Eigenvectors linear independence

The matrix X is easily seen to be unitary. Since the n eigenvectors are linearly independent, the matrix X is non-singular and its inverse X exists. If we multiply equation (1.57) from the left by X , we obtain... 

To show that the eigenvectors are linearly independent assume that a linear relationship does exist. Then... 

A. 1 Definitions / A.2 Basic Matrix Operations / A.3 Linear Independence and Row Operations / A.4 Solution of Linear Equations / A. 5 Eigenvalues, Eigenvectors / References /... 

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

Of course, any linear combination of null eigenvectors also satisfies (12.54a, b), so we are able to choose individual vectors of this null manifold with considerable freedom. We know in general that a / -dimensional manifold of linearly independent null eigenvectors must exist, in order that the rank of the metric matrix,... 

The nonsingularity of the scale-factor matrix D is a necessary consequence of the linear independence of the chosen axis vectors R ). To see this, suppose instead that D has a null eigenvector t], satisfying... 

A and B are W-matrices, C is square, and at least some elements of D and E are non-zero. A splitting matrix is reducible, with the additional property that when the rows and columns corresponding to the transient states are erased a decomposable matrix is left. There are three sets of states, labelled with a, h, c. The states c are transient and deplete into a and b. There are (at least) two linearly independent eigenvectors with eigenvalue zero, viz. 

Exercise. Let W have, at the eigenvalue zero, two linearly independent eigenvectors with positive components. Then W is decomposable. 

Some matrices are diagonalizable over C in the sense that they have a basis of - possibly complex - eigenvectors and these matrices can therefore be diagonalized by matrix similarity. For example, this is so for all n by n matrices with n distinct eigenvalues since eigenvectors for different eigenvalues are linearly independent. Other matrices such... 

The eigenvalues q)2 in eq. (2) are not necessarily all distinct, so the index j will now be replaced by the double index ak where a labels the distinct eigenvalues of D(q) and k= 1, 2,. .., 1(a) labels the linearly independent (LI) eigenvectors associated with the degenerate eigenvalue a. In this notation,... 

Smith and Rocchesso have shown that a DWN scattering matrix (or a FDN feedback matrix) is lossless if and only if its eigenvalues have unit modulus and its eigenvectors are linearly independent. Therefore, lossless scattering matrices may be fully parameterized as... 

Eigenvectors for different characters are linearly independent, since p(v) = v Xv and we know by (2.2) that the different Xv are independent Hence there are only finitely many different jf > and the subgroup H = g xtv — Xv bas finite index in S. But for each n in the equality Xv n) = Xv(ff ifW) is a polynomial equation in g, and. thus H is closed. A connected S cannot have a proper closed subgroup of finite index, since by (S.2) the cosets would disconnect S. Thus H = S, and S acts on all gv by the same character. 

Any real square matrix has N (complex) eigenvalues (not necessarily different), and at most N linearly independent eigenvectors. It can be proved that all eigenvalues of a symmetric matrix A are real numbers, and A has exactly N different eigenvectors. The normalized eigenvectors of a symmetric matrix form an orthonormal set of vectors... 

The orthogonality extends to the linearly independent members of a degenerate set. An eigenvector is labelled with its eigenvalue for the property Ci because, when the system is in a state described by this vector, one is certain that the measurement of the property 2 will yield the particular eigenvalue COf. Since the cUj are the only observable values of the property 2, actual measurement of this property will carry the state vector IT ) of a system over into one of the eigenvectors caj>. In this way the measurement of a physical property places the system in a specific state. Once the measurement has been carried out to yield the value co one can be certain that the state vector... 

The typical situation is for the eigenvalues to be distinct A, Aj. In this case, a theorem of linear algebra states that the corresponding eigenvectors v, and are linearly independent, and hence span the entire plane (Figure 5.2.1). In particular, any initial condition Xq can be written as a linear combination of eigenvectors, say Xo =c,v,+CjVj. 

Simple molecular orbitals (eigenvectors) corresponding to x > 0, x < 0, and x = 0 are called bonding, antibonding, and non-bonding, respectively. Note that the number of linearly independent non-bonding molecular orbitals (NBMOs) is equal to the multiplicity of the number zero in the graph spectrum. This fact can be related to the stability of the... 

In general, if there exist two unit eigenvalues with two linearly independent eigenvectors, respectively, then the system of variational equations has two independent periodic solutions. 

To each eigenvalue of A(T) there corresponds a solution (f), so we have four linearly independent solutions. However, to the double unit eigenvalue Ai = A2 = 1, there exists only one unit eigenvector (0) = x(0). The two linearly independent solutions corresponding to the double unit eigenvalue are... 

The vectors and belong to the eigenvalue +1, the vectors and belong to the eigenvalue —I of 0 (in the standard representation). Notice that each eigenvalue of the matrix 0 has two linearly independent eigenvectors. 

As each eigenvalue has two linearly independent eigenvectors, there is some freedom in the choice of an eigenvector basis. Another frequently used basis is discussed in the following section. 

From the eigenvectors of h(p) we obtain the solutions of the Dirac equation that correspond to the plane waves of the Schrodinger equation. The four linearly independent functions... 

K > 0 of the real symmetric matrix H, there exists n orthonormal and linearly independent eigenvectors. 

Since these eigenvectors are linearly independent, any arbitrary vector X with a euclidean norm of unity may be stated as a linear combination of this orthonormal set of eigenvectors... 

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