Vector linear independence

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

What we fomierly called the nonhomogeneous vector (Chapter 2) is zero in the pair of simultaneous nomial equations Eq. set (6-38). When this vector vanishes, the pair is homogeneous. Let us try to construct a simple set of linearly independent homogeneous simultaneous equations. 

Any linearly independent set of simultaneous homogeneous equations we can construct has only the zero vector as its solution set. This is not acceptable, for it means that the wave function vanishes, which is contrai y to hypothesis (the electron has to be somewhere). We are driven to the conclusion that the normal equations (6-38) must be linearly dependent. 

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... 

Equation (8.90) is non-singular since it has a non-zero determinant. Also the two row and column vectors can be seen to be linearly independent, so it is of rank 2 and therefore the system is controllable. 

Just as a known root of an algebraic equation can be divided out, and the equation reduced to one of lower order, so a known root and the vector belonging to it can be used to reduce the matrix to one of lower order whose roots are the yet unknown roots. In principle this can be continued until the matrix reduces to a scalar, which is the last remaining root. The process is known as deflation. Quite generally, in fact, let P be a matrix of, say, p linearly independent columns such that each column of AP is a linear combination of columns of P itself. In particular, this will be true if the columns of P are characteristic vectors. Then... 

In ra-space any n + 1 of these are linearly dependent. But unless the matrix is rather special in form (derogatory), there exist vectors vx for which any n consecutive vectors are linearly independent (in possible contrast to the behavior in the limit). In fact, this is true of almost every vector vx. Hence, if... 

The vectors 0, , e , and are linearly independent. In this coordinate system an arbitrary vector can be written as follows... 

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... 

The following three vectors z, z and Zj with dimension four are linearly independent ... 

A set of n vectors of dimension n which are linearly independent is called a basis of an -dimensional vector space. There can be several bases of the same vector... 

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. 

Constraint Qualification For a local optimum to satisfy the KKT conditions, an additional regularity condition is required on the constraints. This can be defined in several ways. A typical condition is that the active constraints at x be linearly independent i.e., the matrix [Vh(x ) I VgA(x )] is full column rank, where gA is the vector of inequality constraints with elements that satisfy g x ) = 0. With this constraint qualification, the KKT multipliers (X, v) are guaranteed to be unique at the optimal solution. 

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... 

On the other hand if Xj(l < j < n) is a set of n linearly independent column vectors (matrices of order nx 1), then any column matrix (vector) y can be expressed as a linear combination of the vectors x so that the coefficients c exist such that... 

This result can be generalized into the statement that any arbitrary vector in n dimensions can always be expressed as a linear combination of re basic vectors, provided these are linearly independent. It will be shown that the latent solutions of a singular matrix provide an acceptable set of basis vectors, just like the eigen-solutions of certain differential equations provide an acceptable set of basis functions. 

In the theory of optics this phenomenon is accounted for in terms of geometrical construction, but the physical picture is less convincing. Double refraction is a well-documented property of most crystals, at its most spectacular in Iceland spar. The double image of an object viewed through the crystal indicates the existence of two independent rays and not the components of a single ray. In mathematical terms the two rays are linearly independent and therefore orthogonal. Any intermediate situation represents a linear combination of the two orthogonal basis vectors and can be resolved into two components. What happens to an individual photon is however, not clear. 

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. 

Let x be a local minimum or maximum for the problem (8.15), and assume that the constraint gradients Vhj(x ),j — 1,m, are linearly independent. Then there exists a vector of Lagrange multipliers A = (Af,..., A ) such that (x A ) satisfies the first-order necessary conditions (8.17)-(8.18). 

Formally, we are free to choose any linear combination of the three chemical species as the reacting scalar under the condition that the combination is linearly independent of the rows of A.12 Arbitrarily choosing C3, a new scalar vector can be defined by the linear transformation13... 

Note that 7 depends on the choice of both the reference vector and the linearly independent columns. 

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. 

If AW AW the process of finding a linear-mixture basis can be tedious. Fortunately, however, in practical applications Nm is usually not greater than 2 or 3, and thus it is rarely necessary to search for more than one or two combinations of linearly independent columns for each reference vector. In the rare cases where A m > 3, the linear mixtures are often easy to identify. For example, in a tubular reactor with multiple side-injection streams, the side streams might all have the same inlet concentrations so that c(2) = = c(iVin). The stationary flow calculation would then require only AW = 1 mixture-fraction components to describe mixing between inlet 1 and the Nm — I side streams. In summary, as illustrated in Fig. 5.7, a turbulent reacting flow for which a linear-mixture basis exists can be completely described in terms of a transformed composition vector ipm( defined by... 

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... 

However, care must be taken to avoid the singularity that occurs when C is not full rank. In general, the rank of C will be equal to the number of random variables needed to define the joint PDF. Likewise, its rank deficiency will be equal to the number of random variables that can be expressed as linear functions of other random variables. Thus, the covariance matrix can be used to decompose the composition vector into its linearly independent and linearly dependent components. The joint PDF of the linearly independent components can then be approximated by (5.332). 

If det A =0, the column-vectors A are not linearly independent (nor are the column-rows) and the matrix is singular. At least one edge-vector of the hyper-prism made of the column-vectors is in the subspace of the remaining edge-vectors the volume of the hyper-prism vanishes and det A = 0. 

Unless the initial vector is already an eigenvector, the Krylov vectors are linearly independent and they eventually span the eigenspace of H ... 

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]... 

It can be straightforwardly verified that indeed NK = 0. Each feasible steady-state flux v° can thus be decomposed into the contributions of two linearly independent column vectors, corresponding to either net ATP production (k ) or a branching flux at the level of triosephosphates (k2). See Fig. 5 for a comparison. An additional analysis of the nullspace in the context of large-scale reaction networks is given in Section V. 

Let s assume the elements ci, a and cz of vector c are the unknowns. Thus, the system is comprised of three equations with three unknowns. Such systems of n equations with n unknowns have exactly one solution if none of the individual equations can be expressed by linear combinations of the remaining ones, i.e. if they are linearly independent. Then, the coefficient matrix A is of full rank and non-singular and its inverse, A1, exists such that right multiplication of equation (2.20) with A 1 allows the determination of the unknowns. 

For the computation of the pseudo-inverse, it is crucial that the vectors f j are not parallel, or more correctly, that they are linearly independent. Otherwise, the matrix FlF is singular and cannot be inverted. Matlab issues a warning. We can gain a certain level of understanding by adapting Figure 4-10 ... 

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