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5. On solving linear systems of equations

5.1. Norm and convergence, state-transition matrix e At

A Vector space V is a normed space, if there is a norm || || : V -> R + = [ 0, ), defined for elements of V, satisfying:

(1)   |||| = 0 <=> =

(2)   || + || |||| + ||||    , , V

(3)   ||a|| = |a| ||||  , V,    a K   ( = R tai C)

Proposition 1. If dim V = n < , then all norms on V are equivalent, i.e., if || || and || ||' are norms defined on V, then a, b > 0:

a ||||' |||| b ||||',      V.

The set of M of all m x n matrices is a finite dimensional vector space. The following are all norms on M and A = (aij ) is an m x n matrix.

||A||1 = | aij | ,   the maximum of column sums,

||A|| = | aij | , the maximum of row sums,
||A||2 =
(the spectral norm), is the largest eigenvalue of the matrix TA, and
||A||Fr = , Frobenius norm,
which all are special cases of a norm or the form

(4)   ||A|| = [ ||A|| / |||| ],

where || || is a norm defined on Kn.

Example 1: Norms of a matrix

Note. All the norms defined above satisfy ||AB|| ||A|| ||B||.

In numerical computations one often needs the condition number of a regular matrix A:

(A) = ||A|| · ||A -1||

A sequence 1 , ..., n , ... of vectors of V converges to V in the norm || || if

||n - || = 0.

The convergence a sequence of matrices is defined by the same manner.

It follows from Proposition 1 that if dim V = n < (in particular, if V = Rn, M) and if n -> with respect to a norm || ||, then n -> also in every other norm defined on V (equivalence).

A matrix series Ak , Ak M, converges (in a matrix norm) if the sequence of partial sums

Sn = Ak

converges. It can be shown that Ak = (a (k)ij ) -> A = (aij ) <=> a (k)ij -> aij i,j kun k -> .

Let A be a square matrix. We define the matrix eA by setting

(5)

eA = (1 / k!)Ak     (A0 = I).

If A is diagonalizable, i.e., if A = TDT - 1 where D = diag (1 , ..., n ), then

Ak = TDkT - 1
= Tdiag (1 k, ..., kn )T - 1
=>   (1 / k!) Ak = Tdiag ( (1 k / k!), ..., (n k / k!) )T - 1
= T [ diag ( (1 k / k! ), ..., (n k / k! ) ) ] T - 1

Since

diag ( (1k / k!) , ..., (nk / k!) ) -> diag (e1 , ..., en ) = eD

it follows that

|| ( 1 / k!) Ak - TeDT -1||
= ||T [ diag ( (1k / k!) , ..., (nk / k!) ) ] T -1 - TeDT -1||
= ||T [ diag ( (1k / k!) , ..., (nk / k!) ) - eD ] T -1||
||T|| || diag ( (1k / k!) , ..., (nk / k!) ) - eD|| ||T -1||
-> 0, kun k -> .

Hence: If A = TDT - 1, then

(6)

eA = TeDT - 1.

The solution of the differential equation '(t) = A(t) (A does not need to diagonalizable) with an initial condition (0) = 0 is

(t) = etA0 .

etA is the so-called state-transition matrix. If kontrolli u(t) on mukana eli kun

'(t) = A(t) + Bu(t),

then (prove!)

(t) = etA0 + etAe - sA Bu(s) ds

(B is a column vector).

Example 2: eA


Exercises: E50, E51, E53
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