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Lecture 23: 6.1 Inner Products

Wei-Ta Chu

2008/12/17 Definition

 An inner product on a real vector V is a function that associates a real u, vwith each pair of vectors u and v in V in such a way that the following axioms are satisfied for all vectors u, v, and w in V and all scalars k.  u, v= v, u  u + v, w= u, w+ v, w  ku, v= k u, v  u, u0 and u, u=0if andonlyif u = 0 A real with an inner product is called a real .

2008/12/17 Elementary Linear 2 Euclidean Inner Product on Rn

 If u = (u1, u2, …, un) and v = (v1, v2, …, vn) are vectors in Rn, then the formula

v, u= u · v = u1v1 + u2v2 + … + unvn defines v, uto be the Euclidean product on Rn.  The four inner product axioms hold by Theorem 4.1.2.

2008/12/17 Elementary 3 Properties of Euclidean Inner Product

 Theorem 4.1.2 n  If u, v and w are vectors in R and k is any , then  u · v = v · u  (u + v) · w = u · w + v · w  (k u) · v = k (u · v)  v · v ≥0; Further, v · v = 0 if and only if v = 0  Example

 (3u + 2v) · (4u + v) = (3u) · (4u + v) + (2v) · (4u + v ) = (3u) · (4u) + (3u) · v + (2v) · (4u) + (2v) · v =12(u · u) + 11(u · v) + 2(v · v)

2008/12/17 Elementary Linear Algebra 4 Euclidean Inner Product vs. Inner Product  The Euclidean inner product is the most important inner product on Rn. However, there are various applications in which it is desirable to modify the Euclidean inner product by weighting its terms differently.

 More precisely, if w1,w2,…,wn are positive real , and if u=(u1,u2,…,un) and v=(v1,v2,…,vn) are vectors in Rn, then it can be shown

which is called weighted Euclidean inner product with

weights w1,w2,…,wn …

2008/12/17 Elementary Linear Algebra 5 Example

 Suppose that some physical experiment can produce any

of n possible numerical values x1,x2,…,xn.  We perform m repetitions of the experiment and yield

these values with various frequencies; that is, x1 occurs f1 times, x2 occurs f2 times, and so forth. f1+f2+…+fn=m.  Thus, the arithmetic average, or mean, is

2008/12/17 Elementary Linear Algebra 6 Example

 If we let f=(f1,f2,…,fn), x=(x1,x2,…,xn), w1=w2=…=wn=1/m  Then this equation can be expressed as the weighted inner product

2008/12/17 Elementary Linear Algebra 7 Weighted Euclidean Product

2  Let u = (u1, u2) and v = (v1, v2) be vectors in R . Verify that the weighted Euclidean inner product u, v= 3u1v1 + 2u2v2 satisfies the four product axioms.

 Solution:

 Note first that if u and v are interchanged in this equation, the right side remains the same. Therefore, u, v= v, u.

 If w = (w1, w2), then

u + v, w= 3(u1+v1)w1 + 2(u2+v2)w2 =

(3u1w1 + 2u2w2) + (3v1w1 + 2v2w2)= u, w+ v, w which establishes the second axiom.

2008/12/17 Elementary Linear Algebra 8 Weighted Euclidean Product

 ku, v=3(ku1)v1 + 2(ku2)v2 = k(3u1v1 + 2u2v2) = k u, v which establishes the third axiom. 2 2 2  v, v= 3v1v1+2v2v2 = 3v1 + 2v2 .Obviously , v, v= 3v1 + 2 2 2 2v2 ≥0 . Furthermore, v, v= 3v1 + 2v2 = 0 if and only if v1 = v2 = 0, That is , if and only if v = (v1,v2)=0. Thus, the fourth axiom is satisfied.

2008/12/17 Elementary Linear Algebra 9 Definition

 If V is an inner product space, then the (or length) of a vector u in V is denoted by ||u|| and is defined by ||u|| = u, u½  The between two points (vectors) u and v is denoted by d(u,v) and is defined by d(u, v) = ||u –v||

 If a vector has norm 1, then we say that it is a .

2008/12/17 Elementary Linear Algebra 10 Norm and Distance in Rn

 If u = (u1, u2, …, un) and v = (v1, v2, …, vn) are vectors in Rn with the Euclidean inner product, then

d(,),()()uv uv  uvuv  1/ 2  uvuv   1/ 2

2 2 2  (u1 v 1 ) ( u 2 v 2 ) ... ( un v n )

2008/12/17 Elementary Linear Algebra 11 Weighted Euclidean Inner Product

 The norm and distance depend on the inner product used.

 If the inner product is changed, then the norms and between vectors also change. 2  For example, for the vectors u = (1,0) and v = (0,1) in R with the Euclidean inner product, we have u 1 and d(u, v) u v (1,1)  12 (1)2  2

 However, if we change to the weighted Euclidean inner product

u, v= 3u1v1 + 2u2v2 , then we obtain u u,u 1/ 2 (311200)1/ 2  3 d(u, v) u v (1,1),(1,1) 1/ 2 [3112(1)(1)]1/ 2  5

2008/12/17 Elementary Linear Algebra 12 Unit Circles and Spheres in Inner Product Space

 If V is an inner product space, then the of points in V that satisfy ||u|| = 1 is called the unite sphere or sometimes the unit circle in V. In R2 and R3 these are the points that lie 1 unit away form the origin.

2008/12/17 Elementary Linear Algebra 13 Unit Circles in R2

 Sketch the unit circle in an xy-coordinate system in R2 using the

Euclidean inner product u, v= u1v1 + u2v2  Sketch the unit circle in an xy-coordinate system in R2 using the

Euclidean inner product u, v= 1/9u1v1 + 1/4u2v2  Solution ½ 2 2 ½  If u = (x,y), then ||u|| = u, u = (x + y ) , so the equation of the unit circle is x2 + y2 = 1. ½ 2 2 ½  If u = (x,y), then ||u|| = u, u = (1/9x + 1/4y ) , so the equation of the unit circle is x2 /9 + y2/4 = 1.

2008/12/17 Elementary Linear Algebra 14 Inner Products Generated by Matrices

u1  v 1      u2 v 2  Letu and v =   be vectors in Rn (expressed as n1 ::       un  v n  matrices), and let A be an invertible nn .  If u · v is the Euclidean inner product on Rn, then the formula u, v= Au · Av defines an inner product; it is called the inner product on Rn generated by A.  Recalling that the Euclidean inner product u · v can be written as the matrix product vTu, the above formula can be written in the alternative form u, v= (Av) TAu, or equivalently, u, v= vTATAu

2008/12/17 Elementary Linear Algebra 15 Inner Product Generated by the

 The inner product on Rn generated by the nn identity matrix is the Euclidean inner product: Let A = I, we have u, v= Iu · Iv = u · v

 The weighted Euclidean inner product u, v= 3u1v1 + 2u2v2 is the inner product on R2 generated by 3 0  A   0 2 since

3 0 3 0 u1  3 0u1  u,v v1 v2    v1 v2   3u1v1 2u2v2 0 20 2u2  0 2u2 

2008/12/17 Elementary Linear Algebra 16 Inner Product Generated by the Identity Matrix

 In general, the weighted Euclidean inner product u,

v= w1u1v1 + w2u2v2 + … + wnunvn is the inner product on Rn generated by

2008/12/17 Elementary Linear Algebra 17 An Inner Product on M22

u1 u 2  v 1 v 2   If UV and   u3 u 4  v 3 v 4  are any two 22 matrices, then T T U, V= tr(U V) = tr(V U) = u1v1 + u2v2 + u3v3 + u4v4 defines an inner product on M22  For example, if 1 2  1 0 UV and   3 4  3 2 then U, V= 1(-1)+2(0) + 3(3) + 4(2) = 16  The norm of a matrix U relative to this inner product is

1/ 2 2 2 2 2 U U, U  u1 u 2 u 3 u 4 and the unit sphere in this space consists of all 22 matrices U whose entries satisfy the equation ||U|| = 1, which on squaring yields 2 2 2 2 u1 + u2 + u3 + u4 = 1

2008/12/17 Elementary Linear Algebra 18 An Inner Product on P2

2 2  If p = a0 + a1x + a2x and q = b0 + b1x + b2x are any two vectors in P2, then the following formula defines an inner product on P2: p, q= a0b0 + a1b1 + a2b2  The norm of the p relative to this inner

product is 1/ 2 2 2 2 p p, p   a0 a 1 a 2 and the unit sphere in this space consists of all p in P2 whose coefficients satisfy the equation || p || = 1, which on squaring yields 2 2 2 a0 + a1 + a2 =1

2008/12/17 Elementary Linear Algebra 19 Theorem 6.1.1 (Properties of Inner Products)

 If u, v, and w are vectors in a real inner product space, and k is any scalar, then:

 0, v= v, 0= 0

 u, v + w= u, v+ u, w

 u, kv=k u, v

 u –v, w= u, w–v, w

 u, v –w= u, v–u, w

2008/12/17 Elementary Linear Algebra 20 Example

 u –2v, 3u + 4v = u, 3u + 4v–2v, 3u+4v = u, 3u+ u, 4v–2v, 3u–2v, 4v = 3 u, u+ 4 u, v–6 v, u–8 v, v = 3 || u ||2 + 4 u, v–6 u, v–8 || v ||2 = 3 || u ||2 –2 u, v–8 || v ||2

2008/12/17 Elementary Linear Algebra 21 Example

 We are guaranteed without any further proof that the five properties given in Theorem 6.1.1 are true for the inner product on Rn generated by any matrix A.  u, v + w= (v+w)TATAu =(vT+wT)ATAu [Property of ] =(vTATAu) + (wTATAu) [Property of matrix ] = u, v+ u, w

2008/12/17 Elementary Linear Algebra 22 Lecture 23: 6.2 and

Wei-Ta Chu

2008/12/17 Theorem 6.2.1

 Theorem 6.2.1 (Cauchy-Schwarz )  If u and v are vectors in a real inner product space, then |u, v| ||u|| ||v||

 The inequality can be written in the following two forms

n  The Cauchy-Schwarz inequality for R (Theorem 4.1.3) follows as a special case of this theorem by taking u, vto be the Euclidean inner product u‧v.

2008/12/17 Elementary Linear Algebra 24 Theorem 6.2.2

 Theorem 6.2.2 (Properties of Length)  If u and v are vectors in an inner product space V, and if k is any scalar, then :  || u || 0  || u || = 0 if and only if u = 0  || ku || = | k | || u ||  || u + v || || u || + || v || ()  Proof of (d)

(Property of ) (Theorem 6.2.1)

2008/12/17 Elementary Linear Algebra 25 Theorem 6.2.3

 Theorem 6.2.3 (Properties of Distance)  If u, v, and w are vectors in an inner product space V, and if k is any scalar, then:  d(u, v) 0  d(u, v)=0ifandonlyif u = v  d(u, v) = d(v, u)  d(u, v) d(u, w) + d(w, v) (Triangle inequality)

2008/12/17 Elementary Linear Algebra 26 Angle Between Vectors

 Cauchy-Schwarz inequality can be used to define in general inner product cases.

 We define to be the angle between u and v.

2008/12/17 Elementary Linear Algebra 27 Example

 Let R4 have the Euclidean inner product. Find the cosine of the angle between the vectors u = (4, 3, 1, -2) and v = (-2, 1, 2, 3).

2008/12/17 Elementary Linear Algebra 28 Orthogonality

 Definition

 Two vectors u and v in an inner product space are called orthogonal if u, v= 0.

T T  Example (U, V= tr(U V) = tr(V U) = u1v1 + u2v2 + u3v3 + u4v4)

 If M22 has the inner project defined previously, then the matrices 1 0 0 2 U  and V   1 1 0 0 are orthogonal, since U, V= 1(0) + 0(2) + 1(0) + 1(0) = 0.

2008/12/17 Elementary Linear Algebra 29 Orthogonal Vectors in P2

1  Let P2 havetheinnerproductp,()() q  p x q x dx andlet p = x and 2  q = x . 1

 Then 11/ 2 1 1/ 2 1/ 2   2  2 p p, p  xxdx  x dx  1  1  3

11/ 2 1 1/ 2 1/ 2 2 2  4  2 q q, q  x x dx  x dx  1  1  5 1 1 p, q xx2 dx x 3 dx  0 1 1 because p, q= 0, the vectors p = x and q = x 2 are orthogonal relative to the given inner product.

2008/12/17 Elementary Linear Algebra 30 Theorem 6.2.4 (Generalized Theorem of Pythagoras)

 If u and v are orthogonal vectors in an inner product space, then || u + v ||2 = || u ||2 + || v ||2  Proof

2008/12/17 Elementary Linear Algebra 31 Example

2  Since p =1 x and q = x are orthogonal relative to the inner product p,()() q p x q x dx on P2. 1  It follows from the Theorem of Pythagoras that || p + q ||2 = || p ||2 + || q ||2  Thus, from the previous example:

2 2 2 2 2 16 p+ q  ( )2 ( ) 2    3 5 3 5 15

 We can check this result by direct integration: 1 p+ q2  p + q , p + q  (x x2 )( x x 2 ) dx 1 1 1 1 2 2 16 x2 dx2 x 3 dx x 4 dx   0   1 1 1 3 5 15

2008/12/17 Elementary Linear Algebra 32 Orthogonality

 Definition  Let W be a subspace of an inner product space V. A vector u in V is said to be orthogonal to W if it is orthogonal to every vector in W, and the set of all vectors in V that are orthogonal to W is called the (正交 補餘) of W. 3  If V is a plane through the origin of R with Euclidean inner product, then the set of all vectors that are orthogonal to every vector in V forms the L through the origin that is to V. L

V

2008/12/17 Elementary Linear Algebra 33 Theorem 6.2.5

 Theorem 6.2.5 (Properties of Orthogonal Complements)  If W is a subspace of a finite-dimensional inner product space V, then:  Wis a subspace of V. (read “W perp”)  The only vector common to W and Wis 0; that is ,W W= 0.  The orthogonal complement of Wis W; that is , (W)= W.

2008/12/17 Elementary Linear Algebra 34 Proof of Theorem 6.2.5(a)

 Note first that 0, w=0 for every vector w in W, so W  contains at least the zero vector.  We want to show that the sum of two vectors in W is orthogonal to every vector in W (closed under ) and that any scalar multiple of a vector in W is orthogonal to every vector in W (closed under ).

2008/12/17 Elementary Linear Algebra 35 Proof of Theorem 6.2.5(a)

 Let u and v be any vector in W , let k be any scalar, and let w be any vector in W. Then from the definition of W , we have u, w=0 and v, w=0.  Using the basic properties of the inner product, we have u+v, w= u, w+ v, w= 0 + 0 = 0 ku, w= ku, w= k(0) = 0  Which proves that u+v and ku are in W .

2008/12/17 Elementary Linear Algebra 36 Proof of Theorem 6.2.5(b)

The only vector common to W and Wis 0; that is ,W W= 0.

 If v is common to W and W , then v, v=0, which implies that v=0 by Axiom 4 for inner products.

2008/12/17 Elementary Linear Algebra 37 Theorem 6.2.6

 Theorem 6.2.6  If A is an mn matrix, then:  The nullspace of A and the row space of A are orthogonal complements in Rn with respect to the Euclidean inner product.  The nullspace of AT and the column space of A are orthogonal complements in Rm with respect to the Euclidean inner product.

2008/12/17 Elementary Linear Algebra 38 Proof of Theorem 6.2.6(a)

 We must show that if a vector v is orthogonal to every vector in the row space, then Av=0, and conversely, that if Av=0, then v is orthogonal to every vector in the row space.  Assume that v is orthogonal to every vector in the row space of A. Then in particular, v is orthogonal to the row

vectors r1, r2, …, rm of A; that is

2008/12/17 Elementary Linear Algebra 39 Proof of Theorem 6.2.6(a)

 The linear system Ax=0 can be expressed in notation as

 And it follows that v is a solution of this system and hence lies in the nullspace of A.

2008/12/17 Elementary Linear Algebra 40 Proof of Theorem 6.2.6(a)

 Conversely, assume that v is a vector in the nullspace of A, so Av=0. It follows that  But if r is any vector in the row space of A, then r is expressible as a of the row vectors of A, say  Thus

 Which proves that v is orthogonal to every vector in the row space of A.

2008/12/17 Elementary Linear Algebra 41 Example ( for an Orthogonal Complement)

 Let W be the subspace of R5 spanned  By Theorem 6.2.6, the nullspace of A by the vectors w1=(2, 2, -1, 0, 1), is the orthogonal complement of W. w2=(-1, -1, 2, -3, 1), w3=(1, 1, -2, 0, -  In Example 4 of Section 5.5 we 1), w4=(0, 0, 1, 1, 1). Find a basis for showed that the orthogonal complement of W. 1  1  Solution 1  0   The space W spanned by w1, w2, w3,     and w4 is the same as the row space of v10and v 2  1 the matrix     0 0 2 2 1 0 1      0  1  1 1 2 3 1      A   form a basis for this nullspace. 1 1 2 0 1    Thus, vectors v = (-1, 1, 0, 0, 0) 0 0 1 1 1 1   and v2 = (-1, 0, -1, 0, 1) form a basis for the orthogonal complement of W.

2008/12/17 Elementary Linear Algebra 42 Remarks

 In any inner product space V, the zero space {0} and the entire space V are orthogonal complements.  If A is an matrix, to say that Ax=0 has only the trivial solution is equivalent to saying that the orthogonal complement of the nullspace of A is all of Rn, or equivalently, that the row space of A is all of Rn.

2008/12/17 Elementary Linear Algebra 43 Theorem 6.2.7 (Equivalent Statements) n n  If A is an mn matrix, and if TA : R  R is multiplication by A, then the following are equivalent:  A is invertible.  Ax = 0 has only the trivial solution.

 The reduced row-echelon form of A is In.  A is expressible as a product of elementary matrices.  Ax = b is consistent for every n1 matrix b.  Ax = b has exactly one solution for every n1 matrix b.  det(A)≠0. n  The range of TA is R .  TA is one-to-one.  The column vectors of A are linearly independent.  The row vectors of A are linearly independent. n  The column vectors of A span R . n  The row vectors of A span R . n  The column vectors of A form a basis for R . n  The row vectors of A form a basis for R .  A has n.  A has nullity 0. n  The orthogonal complement of the nullspace of A is R .  The orthogonal complement of the row of A is {0}.

2008/12/17 Elementary Linear Algebra 44