Scientific Computing

Matrix Norms, Convergence, and Matrix Condition Numbers

Vector Norms

• A vector norm is a quantity that measures how large a vector is (the magnitude of the vector).

• For a number x, we have |x| as a measurement of the magnitude of x.

• For a vector x, it is not clear what the “best” measurement of size should be.

• Note: we will use bold-face type to denote a vector. ( x )

Vector Norms

• Example: x = ( 4, -1 ) is the standard Pythagorean length of x. This

is one possible measurement of the size of x. 17)1(4 22

x

Vector Norms

• Example: x = ( 4, -1 ) |4| + |-1|=5 is the “Taxicab” length of x. This is another

possible measurement of the size of x.

x

Vector Norms

• Example: x = ( 4, -1 ) max(|4|,|-1|) =4 is yet another possible measurement of

the size of x.

x

Vector Norms

• A vector norm is a quantity that measures how large a vector is (the magnitude of the vector).

• Definition: A vector norm is a function that takes a vector and returns a non-zero number. We denote the norm of a vector x by

The norm must satisfy:– Triangle Inequality:– Scalar: – Positive: ,and = 0 only when x is the zero vector.

• Our previous examples for vectors in Rn :

• All of these satisfy the three properties for a norm.

Vector Norms

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)(:1

2/12

12

11

Chebyshevxxnorm

Euclideanxxnorm

Manhattanxxnorm

ii

n

ii

n

ii

Vector Norms Example

• Definition: The Lp norm generalizes these three norms. For p > 0, it is defined on Rn by:

• p=1 L1 norm• p=2 L2 norm • p= ∞ L∞ norm

Vector Norms

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iipxx

/1

1

Distance

• The answer depends on the application. • The 1-norm and ∞-norm are good whenever

one is analyzing sensitivity of solutions. • The 2-norm is good for comparing distances

of vectors.• There is no one best vector norm!

Which norm is best?

• In Matlab, the norm function computes the Lp norms of vectors. Syntax: norm(x, p)>> x = [ 3 4 -1 ];>> n = norm(x,2)n = 5.0990>> n = norm(x,1)n = 8>> n = norm(x, inf)n = 4

Matlab Vector Norms

• Definition: Given a vector norm ||x|| the matrix norm defined by the vector norm is given by:

• What does a matrix norm represent? • It represents the maximum “stretching” that A does

to a vector x -> (Ax).

Matrix Norms

x

AxA

x 0max

• Note that, since ||x|| is a scalar, we have

• Since is a unit vector, we see that the matrix norm is the maximum value of Az where z is on the unit ball in Rn.

• Thus, ||A|| represents the maximum “stretching” possible done by the action Ax.

Matrix Norm “Stretch”

x

xA

x

Ax

x

AxA

xxx 000maxmaxmax

x

x

Theorem A: The matrix norm corresponding to 1-norm is maximum absolute column sum:

Proof: From the previous slide, we have Also,

where Aj is the j-th column of A.

•

Matrix 1- Norm

n

iij

jaA

11max

111max AxAx

€

Ax = x1A1 + x2A2 +L + xnAn = x jA jj=1

n

∑

Proof (continued): Then,

Let x be a vector with all zeroes, except a 1 in the spot where ||Aj|| is a max. Then, we get equality above. □

Matrix 1- Norm

€

A1≤ max

x 1 =1x jA j 1

j=1

n

∑ =maxx 1 =1

x j A j 1j=1

n

∑

≤ maxx 1 =1(max

jA j 1) x j =

j=1

n

∑ maxx 1 =1(max

jA j 1) x 1

=maxjA j 1 =maxj

i=1

n

∑ aij

€

as x 1 = x ii=1

n

∑

Theorem B: Matrix norm corresponding to ∞ norm is maximum absolute row sum:

Proof (similar to Theorem A).

Matrix Norms

n

jij

iaA

1

max

• || A || > 0 if A ≠ O• || A || = 0 iff A = O• || c A || = | c| * ||A || if A ≠ O• || A + B || ≤ || A || + || B ||• || A B || ≤ || A || * ||B || • || A x || ≤ || A || * ||x ||

Matrix Norm Properties

• The eigenvectors of a matrix are vectors that satisfy Ax = λx Or, (A – λI)x = 0 So, λ is an eigenvalue iff det(A – λI) = 0

Example:

Eigenvalues-Eigenvectors

2/100

64/30

541

A

)2/1)(4/3)(1(

2/100

64/30

541

)det(

IA

2/1,4/3,1

• The spectral radius of a matrix A is defined as ρ(A) = max |λ| where λ is an eigenvalue of A

• In our previous example, we had

So, the spectral radius is 1.

Spectral Radius

2/1,4/3,1

Theorem 1: If ρ(A)<1, then Proof: We can find a basis for Rn by unit eigenvectors (result from linear algebra), say {e1, e2, …, en}. Then,

For any unit vector x, we have x = a1 e1 + a2 e2 + … + an en

Then, An x = a1 Ane1 + a2 Ane2 + … + an Anen

= a1 λ1ne1 + a2 λ2

ne2 + … + an λnnen

Thus,

Since ρ(A)<1, then the result must hold. □

Convergence

0lim

p

pA

CAaAeaxAA nn

ii

nn

ii

nii

n

x

n )()(max11

1

Theorem 2: If ρ(B)<1, then (I-B)-1 exists and (I-B)-1 = I + B + B2 + · · · Proof: Since we have Bx = λx exactly when (I-B)x = (1- λ)x, then λ is an eigenvalue of B iff (1- λ) is an eigenvalue of (I-B). Now, we know that |λ|<1, so 0 cannot be an eigenvalue of (I-B). Thus, (I-B) is invertible (why?). Let Sp = I + B + B2 + · · ·+Bp Then, (I-B) Sp = (I + B + B2 + · · ·+Bp ) – (B + B2 + · · ·+Bp+1 )

= (I- Bp+1) Since ρ(A)<1, then by Theorem 1, the term Bp+1 will go to the zero matrix as p goes to infinity. □

Convergent Matrix Series

Recall: Our general iterative formula to find x was Q x(k+1) = ωb + (Q-ωA) x(k)

where Q and ω were variable parameters.

We can re-write this as x(k+1) = Q-1 (Q-ωA) x(k) + Q-1ωb

Let B = Q-1 (Q-ωA) and c = ωb Then, our iteration formula has the general form:

x(k+1) = B x(k) + c

Convergence of Iterative solution to Ax=b

Theorem 3: For any x0 in Rn , the iteration formula given by x(k+1) = Bx(k) + c will converge to the unique solution of x=Bx+c (i.e fixed point) iff ρ(B)<1.Proof:

If ρ(B)<1, the term Bk+1x0 will vanish. Also, the remaining term will converge to (I-B)-1. Thus, {x(k+1)} converges to z = (I-B)-1c, or z-Bz = c or z = Bz + c.The converse proof can be found in Burden and Faires, Numerical Analysis. □

Convergence of Iterative solution to Ax=b

cIBBBxB

cIBBxB

cIBxB

ccBxB

cBxx

kk

k

k

k

kk

)(

)(

)(

)(

201

223

12

1

1

Def: A matrix A is called Diagonally Dominant if the magnitude of the diagonal element is larger than the sum of the absolute values of the other elements in the row, for all rows.

Example:

Diagonally Dominant Matrices

iallforaan

iji

ijii

1

521

272

126

Recall: Jacobi Method x(k+1) = D-1(b + (D-A) x(k)) = D-1(D-A) x(k) + D-1b

Theorem 4: If A is diagonally dominant, then the Jacobi method converges to the solution of Ax=b.

Proof: Let B = D-1(D-A) and c = D-1b. Then, we have x(k+1) = B x(k) + c. Consider the L∞ norm of B, which is equal to

Jacobi Method

n

jij

iaA

1

max

Proof: (continued) Then,

If A is diagonally dominant, then the terms we are taking a max over are all less than 1. So, the L∞ norm of B is <1. We will now show that this implies that the spectral radius is <1.

Jacobi Method

n

jijij

iii

n

jijij

iii

n

jij

iii

n

jij

i

n

jij

i

n

jij

i

aa

aa

aa

ADI

ADDbB

,1,1

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1

1

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1max|))

11(1(|max

|)1

1(|max)(max

)(maxmax

Lemma: ρ(A)<||A|| for any matrix norm.Proof: Let λ be an eigenvalue with unit eigenvector x. □

Proof of Theorem 4 (cont): Since we have shown that then, by the Lemma, we have that ρ(B) < 1.By Theorem 3, the iteration method converges. □

Jacobi Method

AxAAxxx

1

B

Through similar means we can show (no proof):

Theorem 5: If A is diagonally dominant, then the Gauss-Seidel method converges to the solution of Ax=b.

Gauss-Seidel Method