Contents
Appendix D (Inner Product Spaces) W-51
Index W-63 W-49
Inner city space W-50 Chapter 0: Appendix D
Inner Product Spaces
The inner product, taken of any two vectors in an arbitrary vector space, generalizes the dot product of two vectors in Rn or Cn.
For two column vectors x and y ∈ Rn we can form two different vector products, namely
• the outer product
x1 x1y1 ... x1yn T . . . Rn,n xy := . y1, ..., yn = . . ∈ xn xny ... xnyn 1 and
• the standard inner product
y1 T . R x y := x1, ..., xn . = x1y1 + ... + xnyn ∈ . yn
Here we interpret the two respective vectors as 1 by n or n by 1 matrices and multiply according to the rules of matrix multiplication. The outer product is a dyadic product since it creates an n by n dyad from two vectors, of which the first appears in column and the second in row form. It allows us to express matrix multiplication as a sum of rank 1 dyadic generators; see Sections 6.2, 10.2, and the proof of Lemma 3 in Section 12.2. A matrix product can be written as the sum of dyads of the columns and rows of the two matrix factors as follows:
0 c Copyright 2002 by Frank Uhlig; “Work in progress”; latest update of 19th November 2002.
W-51 W-52 Appendix D:
||−b1− . AmnBnk = a1... an . ||−b− k | | m,k =a1−b1−+... + an − bk − ∈ R , | | where the vectors ai denote the columns of the first factor A and the bj denote the rows of the second factor B. The standard inner product xT y of two vectors in Rn is the same as the dot product x · y ∈ R of the two vectors. It was introduced in Chapter 1 and interprets the first factor as a row and the second one as a column vector. The inner or dot product is also handy to express matrix multiplication, namely
− a1 − || a1·b1... a1 ·bk . . . m,k AmnBnk = . b... b =. . ∈ R , e. 1 k e. ee . e − am − ||a ·b ... a ·b ee m 1 m k f ˜ f e f e where thea ˜i now denote the rows of A and the bj the columns of the second factor B.In addition, the inner or dot product helps define angles and orthogonality of two vectors in Rn, see Chapters 10 through 12.
We start by listing four fundamental properties of the standard inner product of two vectors .. · .. : Rn × Rn → R.
Proposition 1: (Real Inner Product) The standard inner product of two vectors x and y ∈ Rn is defined as
y1 n T . R x · y := x y = x1, ..., xn . = xiyi ∈ . Xi=1 yn It satisfies the following four properties:
(a) x · y = y · x for all x, y ∈ Rn; (b) x · (y + z)=x·y+x·yfor all x, y, z ∈ Rn; (c) (αx) · y = x · (αy)=α(x·y) for all x, y ∈ Rn and all α ∈ R. n n (d) x · x ≥ 0 for all x ∈ R ,and x·x=0∈R if and only if x =0∈R . J
For two complex vectors x, y ∈ Cn, several modifications are in order in the definition and the properties of an inner product due to the effects of complex conjugation, see Appendix A. Inner Product Spaces W-53
Proposition 2: (Complex Inner Product) The standard inner product of two vectors x and y ∈ Cn is defined as
y1 n ∗ . C x · y := x y = x1, ..., xn . = xiyi ∈ . Xi=1 yn It satisfies the following four properties:
(a) x · y = y · x for all x, y ∈ Cn; (b) x · (y + z)=x·y+x·yfor all x, y, z ∈ Cn; (c) (αx) · y = x · (αy)=α(x·y) for all x, y ∈ Cn and all α ∈ C. n n (d) x · x ≥ 0 for all x ∈ C ,and x·x=0∈C if and only if x =0∈C . J
Proof: We deduce the four properties for the complex inner product only. The properties of the real inner product in Proposition 1 follow immediately by dropping all complex conjugation bars in this proof.
(a) x · y = i xiyi = i xiyi = i yixi = y · x since double conjugation c gives c back for anyP c in C. P P
(b) x · (y + z)= ixi(yi+zi)= ixiyi+ xizi =x·y+x·z. P P P (c) (αx) · y = i (αxi)yi = α i xiyi = αx · y and i (αxi)yi = i xi(αyi)= x·(αy). P P P P 2 (d) x·x = i |xi| ≥ 0 as the sum of real squares. And equality holds precisely when n |xi| =0foreachP i=1, ..., n,orwhenx=0∈C .
The standard dot or inner product of Rn (or Cn) serves very well in many aspects of linear algebra, such as when defining angles, orthogonality, and the length of vectors. More generally, an inner product can be defined in an arbitrary vector space V by requiring the four properties of a dot product; see Appendix C for abstract vector spaces.
Definition 1: Let V be an arbitrary vector space over a field of scalars F.
(1) A function h.., ..i : V × V → F that maps any two vectors f and g ∈ V to the scalar hf,gi in F is bilinear if h.., ..i is linear in each of its arguments, i.e., if hαf +βg,hi = hαf, hi+hβg,hi and hx, δu+vi = hx, δui+hx, vi for all scalars α, β, δ, ∈ F and all vectors x, u, v, f, g, h ∈ V . (2) A bilinear function h.., ..i : V × V → C (or R), operating on a complex (or real) vector space V ,isaninner product on V if it satisfies the following four properties. (a) hx, yi = hy,xi for all x, y ∈ V ; (b) hx, (y + z)i = hx, yi + hx, yi for all x, y, z ∈ V ; (c) h(αx),yi = hx, (αy)i = αhx, yi for all x, y ∈ V and all α ∈ C. W-54 Appendix D:
(d) hx, xi≥0 for all x ∈ V ,and hx, xi =0∈C if and only if x =0∈V. J
Note that if V is a vector space with the scalar field R, then the complex conjugation in parts 2(a) and 2(c) above should simply be dropped.
For V = R n and two vectors x, y ∈ Rn, the standard dot product hx, yi := x · y clearly defines an inner product on Rn. We can express the dot product as x · y = xT Iy via the n by n identity matrix I.ThematrixIn is symmetric with n positive eigenvalues equal to 1 on its diagonal. Positive definite matrices generalize the properties of I; see Section 11.3. For example, every positive definite matrix P = P T ∈ Rn,n can be expressed as a matrix product P = AT A with a nonsingular real square matrix A. For any P = AT A that T T T T is positive definite, we may set hx, yiP := x Py = x A Ay =(Ax) · (Ay) and thereby n n obtain an inner product hx, yiP := R ×R → R that differs from the ordinary dot product x·y = hx, yiI . All one needs to do to verify this statement, is to show that h.., ..iP is bilinear and satisfies the four standard properties of an inner product of Definition 1, see Problem 2below.
Proposition 3: (a) Every bilinear form f(x, y):Rn×Rn →Rcan be expressed as f(x, y)=xTSy for a real symmetric n by n matrix S. T n,n T (b) If P = P ∈ R is a positive definite real matrix, then hx, yiP := x Py ∈ R defines an inner product on Rn. ∗ n,n ∗ (c) If P = P ∈ C is a positive definite complex matrix, then hx, yiP := x Py ∈ n C defines an inner product on C . J
4 4 Example 1: (a) To write f(x, y)=3x1y1−2x1y2+3x3y2−x2y2+4x4y4 : R ×R → R T T 4,4 in the form x Sy with S = S ∈ R , we define the diagonal entries sii of S as the coefficients of xiyi in f,orass11 =3,s22 = −1,s33 =0,ands44 =4. For i>jwe set the off-diagonal entries sij in S equal to half the coefficient of xiyj and then define sji = sij.Thuss12 = −1=s21 and s32 =1.5=s23.Thus 1 −1 −101.5 S= =ST. 1.5−1 4 10 3 (b) Determine whether hx, yi := xT yis an inner product on R2. 31 Clearly hx.yi is bilinear in both x and y ∈ R2 since it is matrix generated. The 10 3 generating matrix S := with hx.yi = hx.yi = xT Sy is symmetric, 31 S i.e., S = ST , and hence its eigenvalues are real according to Section 11.1. Using the trace and determinant conditions of Theorem 9.5, we observe that the two eigenvalues of S add to 11 and multiply to 10 − 9 = 1. Thus both eigenvalues T of S must be positive real, making S = S positive definite and hx.yi = hx.yiS an inner product on R2 according to Proposition 3. Inner Product Spaces W-55
01 (c) Determine whether hu, vi := uT vis an inner product on C2. 10 Clearly the function hu.vi maps any two complex 2-vectors to a complex number 01 1 1 and it is bilinear. For X = =X∗we observe that X = 10 1 1 −1 1 −1 and X = = − . Therefore X has the eigenvalues 1 1 −1 1 and –1 and X is not positive definite. Thus hu, vi = hu, viX is not necessarily n an inner product on C since Proposition 3 does not apply. In fact, hu, viX 2 violates the fourth property (d) of inner products for u = v = e1 6=0∈C : 01 0 1 he ,e iX = 01 = 01 =0∈C. 1 1 10 1 0 C2 Therefore hu, viX is not an inner product on . J Inner products can help us measure and navigate in abstract vector spaces, such as in spaces of functions. As an example, we now consider the space of continuous functions R R F[0,1] := {f :[0,1] → | f continuous} defined on the interval [0, 1] ⊂ . This space is 1 infinite dimensional, see Section 7.2(b). By setting hf,gi := f(x)g(x) dx for all functions R0 f,g ∈F[0,1],wehavemadeF[0,1] into an inner product space. Clearly hf,gi is linear in both of its function variables f and g since integration is linear in the sense that u + vdx= udx+ vdx. Identities such as R R R hαu + βv,wi = (αu + βv)wdx=α uw dx + β vw dx = αhu, wi + βhv, wi Z Z Z prove the properties (b) and (c) of an inner product. Next we observe that property (a) holds since hf,gi = fg dx = gf dx = hg, fi. And the first part of property (d) 1 R R hf,fi = f2(x) dx ≥ 0 holds for any integrable function f since f 2(x) ≥ 0. To show that R0 hf,fi =0forf∈F[0,1] implies that f =0on[0,1] requires more thought: Every function f ∈F[0,1] is continuous. If f is not the zero function on [0, 1], then f(x0) 6=0forsome x0 ∈[0, 1]. By continuity there is an interval [a, b], 0 ≤ a
1 a b 1 hf,fi = f2(x) dx = f 2 dx + f 2 dx + f 2 dx Z Z Z Z 0 0 a b b ≥ f 2 dx ≥ (b − a)2 > 0 . Z a Consequently if f is continuous and hf,fi =0,thenf=0on[0,1]. Thus W-56 Appendix D:
1 hf,gi := f(x)g(x) dx Z 0 is an inner product on F[0,1] = {f :[0,1] → R | f continuous }. Different inner product can be defined on F[0,1] by setting
1
hf,giw := f(x)g(x)w(x) dx Z 0 for an arbitrary continuous weight function w ∈F[0,1] that is positive on [0, 1], see Prob- lem 5.
Inner products define angles and orthogonality in abstract vector spaces V just as the standard dot products do in Rn and Cn; recall Section 10.1. If both f and g 6=0∈V and V is an inner product space with the inner product h.., ..i, then the angle between f and g is defined with respect to a given inner product h.., ..i by the formula
hf,gi cos ∠(f,g):= 1 1 . hf,fi2 hg, gi 2
And f ∈ V is orthogonal to g ∈ V ,orf⊥g∈V,ifhf,gi = 0 for the inner product h.., ..i of V .Herehf,gi may be complex for two functions f and g when V is a complex vector space. In this case the complex valued cosine function cos(z)isused.
1 Example 2: (a) In V = F[0,1] with the inner product hf,gi := 2f(x)g(x) dx, find the R0 angle between the two functions f(x)=1andg(x)=x∈V. To find the angle we have to evaluate three different inner products: hf,gi = 1 1 1 2 1 1 2 2xdx=x |0 =1, hf,fi = 2 dx =2x|0 =2,andhg, gi = 2x dx = R0 R0 √R0 3 1 hf,gi 1 3 2x = 2 . Therefore cos (f,g)= = = .Andthe 3 |0 3 ∠ 1 1 √ hf,fi2 hg, gi 2 2 2 2 3 q√ 3 angle between f and g has the radian measure of arccos 2 . Note that the inner product contains the weight function w(x)=2. 1 (b) Show that the two functions h(x)=1andk(x)=x−2 are orthogonal in V of part (a) with its given inner product. 1 1 1 We evaluate h, k = 2(x 1 ) dx = 2xdx dx = x2 1 x 1 =1 1=0. h i − 2 − |0 − |0 − R0 R0 R0 2 (c) Find an orthogonal basis for the subspace span{x, x }⊂F[0,1] with respect to 1 the inner product hf,gi := f(x)g(x) dx. R0 Inner Product Spaces W-57
Here we use the modified Gram–Schmidt process of Chapter 10 for the functions 2 u1 = x and u2 = x .
v1 := u1 = x ;
v2 := hv1,v1iu2 −hu2,v1iv1 .
1 1 3 4 We have v ,v = x2 dx = x 1 = 1 and u ,v = x3 dx = x 1 = 1 . h 1 1i 3 |0 3 h 2 1i 4 |0 4 0 0 R 1 1 1 2 1 R Thus v1 = x and v2 = 3 u2 − 4 v1 = 3 x − 4 x are orthogonal in F[0,1] with respect to the particular inner product. Normalizing the vi with respect to the given 1 √ 1 inner product h.., ..i makes w1 = v1 = 3xand w2 = v2 = v ,v v ,v √ √ h 1 1i h 2 2i 2 2 4 5 x −3 5 x since hv1,v1i =1p/3andhv2,v2i =1/(12 · 5).p The students 1 should check this assertion by evaluating hv1,v2i = v1(x)v2(x) dx, hv1,v1i, R0 and hv2,v2i. J
Inner products define vector norms in a natural way.
Proposition 4: If h.., ..i is an inner product on a real vector space V ,then
1/2 R kxkh..,..i := hx, xi : V →
defines a vector norm for every x ∈ V with the following properties: R (1) kxkh..,..i ≥ 0 for all x ∈ V ,and kxkh..,..i =0∈ if and only if x =0∈V. R (2) kαxkh..,..i = |α|kxkh..,..i for all vectors x ∈ V and all scalars α ∈ .
(3) |hx, yi| ≤ kxkh..,..ikykh..,..i for all vectors x, y ∈ V . (Cauchy–Schwarz inequality)
(4) kx + ykh..,..i ≤kxkh..,..i + kykh..,..i for all vectors x, y ∈ V . (triangle inequality)
J
A vector norm k..k is called induced by the inner product h.., ..i if k..k = h.., ..i1/2 as it is in Proposition 4. General vector norms without an underlying inner product can, however, be solely defined by the three norm defining properties (1), (2), and (4) of Proposition 4. Proofs of both√ the Cauchy–Schwarz and the triangle inequality for the standard euclidean norm kxk = x∗x of Rn or Cn are outlined in Section 10.1 and Problems 23 and 26 in Section 10.1.P.
Definition 2: A function g(x):V →Ris a vector norm on a real vector space V if it satisfies the properties (1), (2), and (4) of Proposition 4. W-58 Appendix D:
n n n Example 3: The function g(x):=max{|xi|} : R → R is a vector norm on R . i=1 To see this we check the three conditions of a vector norm: Clearly g(x) ≥ 0 n for all x ∈ R ,andg(x) = 0 if and only if max |xi| =0∈R,orifandonly if x =0∈Rn, establishing property (1). Next g(x) satisfies property (2) since g(αx)=max{|αxi|} =max{|α||xi|} = |α| max{|xi|} = |α|g(x). Finally, the triangle inequality |α + β|≤|α|+|β|for scalars α, β ∈ R helps us prove property (4):
g(x + y)=max{|xi + yi|} ≤ max{|xi| + |yi|} ≤ max{|xi|} +max{|yi|} = g(x)+g(y) .
Thus g(x) is a vector norm. n In Example 5 we learn that g is not induced by any inner product of R . J
A vector norm√ k..k measures the length of vectors in V , just as the standard euclidean norm kuk := uT u measures the length of vectors in Rn via the standard dot product.
2 Example 4: (a) Find the length of the standard unit vector e1 ∈ R in terms of the 10 3 vector norm that is induced by the inner product hx, yi = xT yof 31 Example 1(a). We compute
10 3 1 10 ke k2 = he ,e i = 10 = 10 =10. 1 h..,..i 1 1 h..,..i 31 0 3 √ Therefore ke1kh..,..i = 10. (b) Find the norm of the function f(x)=x2 in the function space V of Example 2(a). 1 We have f,f = 2x4 dx = 2x5 1 = 2 , giving f the length, or norm f = h i 5 |0 5 k kh..,..i R0 1/2 2 hf,fi = 5. q 1 (c) Find the length of the vector w = −1 for the norm that is induced by 2 200 the inner product hx, yi := xT 010yon R3. 003 200 Clearly the matrix 010is positive definite as a positive diagonal ma- 003 trix. Therefore h.., ..i is an inner product according to Proposition 3. Next we compute the induced vector norm of w: Inner Product Spaces W-59
200 1 2 kwkh..,..i = hw, wi = 1 −12 010−1 003 2 2 = 1−12 −1= 2+1+12 = 15, 6 √ √ or kwk = 15. Note that in the euclidean norm kwk := wT Iw = √ h..,..√i 2 wT w = 6. x (d) Show that for x = 1 ∈ R2, the function f(x)= 3x2+4x x +3x2 : x 1 1 2 2 2 p R2 →Ris an induced vector norm. 32 x We have f 2 (x)=3x2+4xx +3x2 = x x 1 = 1 1 2 2 1 2 23 x 2 32 xT x; see Example 6 in Section 11.3 for more on quadratic forms 23 32 such as f.ThematrixA:= =ATis real symmetric with eigenvalues 23 that sum to its trace 6 and that multiply to its determinant 9 − 4 = 5, accord- ing to Theorem 9.5. Thus the eigenvalues of A are 5 and 1, making A positive 32 definite and hx, yi := xT yan inner product on R2 due to Proposition 23 2 3. This inner product induces f(x)asanormonR. J
All inner products k..kh..,..i that are induced by an inner product h.., ..i satisfy the parallelogram identity.
Proposition 5: If k..kh..,..i is the induced vector norm for the inner product h.., ..i of an arbitrary real vector space V , then the parallelogram identity
1 2 2 2 2 kx + ykh..,..i + kx − ykh..,..i = kxkh..,..i + kykh..,..i 2 holds for all x, y ∈ V . J
The parallelogram identity states that the sum of the squared lengths of the two sides x and y of any parallelogram equals the average of the lengths of its two diagonals x + y and x − y squared for any induced vector norm.
Proof: To prove the parallelogram identity in a real vector space we expand its left hand side by using the properties of the norm inducing inner product. W-60 Appendix D:
1 2 2 1 kx + ykh..,..i + kx − ykh..,..i = (hx + y,x + yi + hx − y,x − yi) 2 2 1 1 1 1 = hx, xi + hx, yi + hy,yi + hx, xi−hx, yi + hy,yi 2 2 2 2 2 2 = hx, xi + hy,yi = kxkh..,..i + kykh..,..i .
n n Example 5: The maximum vector norm kxk∞ := max{|xi|} of R from Example 3 is i=1 not induced by any inner product of Rn since it does not satisfy the parallelogram identity. 1 0 For example, in R2 we have for x = and y = that kxk = kyk = 1 1 ∞ ∞ 1 1 1=kx−yk and kx + yk =2sincex+y= and x − y = .Thus ∞ ∞ 2 0 1 1 5 kx+yk2 +kx−yk2 = (4 + 1) = 6=2=1+1=kxk2 +kyk2 . 2 ∞ ∞ 2 2 ∞ ∞ J
Proposition 4 makes every inner product space a normed vector space. However, in Example 5 and more generally in functional analysis, it has been shown that not all vector norms derive from inner products. To complete our elementary explorations of inner product spaces and normed vector spaces, we mention without proof that all normed vector spaces whose norm k..k satisfies the parallelogram identity of Proposition 5 can be made into an inner product space by setting
kx + yk2 −kxk2 −kyk2 hx, yi := . 2 Moreover, the parallelogram identity can be generalized to complex vector spaces, but this is beyond the scope of this appendix and elementary linear algebra.
A.D.P Problems
T 1. Show that the function f(x, y)=2x1y1− P = A A for some nonsingular real ma- 3 3 x2y2+4x3y3+x2y3 : R ×R → R is a bilin- trix A. R3 ear function for x, y ∈ . Is this function an (Hint: Use Chapter 11: Diagonalize√ √P or- inner product on R3?Doesfinduce a norm thogonally as U T PU = D = D D for on R3? a positive diagonal matrix D and extract P from this matrix equation.) 2. (a) Show that hx, yi := xT Ay is a bilinear A (c) Show: If P = P T ∈ Rn,n is positive def- function for every real n by n matrix A. T inite, then hx, yiP := x Py is an inner (b) Show that if P = P T ∈ Rn,n is posi- product on Rn. tive definite, then P can be expressed as (Hint: Use part (b).) Inner Product Spaces W-61
3. Test whether the following functions are (a) 8. Orthogonalize the two functions f(x)= bilinear and (b) inner products on their re- 1andg(x)=x∈F[0,1] with respect spective spaces: to the weighted inner product hf,giw := 000 1 f(x)g(x)x2 dx. (1) h(x, y)=xT 042 yfor x, y ∈ 0 011 R R3 9. Orthogonalize the three functions f(x)= . 2 42 1,g(x)=x,andh(x)=x in F[0,1] (2) k(x, y)=xT yfor x, y ∈ R2. 11 with respect to the inner product hf,gi := 42 1 (3) `(x, y)=xT yfor x, y ∈ R2. f(x)g(x) dx. 1 1 − R0 41 (4) m(x, y)=xT yfor x, y ∈ R2. 10. Show that the√ standard euclidean vector 11 norm kxk := xT x for x ∈ Rn satisfies the 2 2 R2 (5) n(x, y)=2x1−3x1y1+4y1 on . parallelogram identity. R2 (6) p(x, y)=−x1y2 on . n 11. Show that kxk1 := |xi| is a vector norm 4. (a) Find the length of the two vec- i=1 Rn P tors x = 1 ... 1 and y = on . Is it induced by an inner product or not? 10... 0 −1 ∈ Rn for both the standard euclidean vector norm and for 12. Examine whether the vector norm kxk1 of the maximum vector norm. Rn in the previous problem is an induced (b) Find the cosine of the angle between x vector norm. and y ∈ Rn in part (a) for 13.(a) Assume that the norm k..k is an induced (1) the standard inner product and the norm on V .Ifkuk=4, ku+vk=6, 31 0 and ku − vk = 5, what is the length of v? . . What is the distance between u and v? 1.. .. (2) inner product xT y . . (b) Repeat part (a) for kuk =7. .. ..1 01314. Let V be a real inner product space. Let k..k of Rn. be the vector norm that is induced by the inner product h.., ..i on V . Show that 5. If w(x) > 0 is continuous on the kx + yk2 −kx−yk2 interval [0, 1], prove that hf,giw := hx, yi = 1 4 f(x)g(x)w(x) dx is an inner product on holds for all x, y V . R0 ∈ the space of continuous functions F[0,1]. 15. Let {u1, ..., uk} be an orthonormal set of 6. Construct a positive and continuous weight vectors in an inner product space of fi- function w(x) so that the two functions nite dimension n ≥ k. Prove that x ∈ 2 f(x)=1andg(x)=x∈F[0,1] become or- span{u1, ..., uk} if and only if kxk = 2 2 thogonal with respect to the inner product |hx, u1i| + ... + |hx, uki| for the induced 1 vector norm. hf,giw := f(x)g(x)w(x) dx, if possible. 2 −i R0 16.(a) Does hx, yi := x∗ y define an i 1 7. Construct a positive and continuous weight inner product on C2? function w(x) so that the two functions f(x)=1andg(x)=x−1 ∈F are not (b) Repeat part (a) for hx, yi := 2 [0,1] ii orthogonal with respect to the inner product x∗ y. 1 i1 hf,giw := f(x)g(x)w(x) dx, if possible. (c) Repeat part (a) for hx, yi := x1y1−2x2y2. R0 W-62 Appendix D:
(d) Are any of the functions in parts (a) (a) kxk and kyk, (b) the distance d(x, y), through (c) bilinear? (c) ky − xk,(d)kx+yk2,and C3 17. In a normed vector space V we define the (e) the angle between x and y ∈ . distance function between any two vectors 20. In F[0,1] with the standard inner prod- as d(x, y):=kx−yk. Show that 1 uct f,g := f(x)g(x) dx, determine the (a) d(x, y)=d(y,x), h i R0 (b) d(x, y) ≤ d(x, z)+d(z,y) for all x, y,and ‘length’ of the two functions f(x)=3x−2 2 z∈V. and g(x)=x +x∈F[0,1],aswellasthe (c) Evaluate the distance between cosine of the angle between f and g. x = 1 −24−3 and y = 21. Consider the function f(p, q):P ×P →R R4 n n 02 −1−3 ∈ for the eu- defined by f(p, q)=p(1)q(1) + 2p(2)q(2) + clidean norm and the maximum norm. 3p(3)q(3), where Pn denotes the real variable T 18.(a) Assume that hx, yiH := x Hx and polynomials of degree not exceeding n. T hx, yiK := x Kx are two bilinear forms on Rn with H and K both n by n and (a) Show that f is bilinear on Pn ×Pn. positive definite. (b) Show that f induces a norm k..k on Pm n If hx, yiH = hx, yiK for all x, y ∈ R , for all m ≤ 2. show that H = K as matrices. (c) Show that f does not induce a norm on (b) Show: If xT Ay = 0 for a real symmetric any Pm if m>2. matrix A and all vectors x, y ∈ Rn,then (d) Find k2x2 − 3xk and k1 − x2 − x3 +4xk A=On. in the vector norm that is induced by f. 19. Let x = 2 − i 1+i 3 and y = i 2 −i ∈ C3. For the standard eu- (e) Write down a formula for the induced clidean vector norm of C3, compute norm kpk and p ∈P2. Index
A G angle W-56 Gram–Schmidt process W-57
B I induced norm W-57 bilinear form W-53 inequality Cauchy–Schwarz W-57 C triangle W-57 inner product W-52, W-53 Cauchy–Schwarz inequality W-57 complex conjugation W-54 M D matrix multiplication W-51 dimension positive definite W-54 infinite W-55 symmetric W-54 distance W-62 multiplication dot product W-52 matrix W-51 dyad W-51 N E norm euclidean norm W-57, W-59 euclidean W-57, W-59 induced W-57 maximum W-60 F vector W-57 form bilinear W-53 O quadratic W-59 orthogonal W-56 function bilinear W-53 P distance W-62 weight W-56 parallelogram identity W-59
W-63 W-64 Index
product dot W-52 V dyadic W-51 vector inner W-51–W-53 complex W-52 outer W-51 inner product W-51 norm W-57 Q maximum W-60 orthogonal W-56 quadratic form W-59 outer product W-51 space W-57 T W triangle inequality W-57 weight function W-56 Frank Uhlig Department of Mathematics Auburn University Auburn, AL 36849–5310 [email protected] http://www.auburn.edu/˜uhligfd