Module MA2321: Analysis in Several Real Variables Michaelmas Term 2016 Section 7: Norms on Finite-Dimensional Vector Spaces

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1 Module MA2321: Analysis in Several Real Variables Michaelmas Term 2016 Section 7: Norms on Finite-Dimensional Vector Spaces D R Wilkins Copyright c David R Wilkins Contents 7 Norms on Finite-Dimensional Vector Spaces Norms Linear Transformations The Operator Norm of a Linear Transformation The Hilbert-Schmidt Norm of a Linear Transformation 113 i

2 7 Norms on Finite-Dimensional Vector Spaces 71 Norms Definition A norm on a real or complex vector space X is a function, associating to each element x of X a corresponding real number x, such that the following conditions are satisfied: (i x 0 for all x X, (ii x + y x + y for all x, y X, (iii λx = λ x for all x X and for all scalars λ, (iv x = 0 if and only if x = 0 A normed vector space (X, consists of a a real or complex vector space X, together with a norm on X The Euclidean norm is a norm on R n defined so that (x 1, x 2,, x n = x x x 2 n for all (x 1, x 2,, x n There are other useful norms on R n These include the norms 1 and sup, where (x 1, x 2,, x n 1 = x 1 + x x n and (x 1, x 2,, x n sup = maximum( x 1, x 2,, x n for all (x 1, x 2,, x n Definition Let and be norms on a real vector space X The norms and are said to be equivalent if and only if there exist constants c and C, where 0 < c C, such that for all x X c x x C x Lemma 71 If two norms on a real vector space are equivalent to a third norm then they are equivalent to each other 108

3 Proof let and be norms on a real vector space X that are both equivalent to a norm on X Then there exist constants c, c, C and C, where 0 < c C and 0 < c C, such that c x x C x and for all x X But then c x x C x c C x x C c x for all x X, and thus the norms and are equivalent to one another The result follows We shall show that all norms on a finite-dimensional real vector space are equivalent Lemma 72 Let be a norm on R n Then there exists a positive real number C with the property that x C x for all x R n Proof Let e 1, e 2,, e n denote the basis of R n given by e 1 = (1, 0, 0,, 0, e 2 = (0, 1, 0,, 0,, e n = (0, 0, 0,, 1 Let x be a point of R n, where x = (x 1, x 2,, x n Using Schwarz s Inequality, we see that x = x j e j x j e j ( 1 ( e j 2 = C x, where and x 2 j C 2 = e e e n 2 x = ( x 2 j 1 2 for all (x 1, x 2,, x n R n The result follows 109

4 Then there exists a positive con- Lemma 73 Let be a norm on R n stant C such that x y x y C x y for all x, y R n Proof Let x, y R n Then x x y + y, y x y + x It follows that and and therefore x y x y y x x y, y x x y for all x, y R n The result therefore follows from Lemma 72 Theorem 74 Any two norms on R n are equivalent Proof Let be any norm on R n We show that is equivalent to the Euclidean norm Let S n 1 denote the unit sphere in R n, defined by S n 1 = {x R n : x = 1} Now it follows from Lemma 73 that the function x x is continuous Also S n 1 is a compact subset of R n, since it is both closed and bounded It therefore follows from the Extreme Value Theorem (Theorem 55 that there exist points u and v of S n 1 such that u x v for all x S n 1 Set c = u and C = v Then 0 < c C (since it follows from the definition of norms that the norm of any non-zero element of R n is necessarily non-zero If x is any non-zero element of R n then λx S n 1, where λ = 1/ x But λx = λ x (see the the definition of norms Therefore c λ x C, and hence c x x C x for all x R n, showing that the norm is equivalent to the Euclidean norm on R n If two norms on a vector space are equivalent to a third norm, then they are equivalent to each other It follows that any two norms on R n are equivalent, as required 110

5 72 Linear Transformations The space R n consisting of all n-tuples (x 1, x 2,, x n of real numbers is a vector space over the field R of real numbers, where addition and multiplication by scalars are defined by (x 1, x 2,, x n + (y 1, y 2,, y n = (x 1 + y 1, x 2 + y 2,, x n + y n, λ(x 1, x 2,, x n = (λx 1, λx 2,, λx n for all (x 1, x 2,, x n, (y 1, y 2,, y n R n and λ R Definition A map T : R m R n is said to be a linear transformation if T (x + y = T x + T y, T (λx = λt x for all x, y R m and λ R Every linear transformation T : R m R n is represented by an n m matrix (T Indeed let e 1, e 2,, e m be the standard basis vectors of R m defined by e 1 = (1, 0,, 0, e 2 = (0, 1,, 0,, e m = (0, 0,, 1 Thus if x R m is represented by the m-tuple (x 1, x 2,, x m then x = m x j e j Similarly let f 1, f 2,, f n be the standard basis vectors of R n defined by f 1 = (1, 0,, 0, f 2 = (0, 1,, 0,, f n = (0, 0,, 1 Thus if v R n is represented by the n-tuple (v 1, v 2,, v n then v = v i f i Let T : R m R n be a linear transformation Define T for all integers i between 1 and n and for all integers j between 1 and m such that T e j = T f i 111

6 Using the linearity of T, we see that if x = (x 1, x 2,, x m then ( m T x = T x j e j = Thus the ith component of T x is m (x j T e j = ( m T x j f i T i,1 x 1 + T i,2 x T i,m x m Writing out this identity in matrix notation, we see that if T x = v, where x 1 v 1 x 2 v 2 x = x m, v = v n, then v 1 v 2 v n = T 1,1 T 1,2 T 1,m,1,2,m T n,1 T n,2 T m,n x 1 x 2 x m 73 The Operator Norm of a Linear Transformation Definition Given T : R m R n be a linear transformation The operator norm T op of T is defined such that T op = sup{ T x : x R m and x = 1} Lemma 75 Let T : R m R n and U: R m R n be linear transformations from R m to R n, and let λ be a real number Then T op is the smallest nonnegative real number with the property that T x T op x for all x R m Moreover λt op = λ T op and T + U op T op + U op Proof Let x be an element of R m Then we can express x in the form x = µz, where µ = x and z R m satisfies z = 1 Then T x = T (µz = µt z = µ T z = x T z T op x Next let C be a non-negative real number with the property that T x C x for all x R m Then C is an upper bound for the set { T x : x R m and x = 1}, 112

7 and thus T op C Thus T op is the smallest non-negative real number C with the property that T x C x for all x R m Next we note that λt op = sup{ λt x : x R m and x = 1} Let x R m Then It follows that = sup{ λ T x : x R m and x = 1} = λ sup{ T x : x R m and x = 1} = λ T op (T + Ux T x + Ux T op x + U op x This completes the proof ( T op + U op x (T + U op T op + U op 74 The Hilbert-Schmidt Norm of a Linear Transformation Recall that the length (or norm of an element x R n is defined such that x 2 = x x x 2 n Definition Let T : R m R n be a linear transformation from R m to R n, and let (T be the n m matrix representing this linear transformation with respect to the standard bases of R m and R n The Hilbert-Schmidt norm T HS of the linear transformation is then defined so that m T HS = T 2 Note that the Hilbert-Schmidt norm is just the Euclidean norm on the real vector space of dimension mn whose elements are n m matrices representing linear transformations from R m to R n with respect to the standard bases of these vector spaces Therefore it has the standard properties of the Euclidean norm In particular it follows from the Triangle Inequality (Lemma 42 that T + U HS T HS + U HS and λt HS = λ T HS for all linear transformations T and U from R m to R n and for all real numbers λ 113

8 Lemma 76 Let T : R m R n be a linear transformation from R m to R n Then T is uniformly continuous on R n Moreover T x T y T HS x y for all x, y R m, where T HS transformation T is the Hilbert-Schmidt norm of the linear Proof Let v = T x T y, where v R n (v 1, v 2,, v n Then is represented by the n-tuple v i = T i,1 (x 1 y 1 + T i,2 (x 2 y T i,m (x m y m for all integers i between 1 and n It follows from Schwarz s Inequality (Lemma 41 that ( m ( m ( m vi 2 (x j y j 2 = x y 2 Hence v 2 = ( vi 2 m x y 2 = T 2 HS x y 2 Thus T x T y T HS x y It follows from this that T is uniformly continuous Indeed let some positive real number ε be given We can then choose δ so that T HS δ < ε If x and y are elements of R n which satisfy the condition x y < δ then T x T y < ε This shows that T : R m R n is uniformly continuous on R m, as required Lemma 77 Let T : R m R n be a linear transformation from R m to R n and let S: R n R p be a linear transformation from R n to R p Then the Hilbert- Schmidt norm of the composition of the linear operators T and S satisfies the inequality ST HS S HS T HS Proof The composition ST of the linear operators is represented by the product of the corresponding matrices Thus the component (ST k,j in the kth row and the jth column of the p m matrix representing the linear transformation ST satisfies (ST k,j = S k,i T 114

9 It follows from Schwarz s Inequality (Lemma 41 that ( ( (ST 2 k,j Summing over k, we find that ( p p (ST 2 k,j k=1 k=1 S 2 k,i S 2 k,i ( = S 2 HS Then summing over j, we find that ( p m ST 2 HS = (ST 2 k,j S 2 HS k=1 S 2 HS T 2 HS ( On taking square roots, we find that ST HS S HS T HS, as required m 115