x). By the definition of finite-dimensional space, any vector x 6 Rn can uniquely be represented as a linear combination x = Cj + c of linearly independent vectors, ..., which constitute a basis for the space Rn. The numbers ck are called the coordinates of the vector x. One can always choose as a basis an orthogonal and normed system of vectors. .. , n ... 

Remark. Apart from the question whether the set of all eigenfunctions is complete, one is in practice often faced with the following problem. Suppose for a certain operator W one has been able to determine a set of solutions of (7.1). Are they all solutions For a finite matrix W this question can be answered by counting the number of linearly independent vectors one has found. For some problems with a Hilbert space of infinite dimensions it is possible to show directly that the solutions are a complete set, see, e.g., VI.8. Ordinarily one assumes that any reasonably systematic method for calculating the eigenfunctions will give all of them, but some problems have one or more unsuspected exceptional eigenfunctions. 

This procedure depends upon v 0 in each step, which is guaranteed when we start from linearly independent vectors. That it generates orthonormal vectors is shown by recursion, and by the hermiticity (or symmetry) of ( ). 

The rank for a set of vectors is the maximum number of linearly independent vectors. If, from the total set of vectors, one chooses the combination containing the maximum number of linearly independent vectors it will be a basis. The matrix rank remains unchanged if one adds a row that is a linear combination of the other rows and if this row is cancelled. 

First, we need to identify the state variables y describing the chemical equilibrium. Following the notation of Gorban and Karlin (2003), the conservation laws in chemical reactions introduce k linearly independent vectors blr b2,..., bk. The state variables describing the system at the chemical equilibrium are... 

Here, (I, 12,13) denote three linearly independent vectors of R3, serving as a basis of A, whereas n1,n2>n3 = n> say, is a triad of integers. The symbol 0 = 0,0,0 will be arbitrarily chosen to designate the lattice origin. 

Basis vectors Linearly independent vectors a, b, and c that generate the lattice. 

To solve this problem we assume that F j = l,2,...n is a system of linear independent vectors, which forms the subspace L C M. If the dimension of M is greater than L, the element m is not unequally defined by (C.9). So we can find the solution of (C.9) which possesses the additional properties, for example the smallest norm. 

The maximum number of linearly independent vectors in a linear space or subspace is said to be the dimension of the space. Such a linearly independent set of vectors is said to be a basis for that space, by which it is meant that any arbitrary vector in the space can be expressed as a linear combination of the basis set. 

Let V represent a set of linearly independent vectors, v, V2,Vz, , Vn, in a N- dimensional space which can in general be a complex vector space. We can define a general non-singular linear transformation A for the basis V to go to a new basis Z ... 

Equation 2.67 in Exercise 2.8 is begging to be analyzed by the fundamental theorem of linear algebra [7], so we explore that concept here. Consider an arbitrary m x n matrix, B. The null space of matrix B, written 3 B), is defined to be the set of all vectors x such that S x = 0. The dimension of SEiB) is the. number of linearly independent vectors x satisfying B x == 0. One of the remarkable results of the fundamental theorem of linear algebra is the relation... 

The numerical support for computing null spaces is excellent. For example, the Octave command null CB) returns a matrix with columns consisting of linearly independent vectors in the null space of matrix B. ... 

The rank of a matrix is the maximum number of linearly independent vectors (rows or columns) in an X/t matrix denoted as r X). Linearly dependent rows or columns reduce the rank of a matrix. 

Necessity of Lemma Vectors (ya) form a subspace the dimension of which is given by a number p of linear independent vectors. We arrange (ya) in such a way that p linearly independent vectors are at the beginning. 

Then there exists tensor Q unique for these both p-tuples of linear independent vectors giving... 

Result (A.19) proves the necessity of a Lemma for linear independent vectors (and in fact it transforms one basis y-y of p-dimensional vector space to another one by linear transformation Q). To prove this for the remaining linearly dependent vectors y, we express them through those which are independent... 

In Seet. 4.2, we need veetor spaee with abasis whieh is formed by A linear independent vectors gp p =, ..., k) which are not generally perpendicular or of unit length [12, 18, 19]. Sueh nonorthogonal basis, we eall a contravariant one. Covariant components of the so called metric tensor are defined by... 

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