Since the Jordan block for each eigenvalue with modulus 1 is diagonal, we see that there is an invertible matrix such that the -sum of each row of is less than or equal to 1, that is, , where is the maximum row sum matrix norm [ 1 , page ]. Define a matrix norm by. Then we have. Let be an eigenvalue of a matrix.
The index of , denoted by index is the smallest value of for which [ 1 , pages and ]. Thus condition e above can be restated as , and for any eigenvalue of with ,. We call the -average of. As with , we have for every if and only if in , and that is bounded for every if and only if is bounded in. We have the following theorem.
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First we prove the sufficiency part of a. Let be a vector in.
By Theorem 2 for any eigenvalues of either or and. If is written in its Jordan canonical form , then the -average of is , where is the -average of.
For a Jordan block of of the form. Its -average has constant diagonal and upper diagonals. Let be the constat value of its th upper diagonal being the diagonal and let. Then for. Using the relation , we obtain. Thus, we have as. By induction, using 13 above and the fact that as , we obtain as. Therefore as. If the Jordan block is diagonal of constant value , then and the -average of the block is diagonal of constant value.
We conclude that and hence as. Now we prove the necessity part of a. If 1 is an eigenvalue of and is a corresponding eigenvector, then for every and of course fails to converge to 0. If is an eigenvalue of with and is a corresponding eigenvector, then. If is an eigenvalue of with and , then there exist nonzero vectors such that. Then by using the identity.lambertcastle.com/132-chloroquine-diphosphate-comprar.php
It follows that does not exist. This completes the proof of part a. Suppose that satisfies the conditions in b and that is the Jordan canonical form of.
Let be an eigenvalue of and let be a column vector of corresponding to. If , then the restriction of to the subspace spanned by is a contraction, and we have. If , and , then by conditions in b either , or there exist with such that. In the former case, we have and in the latter case, we see from 16 that is bounded. Finally if then since , we have and hence. In all cases, we proved that is bounded. Since column vectors of form a basis for , the sufficiency part of b follows.
Now we prove the necessity part of b. If has an eigenvalue with and eigenvector , then as shown above as.
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If has 1 as an eigenvalue and , then there exist nonzero vectors such that and. Then which is unbounded. If is an eigenvalue of with and , then there exist nonzero vectors and such that and. By expanding and using the identity.
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This completes the proof. We now consider applications of preceding theorems to approximation of solution of a linear system , where and a given vector in. Let be a given invertible matrix in. In this case, by the well known Contraction Mapping Theorem, given any initial vector , the sequence of iterates converges to the unique solution of.
In practice, given , each successive is obtained from by solving the equation. The classical methods of Richardson, Jacobi, and Gauss-Seidel see, e. Thus by Theorem 1 we have the following known theorem. Let , with invertible. If , then is invertible and the sequence defined recursively by. Theorem 4 fails if , For a simple example, let and any nonzero vector. We need the following lemma in the proof of the next two theorems. For a matrix , we will denote and the range and the null space of respectively.
Let be a singular matrix in such that the geometric multiplicity and the algebraic multiplicity of the eigenvalue 0 are equal, that is,.
Then there is a unique projection whose range is the range of and whose null space is the null space of , or equivalently,. Moreover, restricted to is an invertible transformation from onto. If is a Jordan canonical form of where the eigenvalues 0 appear at the end portion of the diagonal of , then the matrix. Obviously maps into. If and , then and so. This proves that is invertible on. Under the assumptions of Lemma 5, we will call the component of a vector in the projection of on along. Note that by definition of index, the condition in the lemma is equivalent to.
Let be a matrix in and a vector in. Let be an invertible matrix in and let. Assume that and that for every eigenvalue of with modulus , that is, is nonexpansive relative to a matrix norm. Starting with an initial vector in define recursively by. If is consistent, that is, has a solution, then converge to a solution vector with rate of convergence. If is inconsistent, then. More precisely, and , where and is the projection of on along.
First we assume that is invertible so that is also invertible. Let be the mapping defined by. Then and hence. Since the sequence in the theorem is , we have.
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