Linear null space

In terms of linear vector space, Buckingham s theorem (Theorem 2) simply states that the null space of the dimensional matrix has a fixed dimension, and Van Driest s rule (Theorem 3) then specifies the nullity of the dimensional matrix. The problem of finding a complete set of B-numbers is equivalent to that of computing a fundamental system of solutions of equation 13 called a complete set of B-vectors. For simplicity, the matrix formed by a complete set of B-vectors will be called a complete B-matrix. It can also be demonstrated that the choice of reference dimensions does not affect the B-numbers (22). 

Siace the columns of any complete B-matrix are a basis for the null space of the dimensional matrix, it follows that any two complete B-matrices are related by a nonsingular transformation. In other words, a complete B-matrix itself contains enough information as to which linear combiaations should be formed to obtain the optimized ones. Based on this observation, an efficient algorithm for the generation of an optimized complete B-matrix has been presented (22). No attempt is made here to demonstrate the algorithm. Instead, an example is being used to illustrate the results. 

Although matrix multiplications, row reductions, and calculation of null spaces can be done by hand for small matrices, a computer with programs for linear algebra are needed for large matrices. Mathematica is very convenient for this purpose. More information about the operations of linear algebra can be obtained from textbooks (Strang, 1988), but this section provides a brief introduction to making calculations with Mathematica (Wolfram, 1999). 

Therefore, the linear inverse problem (10.4) can have an infinite number of equivalent solutions. All these nonradiating currents form a so-called null-space of the linear inverse problem (10.4). In principle, we can overcome this difficulty using a regularization method, which restricts the class of the inverse excess currents j in equation (10.4) to physically meaningful solutions only. We will discuss an approach to the solution of this problem below, considering a quasi-linear method. 

First, one can experimentally measure as many fluxes (or more) as the dimension of the null-space, so as to uniquely calculate the remaining fluxesJ " This approach is called metabolic flux analysis. Alternatively, an objective of the metabolic network can be chosen to computationally explore the best use of the metabolic network by a given metabolic genotype. Herein, we pursue the second option. The solution to Eq. (7) subject to the linear inequality constraints can be formulated as a linear programming (LP) problem, in which one finds fhe flux distribution that minimizes a particular objective. Mathematically, the LP problem is stated as ... 

Exercise 2.9 Null space and fundamental theorem of linear algebra... 

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 set of null-space vectors of the matrix A represents the independent reactions that satisfy the stoichiometric condition of mass conservation. If we select all possible bimolecular reactions among all the species, but permit H2O and H+ to be additional reactants, and impose conservation of atoms and charge, then we find that linear combinations of the independent reactions yield seven elementary steps ... 

Null Space For many purposes in AR theory, it is useful to understand the set of concentrations that lie perpendicular (orthogonal) to S, which are spanned by the stoichiometric coefficient matrix A. For instance, the computation of critical DSR solution trajectories and CSTR effluent compositions that form part of the AR boundary require the computation of this space. It is therefore important that we briefly provide details of this topic here. It is simple to show from linear algebra that all points orthogonal to the space spanned by the columns of A are those that obey the following relation ... 

Here, mattix A is the stoichiometric coefficient matrix formed from the reaction stoichiometry, and C is any concentration vector that satisfies Equation 6.7. Determination of the subspace of concenttations orthogonal to the stoichiometric subspace, is in fact equivalent to computing the mil space of A. We shall denote by N the mattix whose columns form a basis for the null space of A. Linear combination of the columns in N hence generates the set of concentrations orthogonal to the stoichiometric subspace. 

Assume that A contains n rows (there are n components in the system). If A has d linearly independent columns (indicating d linearly independent reactions), then rank(A) = d. If the rank of A is d, then the null space of A must have rank (n - d), or rank(N) = (n - d) (Strang, 2003). Thus, N is a matrix having n rows and (n-d) columns. 

Computing the null space of a matrix is common function in linear algebra. We provide a brief description here for convenience however, many standard texts (Lay, 2012 Strang, 2003) describe this topic in detail. 

We can validate this result as well using a numerical linear algebra package. Using the null () function in MATLAB for the A provided produces the result Empty matrix 1-by- 0, indicating that there are no vectors that form part of the null space (excluding the trivial solution). 

Hence, Xj = -2 if X2 = 1 and X3 = 0. We have therefore found a combination of values for Xj, X2, and X3 that produce zero when multiplied by A. In the language of linear algebra, we have found a basis for the null space of A. The vector... 

Linear combinations of iij and n2 also describe a two-dimensional subspace in and in this instance the dimension of the stoichiometric subspace is the same as the dimension of the null space. 

Notice that if we substitute matrix X discussed in this example with the stoichiometric coefficient matrix A, then the columns in X (Xj and X2) represent two reactions participating in n-dimensional concentration space, R". Hence, to compute N, we simply determine the stoichiometric coefficient matrix A (as in Section 6.2.1.3), and then compute the null space of A. From linear algebra, we can show that if A has size nxd (n components participating in d reactions), the size of matrix N will benx(n-d). 

From Section 7.2.1.1, the dimension of the AR is three (d=3) and there are four components (n = 4). It is expected that the dimension of the subspace orthogonal to the stoichiometric subspace is (4 - 3) = 1. Therefore, for the three-dimensional Van de Vusse system, the null space is given by a one-dimensional subspace (in other words, the basis for the null space is composed of a single, linearly independent vector). This is confirmed when the null space of A is computed, giving... 

The difference between vector projection on a set of linear constraints and the use of null space of constraints to satisfy the constraints was illustrated earlier in Chapter 8. In both cases, the original direction is transformed into a new direction that satisfies the selected linear constraints. 

In Chapter 8, we showed that in the case of linear constraints with the matrix A m < n it is possible to obtain the null space of the matrix using a Gauss factorization with good stability features. 

The difference of any two solutions of the linear system Ax = b is a member of the null space of A. Thus this system has at most one solution if and only if the nullity of A is zero. If the system is square (that is, if A is X n), then there will be a solution for every right-hand side b if and only if the collection of columns of A is linearly independent, which is the same as saying the rank of A is . In this case the nullity must be zero. Thus, for any b, the square system Ax = b has exactly one solution if and only if rank A = . In other words the n x matrix A is invertible if and only if rank A = . 

The null space KerB is a vector subspace of the whole space of Af-vectors. Let us have M-L linearly independent vectors (transposed row vectors) oej e KerB (A = 1, —, M-L), thus a basis of KerB . The basis can be completed by some L linearly independent vectors, say Bk (A = 1, —, L) to a basis of. Then the matrix... 

Now comes an important observation. Because L is not an invertible operator, bounded solutions of (13) exist only if its right hand side is orthogonal to all elements in the null space of the adjoint operator L. The null space of L is easily determined it is the three-dimensional space spanned by the linearly independent functions... 

Theorem Uniqueness of solution for Ax = b Let x e St be a solution to the linear system Ax = b, where b 6 and A is an NxN real matrix. If the null space (kernel) of A contains only the null vector, Ka = 0, this solution is unique. 

Let US examine how SVD aids detecting the existence and uniqueness properties of linear systems. As noted in Chapter 1, the nature of the null space (kernel) of A and of the range are vitally important however, we have not described how we may identify these subspaces for a particular matrix. Let 4 be a real, square N x N matrix, with the SVD A= fVH... 

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