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In section 2.3 we looked at the infinity and one norms and saw how they could be used to estimate the error in the solution of a system of linear equations y = Ax in terms of the errors in the numbers in y. In this section we look at general matrix norms where one may use some other norm for x and y. They can be used in the same way as the infinity and one norm to estimate the error in the solution of a system of linear equations.
Definition 1. Let A be a matrix and consider the mapping y = Ax. Choose a norm || x || for vectors x and choose a norm || Ax || for vectors Ax. Then the norm of A, denoted by || A ||, is the maximum value of the ratio \eq \f(|| Ax ||,|| x ||) as x varies over all non-zero vectors x, i.e.
(1) || A || = \eq \o(max,x ( 0) \f(|| Ax ||,|| x ||)
It follows from (1) that
(2) || Ax || ( || A || || x ||
which is a generalization of (9) and (10) in section 2.3. In particular, || A || is the maximum stretching that A does when applied to vectors x.
Just as when we are dealing with the sup norm and one norm for x and y, (2) turns out to be useful when we want to analyze how errors propagate when we multiply by A. The following proposition generalizes Proposition 2 of section 2.3.
Proposition 1. Let A = EQ \b(\a\al(a11 a12 a1n,a21 a22 a2n,....,am1 am2 amn)) and y = Ax and ya = Axa. Then
(3) || y - ya || ( || A || || x - xa ||
If m = n and A is invertible then
(4) || x - xa || ( || A-1 || || y - ya ||
Proof. (3) follows from (2) and the fact that y ya = A(x xa). (4) follows from (3) and the fact that x = A-1y and xa = A-1ya. //
Unfortunately, the formula (1) is not so nice for calculating the norm of some matrix A. Sometimes it is convenient to restrict the vectors x that we are maximizing over in (1) to unit vectors.
Proposition 2. The norm of A is the maximum of || Au || as u varies over unit vectors, i.e.
(5) || A || = \eq \o(max,|| u || = 1) || Au ||
Proof. Note that in (1) we can write
\eq \f(|| Ax ||,|| x ||) = \eq \f(1,|| x ||) || Ax || = || \eq \f(1,|| x ||) Ax || = || A \eq \b( \f(x,|| x ||) ) || = || Au ||
where u = \eq \f(x,|| x ||) and we have used the homogeniety of the norm. Note that u is a unit vector, since
|| \eq \f(x,|| x ||) || = \eq \f(1,|| x ||) || x || = 1
So, when we maximize over all non-zero vectors x in (1), it is the same as maximizing over all unit vectors in (5). //
Here are some properties of the matrix norms.
Proposition 3. Let || A || be a matrix norm defined by (1). If A and B are matrices and c is a number then
(6) || A + B || ( || A || + || B ||
(7) || cA || = | c | || A ||
(8) || A || = 0 ( A = 0
(9) || AB || ( || A || || B ||
(10) || I || = 1
(11) || A-1 || ( eq \f(1,|| A ||)
Proof. To prove (6) we use (5).
|| A + B || = \eq \o(max,|| x || = 1) || (A + B)x || = \eq \o(max,|| x || = 1) || Ax + Bx || ( \eq \o(max,|| x || = 1) (|| Ax || + || Bx ||)
( \eq \o(max,|| x || = 1) || Ax || + \eq \o(max,|| x || = 1) || Bx || = || A || + || B ||
For (7) one has
|| cA || = \eq \o(max,|| x || = 1) || (cA)x || = \eq \o(max,|| x || = 1) || c(Ax) || ( \eq \o(max,|| x || = 1) | c | || Ax ||
( | c | \eq \o(max,|| x || = 1) || Ax || = | c | || A ||
To prove (8) note that || A || = 0 and (1) imply \eq \o(max,x ( 0) \f(|| Ax ||, || x ||() = 0 which implies ||Ax||=0 for all x which implies A= 0. Note that (1) implies || (AB)x || = || A(Bx) || ( || A || || Bx || ( || A || || B || || x ||. Therefore eq \f(|| (AB)x ||,|| x ||)(||A||||B || which implies \eq \o(max,x ( 0) \f(|| (AB)x ||, || x ||() ( || A || || B || which implies (9). (10) follows from (1) and the fact that Ix = x for all x. For (11) note that I = AA-1. Using (9) and (10) we get 1=||I||(||A||||A-1|| from which (11) follows. //
Even the formula (5) is not convenient for finding the norm of a matrix. For given norms || x || and || Ax || it may take some work to find a formula for the corresponding matrix norm. For the Euclidean norm here are two useful formulas.
Proposition 2. Let || x || and || Ax || be the Euclidean norms and || A || be associated matrix norm given by (1). Then
(6) || A || = max{ | ( |: ( is an eigenvalue of A} if A is symmetric
(7) || A || = max{ eq \r(,(): ( is an eigenvalue of ATA} in general
Proof. First consider the case where A is symmetric. Then the eigenvalues (1, , (n of A are all real and there is an orthonormal basis v1, , vn of eigenvalues of A. Suppose Avi = (ivi and | (1 | ( ( | (n |. For a given vector x one can write x = (1v1 + + (nvn. Since the vi are orthonormal one has || x ||2 = One has Ax =
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