Chapter 6 Inner Product Spacesmathstat.sci.tu.ac.th/~archara/MA131/MA131-116/stnote131-116-ch… ·...

Chapter 6

Inner Product Spaces

6.1 Inner Products

An inner product on a real vector space V is an operator on V that assignsto each pair of vector u and v in V a real number 〈u,v〉 satisfying the followingconditions:

(i) 〈u,v〉 = 〈v,u〉 for all u and v in V .

(ii) 〈u+ v, z〉 = 〈u, z〉+ 〈v, z〉 for all u, v, z in V .

(iii) 〈αu,v〉 = α〈u,v〉 for all u, v in V and all scalars α.

(iv) 〈u,u〉 ≥ 0 with equality if and only if u = 0.

Definition 6.1.

A real vector space V with an inner product is called an real inner product space.

Remark In Section 6.4 we shall study inner products over complex vector spaces. How-ever, until that time we shall use the term “ inner product space ” to mean real innerproduct space.

191

MA131 (Section 750002): Prepared by Asst.Prof.Dr. Archara Pacheenburawana 192

If u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are vectors in Rn, then thestandard inner product for Rn is defined by

〈u,v〉 = u · v = uTv = u1v1 + u2v2 + · · ·+ unvn

This inner product is also known as the Euclidean inner product on Rn.Given a vector w with positive entries, we could also define an inner producton Rn by

〈u,v〉 =n∑

i=1

wiuivi

the entries wi are referred to as weights. ♣

Example 6.1.

Let u = (u1, u2) and v = (v1, v2) be vectors in R2. Verify that the weightedEuclidean inner product

〈u,v〉 = 3u1v1 + 2u2v2

satisfies the four inner product conditions.

Example 6.2.

Solution


Given A and B in Mm×n, we can define an inner product on Mm×n by

〈A,B〉 =m∑

i=1

n∑

j=1

aijbij ♣ (6.1)

Example 6.3.

Let x1, x2, . . . , xn be distinct real numbers. For each pair of polynomials inPn−1, define

〈p, q〉 =n∑

i=1

p(xi)q(xi) (6.2)

It is easily seen that (6.2) satisfies conditions (i), (ii), and (iii) of the definitionof an inner product. To show that (iv) holds, note that

〈p, p〉 =n∑

i=1

(

p(xi))2 ≥ 0

If 〈p, p〉 = 0, then x1, x2, . . . , xn must be roots of p(x) = 0. Since p(x) is ofdegree less than or equal n− 1, it must be the zero polynomial.If w(x) is a positive function, then

〈p, q〉 =n∑

i=1

w(xi)p(xi)q(xi)

also defines an inner product on Pn−1. ♣

Example 6.4.

Let f = f(x) and g = g(x) be two functions in C[−1, 1], the vector space of allreal-valued functions continuous on the interval [−1, 1]. Define

〈f , g〉 =∫

1

−1

f(x)g(x)dx

Verify that 〈f , g〉 is an inner product on C[−1, 1].

Example 6.5.

Solution


Length and Distance in Inner Product Spaces

If V is an inner product space, then the norm (or length) of a vector v in Vis denoted by ‖v‖ and is defined by

‖v‖ = 〈v,v〉1/2 =√

〈v,v〉

The distance between two vectors u and v is denoted by d(u,v) and is definedby

d(u,v) = ‖u− v‖

Definition 6.2.

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

If u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are vectors in Rn with the Eu-clidean inner product, then

‖u‖ = 〈u,u〉1/2 =√

u21 + u2

2 + · · ·+ u2n

and

d(u,v) = ‖u− v‖ = 〈u− v,u− v〉1/2

=√

(u1 − v1)2 + (u2 − v2)2 + · · ·+ (un − vn)2

For example, for the vectors u = (1, 0) and v = (0, 1) in R2 with the Euclideaninner product, we have

‖u‖ =√12 + 02 = 1

andd(u,v) = ‖u− v‖ = ‖(1,−1)‖ =

√

12 + (−1)2 =√2

However, if we change to the weighted Euclidean inner product in Example 6.2,

〈u,v〉 = 3u1v1 + 2u2v2

then we obtain

‖u‖ = 〈u,u〉1/2 =√

3(1)(1) + 2(0)(0) =√3

and

d(u,v) = ‖u− v‖ = 〈u− v,u− v〉1/2= 〈(1,−1), (1,−1)〉1/2

=√

3(1)(1) + 2(−1)(−1)

=√5 ♣

Example 6.6.


Inner Products Generated by Matrices

The Euclidean inner product and the weighted Euclidean inner product are special casesof a general class of inner products on Rn, which we shall now describe. Let

u =

u1

u2

...un

and v =

v1v2...vn

be vectors in Rn, and let A be an invertible n× n matrix. It can be shown that if u · vis the Euclidean inner product on Rn, then the formula

〈u,v〉 = Au ·Av (6.3)

defines an inner product; it is called the inner product on Rn generated by A.Moreover, (6.3) can be written in the alternative form

〈u,v〉 = (Au)TAv

or, equivalently,〈u,v〉 = uTATAv (6.4)

The inner product on Rn generated by the n×n identity matrix is the Euclideaninner product, since substituting A = I in (6.3) yields

〈u,v〉 = Iu · Iv = u · v

The weighted Euclidean inner product 〈u,v〉 = 3u1v1 + 2u2v2 discussed inExample 6.2 is the inner product on R2 generated by

A =

[√3 0

0√2

]

because substituting this matrix in (6.4) yields

〈u,v〉 =[

u1 u2

]

[√3 0

0√2

] [√3 0

0√2

] [

v1v2

]

=[

u1 u2

]

[

3 00 2

] [

v1v2

]

= 3u1v1 + 2u2v2

Example 6.7.

In general, the weighted Euclidean inner product

〈u,v〉 = w1u1v1 + w2u2v2 + · · ·+ wnunvn


is the inner product on Rn generated by

A =

√w1 0 0 · · · 00

√w2 0 · · · 0

......

......

0 0 0 · · · √wn

If

A =

[

a1 a2a3 a4

]

and B =

[

b1 b2b3 b4

]

are any two 2× 2 matrices, then the following formula defines an inner producton M2×2:

〈A,B〉 = tr(ATB) = a1b1 + a2b2 + a3b3 + a4b4

For example, if

A =

[

1 23 4

]

and B =

[

−1 03 2

]

then〈A,B〉 = 1(−1) + 2(0) + 3(3) + 4(2) = 16

The norm of a matrix A relative to this inner product is

‖A‖ = 〈A,A〉1/2 =√

a21 + a22 + a23 + a24 ♣

Example 6.8.

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

(a) 〈0,v〉 = 〈v, 0〉=0

(b) 〈u,v +w〉 = 〈u,v〉+ 〈u,w〉

(c) 〈u, kv〉 = k〈u,v〉

(d) 〈u− v,w〉 = 〈u,w〉 − 〈v,w〉

(e) 〈u,v−w〉 = 〈u,v〉 − 〈u,w〉

Theorem 6.1 (Properties of Inner Products).


〈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〈v,u〉 − 8‖v‖2= 3‖u‖2 − 2〈u,v〉 − 8‖v‖2 ♣

Example 6.9.

Exercise 6.1

1. Let 〈u,v〉 be the Euclidean inner product on R2, and let u = (3,−2), v = (4, 5),w = (−1, 6), and α = −4. Verify that

(a) 〈u,v〉 = 〈v,u〉 (b) 〈u+ v,w〉 = 〈u,w〉+ 〈v,w〉(c) 〈u,v +w〉 = 〈u,v〉+ 〈u,w〉 (d) 〈αu,v〉 = α〈u,v〉 = 〈u, αv〉(e) 〈0,v〉 = 〈v, 0〉 = 0

2. Repeat Exercise 1 for the weighted Euclidean inner product

〈u,v〉 = 4u1v1 + 5u2v2.

3. Compute 〈u,v〉 using the inner product in Example 6.8.

(a) u =

[

3 −24 8

]

, v =

[

−1 31 1

]

(b) u =

[

1 2−3 5

]

, v =

[

4 60 8

]

4. (a) Use Formula (6.3) to show that 〈u,v〉 = 9u1v1 + 4u2v2 is the inner product onR2 generated by

A =

[

3 00 2

]

(b) Use the inner product in part (a) to compute 〈u,v〉 if u = (−3, 2) and v =(1, 7).

5. Let u = (u1, u2) and v = (v1, v2). In each part, the given expression is an innerproduct on R2. Find a matrix that generates it.

(a) 〈u,v〉 = 3u1v1 + 5u2v2

(b) 〈u,v〉 = 4u1v1 + 6u2v2

6. Let u = (u1, u2, u3) and v = (v1, v2, v3). Determine which of the following are innerproducts on R

3. For those that are not, list the conditions that do not hold.

(a) 〈u,v〉 = u1v1 + u3v3 (b) 〈u,v〉 = u21v

21 + u2

2v22 + u2

3v23

(c) 〈u,v〉 = 2u1v1 + u2v2 + 4u3v3 (d) 〈u,v〉 = u1v1 − u2v2 + u3v3


7. In each part, use the given inner product on R2 to find d(u,v) for u = (−1, 2) andv = (2, 5).

(a) the Euclidean inner product

(b) the weighted Euclidean inner product 〈u,v〉 = 3u1v1 + 2u2v2, where u =(u1, u2) and v = (v1, v2)

(c) the inner product generated by the matrix

A =

[

1 2−1 3

]

8. Let M2×2 have the inner product in Example 6.8. In each part, find ‖A‖.

(a) A =

[

−2 53 6

]

(b) A =

[

0 00 0

]

9. Let M2×2 have the inner product in Example 6.8. Find d(A,B).

(a) A =

[

2 69 4

]

, B =

[

−4 71 6

]

(b) A =

[

−2 41 0

]

, B =

[

−5 16 2

]

10. Let the vector space P2 have the inner product

〈p,q〉 =∫

1

−1

p(x)q(x) dx

(a) Find ‖p‖ for p = 1,p = x, and p = x2.

(b) Find d(p,q) if p = 1 and q = x.

11. Let A =

[

a1 a2a3 a4

]

and B =

[

b1 b2b3 b4

]

.

Show that 〈A,B〉 = a1b1 + a2b3 + a3b2 + a4b4 is not an inner product on M2×2.

12. Use the inner product

〈p,q〉 =∫

1

−1

p(x)q(x) dx

to compute 〈p,q〉 for the vectors p = p(x) and q = q(x) in P3.

(a) p = 1− x+ x2 + 5x3, q = x− 3x2

(b) p = x− 5x3, q = 2 + 8x2

Answer to Selected Exercise 6.1

1. (a) 2 (b) 11 (c) −13 (d) −8 (e) 0 3. (a) 3 (b) 56 4. (b) 29

5. (a)

[√3 0

0√5

]

(b)

[

2 0

0√6

]


6. (a) No. Condition 4 fails. (b) No. Conditions 2 and 3 fail. (c) Yes.

(d) No. Condition 4 fails.

7. (a) 3√2 (b) 3

√5 (c) 3

√13 8. (a)

√74 (b) 0 9. (a)

√105 (b)

√47

10. (a)√2, 1

3

√6, 1

5

√10 (b) 2

3

√6 12. (a) −28

15(b) 0

6.2 Angle and Orthogonality in Inner Product Spaces

Recall that if u and v are nonzero vectors in R2 and R3 and θ is the angle between them,then

u · v = ‖u‖‖v‖ cos θor, alternatively,

cos θ =u · v

‖u‖‖v‖ (6.5)

Our first goal in this section is to define the concept of an angle between two vectorsin a general inner product space. For such a definition to be reasonable, we would wantit to be consistent with Formula (6.5) when it applied to the special case of R2 and R3

with the Euclidean inner product.Since | cos θ ≤ 1|, every pair of nonzero vectors in an inner product satisfies the

inequality∣

∣

∣

∣

u · v‖u‖‖v‖

∣

∣

∣

∣

≤ 1

If u and v are vectors in a real inner product space, then

|〈u,v〉| ≤ ‖u‖‖v‖

Theorem 6.2 (Cauchy-Schwarz Inequality).

For reference, we note that the Cauchy-Schwarz Inequality can be written in thefollowing two alternative forms:

〈u,v〉2 ≤ 〈u,u〉〈v,v〉 (6.6)

〈u,v〉2 ≤ ‖u‖2‖v‖2 (6.7)


If u and v are vectors in an inner product space V , and if k is any scalar, then

(a) ‖u‖ ≥ 0

(b) ‖u‖ = 0 if and only if u = 0

(c) ‖ku‖ = |k|‖u‖

(d) ‖u+ v‖ ≤ ‖u‖+ ‖v‖ (Triangle inequality)

Theorem 6.3 (Properties of Length).

If u, v, and w are vectors in an inner product space V , and if k is any scalar,then

(a) d(u,v) ≥ 0

(b) d(u,v) = 0 if and only if u = v

(c) d(u,v) = d(v,u)

(d) d(u,v) ≤ d(u,w) + d(w,v) (Triangle inequality)

Theorem 6.4 (Properties of Distance).

Angle Between Vectors

We shall now show how the Cauchy-Schwarz inequality can be used to define angles ingeneral inner product spaces. Suppose that u and v are nonzero vectors in an innerproduct space V . If we divide both sides of Formula (6.7) by ‖u‖2‖v‖2, we obtain

[ 〈u,v〉‖u‖‖v‖

]2

≤ 1

or, equivalently,

−1 ≤ 〈u,v〉‖u‖‖v‖ ≤ 1 (6.8)

Thus, from (6.8), there is a unique angle θ such that

cos θ =〈u,v〉‖u‖‖v‖ and 0 ≤ θ ≤ π (6.9)

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

Let R4 have the Euclidean inner product. Find the cosine of the angle θ betweenthe vectors u = (4, 3, 1,−2) and v = (−2, 1, 2, 3).

Example 6.10.

Solution


Orthogonality

Two vectors u and v in an inner product are called orthogonal if 〈u,v〉 = 0.

Definition 6.3.

If M2×2 has the inner product of Example 6.8 in the preceding section, then thematrices

A =

[

1 01 1

]

and B =

[

0 20 0

]

are orthogonal, since

〈A,B〉 = 1(0) + 0(2) + 1(0) + 1(0) = 0 ♣

Example 6.11.

Let P2 have the inner product

〈p,q〉 =∫

1

−1

p(x)q(x) dx

and let p = x and q = x2. Then

‖p‖ = 〈p,p〉1/2 =

[∫

1

−1

xx dx

]1/2

=

[∫

1

−1

x2 dx

]1/2

=

√

2

3

‖q‖ = 〈q,q〉1/2 =

[∫

1

−1

x2x2 dx

]1/2

=

[∫

1

−1

x4 dx

]1/2

=

√

2

5

〈p,q〉 =

∫

1

−1

xx2 dx =

∫

1

−1

x3 dx = 0

Because 〈p,q〉 = 0, the vectors p = x and q = x2 are orthogonal relative to thegiven inner product. ♣

Example 6.12.

If u and v are orthogonal vectors in an inner product space, then

‖u+ v‖2 = ‖u‖2 + ‖v‖2

Theorem 6.5 (Generalized Theorem of Pythagoras).


In Example 6.12 we showed that p = x and q = x2 are orthogonal relative tothe inner product

〈p,q〉 =∫

1

−1

p(x)q(x) dx

on P2. It follows from the Theorem of Pythagoras that

‖p+ q‖2 = ‖p‖2 + ‖q‖2

Thus, from the computations in Example 6.12, we have

‖p+ q‖2 =(

√

2

3

)2

+

(

√

2

5

)2

=2

3+

2

5=

16

15

We can check this result by direct integration:

‖p+ q‖2 = 〈p+ q,p+ q〉

=

∫

1

−1

(x+ x2)(x+ x2) dx

=

∫

1

−1

x2 dx+ 2

∫

1

−1

x3 dx+

∫

1

−1

x4 dx

=2

3+ 0 +

2

5=

16

15♣

Example 6.13.

Orthogonal Complements

If V is a plane though the origin of R3 with the Euclidean inner product, then the set ofall vectors that are orthogonal to every vector in V forms the line L through the originthat is perpendicular to V . In the language of linear algebra we say that the line andthe plane are orthogonal complements of one another. The following definition extendsthis concept to general inner product spaces.

V

L


Let W be a subspace of an inner product space V . A vector u in V is said tobe orthogonal to W if it is orthogonal to every vector in W , and the set of allvectors in V that are orthogonal to W is called the orthogonal complement

of W and is denoted by W⊥.

Definition 6.4.

If W is a subspace of a finite-dimensional inner product space V , then

(a) W⊥ is a subspace of V .

(b) The only vector common to W and W⊥ is 0.

(c) The orthogonal complement of W⊥ is W ; that is, (W⊥)⊥ = W .

Theorem 6.6 (Properties of Orthogonal Complements).

Remark Because W and W⊥ are orthogonal complements of one another by part (c) ofthe Theorem 6.6, we shall say that W and W⊥ are orthogonal complements.

If A is an m× n matrix, then

(a) The nullspace of A and the row space of A are orthogonal complementsin Rn with respect to the Euclidean inner product.

(b) The nullspace of AT and the column space of A are orthogonal comple-ments in Rm with respect to the Euclidean inner product.

Theorem 6.7.

Let W be the subspace of R4 spanned by the vectors

w1 = (1,−1, 2, 1) w2 = (0, 1, 1,−2) w3 = (1,−3, 0, 5)

Find a basis for the orthogonal complement of W .

Example 6.14.

Solution


Orthonormal

A set of vectors in an inner product space is called an orthogonal set if allpairs of distinct vectors in the set are orthogonal. An orthogonal set in whicheach vector has norm 1 is called orthonormal.

Definition 6.5.

Letu1 = (0, 1, 0), u2 = (1, 0, 1), u3 = (1, 0,−1)

and assume that R3 has the Euclidean inner product. It follows that the set ofvectors S = {u1,u2,u3} is orthogonal since 〈u1,u2〉 = 〈u1,u3〉 = 〈u2,u3〉 = 0.♣

Example 6.15.

If v is a nonzero vector in an inner product space, then by part (c) of Theorem 6.3,the vector

1

‖v‖ v

has norm 1, since∥

∥

∥

∥

1

‖v‖ v

∥

∥

∥

∥

=

∣

∣

∣

∣

1

‖v‖

∣

∣

∣

∣

‖v‖ =1

‖v‖ ‖v‖ = 1

The process of multiplying a nonzero vector v by the reciprocal of its length to obtain aunit vector is called normalizing v. An orthogonal set of nonzero vectors can alwaysbe converted to an orthonormal set by normalizing each of its vectors.

The Euclidean norm of the vectors in Example 6.15 are

‖u1‖ = 1, ‖u2‖ =√2 , ‖u3‖ =

√2

Consequently, normalizing u1, u2, and u3 yields

v1 =u1

‖u1‖= (0, 1, 0), v2 =

u2

‖u2‖=

(

1√2, 0,

1√2

)

,

v3 =u3

‖u3‖=

(

1√2, 0,− 1√

2

)

We can verify that the set S = {v1,v2,v3} are orthonormal by showing that

〈v1,v2〉 = 〈v1,v3〉 = 〈v2,v3〉 = 0 and ‖v1‖ = ‖v2‖ = ‖v3‖ = 1 ♣

Example 6.16.


In an inner product space, a basis consisting of orthonormal vectors is called a or-

thonormal basis, and a basis consisting of orthogonal vectors is called a orthogonal

basis. A familiar example of an orthonormal basis is the standard basis for R3 with theEuclidean inner product:

i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

More generally, in Rn with the Euclidean inner product, the standard basis

e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1)

is orthonormal.

Coordinates Relative to Orthonormal Bases

If S = {v1,v2, . . . ,vn} is an orthonormal basis for an inner product space Vand u is any vector in V , then

u = 〈u,v1〉v1 + 〈u,v2〉v2 + · · ·+ 〈u,vn〉vn

Theorem 6.8.

Letv1 = (0, 1, 0), v2 =

(

−4

5, 0, 3

5

)

, v3 =(

3

5, 0, 4

5

)

It is easy to check that S = {v1,v2,v3} is an orthonormal basis for R3 with theEuclidean inner product. Express the vector u = (1, 1, 1) as a linear combinationof the vectors in S, and find the coordinate vector (u)S.

Example 6.17.

Solution


If S is an orthonormal basis for an n-dimensional inner product space, and if

(u)S = (u1, u2, . . . , un) and (v)S = (v1, v2, . . . , vn)

then

(a) ‖u‖ =√

u21 + u2

2 + · · ·+ u2n

(b) d(u,v) =√

(u1 − v1)2 + (u2 − v2)2 + · · ·+ (un − vn)2

(c) 〈u,v〉 = u1v1 + u2v2 + · · ·+ unvn

Theorem 6.9.

If R3 has the Euclidean inner product, then the norm of the vector u = (1, 1, 1)is

‖u‖ = (u · u)1/2 =√12 + 12 + 12 =

√3

However, if we let R3 have the orthonormal basis S in Example 6.17, then weknow from that example that the coordinate vector of u relative to S is

(u)S =(

1,−1

5, 75

)

The norm of u can also be calculated from this vector using part (a) of Theo-rem 6.9. This yields

‖u‖ =

√

12 +(

−1

5

)2+(

7

5

)2=√

75

25=

√3 ♣

Example 6.18.

Coordinates Relative to Orthogonal Bases

If S = {v1,v2, . . . ,vn} is an orthogonal basis for a vector space V , then normalizingeach of these vectors yields the orthonormal basis

S ′ =

{

v1

‖v1‖,

v2

‖v2‖, . . . ,

vn

‖vn‖

}

Thus, if u is any vector in V , it follow from Theorem 6.8 that

u =

⟨

u,v1

‖v1‖

⟩

v1

‖v1‖+

⟨

u,v2

‖v2‖

⟩

v2

‖v2‖+ · · ·+

⟨

u,vn

‖vn‖

⟩

vn

‖vn‖which, by part (c) of Theorem 6.3, can be written as

u =〈u,v1〉‖v1‖2

v1 +〈u,v2〉‖v2‖2

v2 + · · ·+ 〈u,vn〉‖vn‖2

vn (6.10)

This formula expresses u as a linear combination of the vectors in the orthogonal basisS.


If S = {v1,v2, . . . ,vn} is an orthogonal set of nonzero vectors in an innerproduct space, then S is linearly independent.

Theorem 6.10.

Orthogonal Projections

In R2 or R3 with the Euclidean inner product, it is evident geometrically that if W is aline or a plane through the origin, then each vector u in the space can be expressed asa sum

u = w1 +w2

where w1 is in W and w2 is perpendicular to W .

Ww1

w2

u

o

If W is a finite-dimensional subspace of an inner product space V , then everyvector u in V can be expressed in exactly one way as

u = w1 +w2

where w1 is in W and w2 is in W⊥.

Theorem 6.11 (Projection Theorem).

The vector w1 in Theorem 6.11 is called the orthogonal projection of u on Wand is denoted by projWu. The vector w2 is called the component of u orthogonal

to W and is denoted by projW⊥u. Thus Formula in the Projection Theorem can beexpressed as

u = projWu+ projW⊥u (6.11)

Since w2 = u−w1, it follows that

projW⊥u = u− projWu

so Formula (6.11) can also be written as

u = projWu+ (u− projWu) (6.12)

WprojWu

u− projWu

u

o


Let W be a finite-dimensional subspace of an inner product space V .

(a) If {v1,v2, . . . ,vr} is an orthonormal basis for W , and u is any vector inV , then

projWu = 〈u,v1〉v1 + 〈u,v2〉v2 + · · ·+ 〈u,vr〉vr (6.13)

(b) If {v1,v2, . . . ,vr} is an orthogonal basis for W , and u is any vector in V ,then

projWu =〈u,v1〉‖v1‖2

v1 +〈u,v2〉‖v2‖2

v2 + · · ·+ 〈u,vr〉‖vr‖2

vr (6.14)

Theorem 6.12.

Let R3 have the Euclidean inner product, and let W be the subspace spannedby the orthonormal vectors v1 = (0, 1, 0) and v2 =

(

−4

5, 0, 3

5

)

. From (6.13) theorthogonal projection of u = (1, 1, 1) on W is

projWu = 〈u,v1〉v1 + 〈u,v2〉v2

= (1)(0, 1, 0) +(

−1

5

) (

−4

5, 0, 3

5

)

=(

4

25, 1,− 3

25

)

The component of u orthogonal to W is

projW⊥u = u− projWu = (1, 1, 1)−(

4

25, 1,− 3

25

)

=(

21

25, 0, 28

25

)

Observe that projW⊥u is orthogonal to both v1 and v2, so this vector is or-thogonal to each vector in the space W spanned by v1 and v2, as it should be.♣

Example 6.19.

Exercise 6.2

1. In each part, determine whether the given vectors are orthogonal with respect tothe Euclidean inner product.

(a) u = (−1, 3, 2), v = (4, 2,−1) (b) u = (−2,−2,−2), v = (1, 1, 1)

(c) u = (u1, u2, u3), v = (0, 0, 0) (d) u = (a, b), v = (−b, a)

(e) u = (0, 3,−2, 1), v = (5, 2,−1, 0)

(f) u = (−4, 6,−10, 1), v = (2, 1,−2, 9)

2. Let R2, R3, and R4 have the Euclidean inner product. In each part, find the cosineof the angle between u and v.


(a) u = (1,−3), v = (2, 4) (b) u = (−1, 0), v = (3, 8)

(c) u = (−1, 5, 2), v = (2, 4,−9) (d) u = (4, 1, 8), v = (1, 0,−3)

(e) u = (1, 0, 1, 0), v = (−3,−3,−3,−3)

(f) u = (2, 1, 7,−1), v = (4, 0, 0, 0)

3. Let

A =

[

2 1−1 3

]

Which of the following matrices are orthogonal to A with respect to the innerproduct in Example 6.8 of Section 6.1?

(a)

[

−3 00 2

]

(b)

[

1 10 −1

]

(c)

[

0 00 0

]

(d)

[

2 15 2

]

4. Let R4 have the Euclidean inner product. Find two unit vectors that are orthogonalto three vectors u = (2, 1,−4, 0), (−1,−1, 2, 2), and w = (3, 2, 5, 4).

5. Let

A =

1 2 −1 23 5 0 41 1 2 0

(a) Find bases for the column space of A and nullspace of AT .

(b) Verify that every vector in the column space of A is orthogonal to every vectorin the nullspace of AT .

6. Find a basis for the orthogonal complement of the subspace of Rn spanned by thevectors.

(a) v1 = (1,−1, 3), v2 = (5,−4,−4), v3 = (7,−6, 2)

(b) v1 = (2, 0,−1), v2 = (4, 0,−2)

(c) v1 = (1, 4, 5, 2), v2 = (2, 1, 3, 0), v3 = (−1, 3, 2, 2)

(d) v1 = (1, 4, 5, 6, 9), v2 = (3,−2, 1, 4,−1), v3 = (−1, 0,−1,−2,−1),

v4 = (2, 3, 5, 7, 8)

7. Indicate whether each statement is always true or sometimes false.

(a) If V is a subspace of Rn and W is a subspace of V , then W⊥ is a subspace ofV ⊥.

(b) ‖u+v+w‖ ≤ ‖u‖+‖v‖+‖w‖ for all vectors u,v, and w in an inner productspace.

(c) If u is in the row space and the nullspace of a square matrix A, then u = 0.

(d) If u is in the row space and the column space of an n × n matrix A, thenu = 0.


8. Which of the following sets of vectors are orthogonal with respect to the Euclideaninner product on R2?

(a) (0, 1), (2, 0) (b) (−1/√2, 1/

√2), (1/

√2, 1/

√2)

(c) (−1/√2,−1/

√2), (1/

√2, 1/

√2) (d) (0, 0), (0, 1)

9. Which of the following sets of vectors are orthonormal with respect to the Euclideaninner product on R3?

(a)

(

1√2, 0,

1√2

)

,

(

1√3,1√3,− 1√

3

)

,

(

− 1√2, 0,

1√2

)

(b)

(

2

3,−2

3,1

3

)

,

(

2

3,1

3,−2

3

)

,

(

1

3,2

3,2

3

)

(c) (1, 0, 0),

(

0,1√2,1√2

)

, (0, 0, 1)

(d)

(

1√6,1√6,− 2√

6

)

,

(

1√2,− 1√

2, 0

)

10. Which of the following sets of polynomials are orthonormal with respect to theinner product on P3 defined by

〈p,q〉 = a0b0 + a1, b1 + a2, b2

where p = a0 + a1x+ a2x2 and q = b0 + b1x+ b2x

2.

(a)2

3− 2

3x+

1

3x2,

2

3+

1

3x− 2

3x2,

1

3+

2

3x+

2

3x2

(b) 1,1√2x+

1√2x2, x2

11. Verify that the given set of vectors is orthogonal with respect to the Euclideaninner product; then convert it to an orthonormal set by normalizing the vectors.

(a) (−1, 2), (6, 3)

(b) (1, 0,−1), (2, 0, 2), (0, 5, 0)

(c) (15, 1

5, 15), (−1

2, 1

2, 0), (1

3, 1

3,−2

3)

12. Verify that the vectors v1 = (−3

5, 4

5, 0), v2 = (4

5, 3

5, 0), v3 = (0, 0, 1) form an

orthonormal basis for R3 with the Euclidean inner product; then use Theorem 6.8to express each of the following as linear combinations of v1,v2, and v3?

(a) (1,−1, 2) (b) (3,−7, 4) (c) (17,−3

7, 5

7)

13. In each part, an orthonormal basis relative to the Euclidean inner product is given.Use Theorem 6.8 to find the coordinate vector of w with respect to that basis.

(a) w = (3, 7); u1 =

(

1√2,− 1√

2

)

, u2 =

(

1√2,1√2

)

(b) w = (−1, 0, 2); u1 =(

2

3,−2

3, 13

)

, u2 =(

2

3, 13,−2

3

)

, u3 =(

1

3, 23, 2

3

)


14. Let R3 have the Euclidean inner product, and let S = {w1,w2,w3} be the or-thonormal basis with w1 =

(

0,−3

5, 4

5

)

, w2 = (1, 0, 0), and w3 =(

0, 45, 3

5

)

.

(a) Find the vectors u,v, and w that have the coordinate vectors(u)S = (−2, 1, 2), (v)S = (3, 0,−2), and (w)S = (5,−4, 1).

(b) Compute ‖v‖, d(u,w), and 〈w,v〉 by applying Theorem 6.9 to the coordi-nate vectors (u)S, (v)S, and (w)S; then check the results by performing thecomputation directly on u,v, and w.

15. (a) Show that the vectors v1 = (1,−2, 3,−4),v2 = (2, 1,−4,−3),v3 = (−3, 4, 1,−2),and v4 = (4, 3, 2, 1) form an orthogonal basis for R4 with the Euclidean inner prod-uct.

(b) Use (6.10) to express u = (−1, 2, 3, 7) as a linear combination of the vectors inpart (a).

16. Find vectors x and y in R2 that are orthonormal with respect to the inner product〈u,v〉 = 3u1v1+2u2v2 but are not orthonormal with respect to the Euclidean innerproduct.

Answer to Exercise 6.2

1. (a) Yes (b) No (c) Yes (d) Yes (e) No (f) No

2. (a) − 1√2

(b) − 3√73

(c) 0 (d) − 20

9√10

(e) − 1√2

(f) 2√55

3. (a) Orthogonal (b) Orthogonal (c) Orthogonal (d) Not orthogonal

4. ± 1

57(−34, 44,−6, 11) 5. (a)

131

,

011

;

2−11

6. (a) (16, 19, 1) (b) (0, 1, 0), (12, 0, 1) (c) (−1,−1, 1, 0), (2

7,−4

7, 0, 1)

(d) (−1,−1, 1, 0, 0), (−2,−1, 0, 1, 0), (−1,−2, 0, 0, 1)

7. (a) False (b) True (c) True (d) False

8. (a), (b), (d) 9. (b), (d) 10. (a)

11. (a)(

− 1√5, 2√

5

)

,(

2√5, 1√

5

)

(b)(

1√2, 0,− 1√

2

)

,(

1√2, 0, 1√

2

)

, (0, 1, 0)

(c)(

1√3, 1√

3, 1√

3

)

,(

− 1√2, 1√

2, 0)

,(

1√6, 1√

6,− 2√

6

)

12. (a) −7

5v1 +

1

5v2 + 2v3 (b) −37

5v1 − 9

5v2 + 4v3 (c) −3

7v1 − 1

7v2 +

5

7v3

13. (a) (w)S = (−2√2, 5

√2) (b) (w)S = (0,−2, 1)

14. (a) u =(

1, 14

5,−2

5

)

, v =(

0,−17

5, 65

)

w =(

−4,−11

5, 23

5

)

(b) ‖v‖ =√13, d(u,v) = 5

√3, 〈w,v〉 = 13

15. (b) u = −4

5v1 − 11

10v2 + 0v3 +

1

2v4

16.(

1/√5, 1/

√5)

,(

2/√30,−3/

√30)


6.3 Least Squares Problems

In this section we shall show how orthogonal projections can be used to solve certainapproximation problems. The results obtained in this section have a wide variety ofapplications in both mathematics and science.

Least Squares Solutions of Linear Systems

Up to now we have been concerned primarily with consistent systems of linear equations.In Chapter 1, we showed how a consistent linear system can be solved by the Gaussianelimination or Gauss-Jordan elimination. However, when a linear system Ax = b doesnot have a solution, one is interested in finding a best approximate solution. By an “ap-proximate solution,” we mean a vector x such that it minimizes the value of ‖Ax− b‖with respect to Euclidean inner product. The quantity ‖Ax − b‖ can be viewed as ameasure of the “error”. If the system is consistent and x is an exact solution, then theerror is zero, since ‖Ax − b‖ = ‖0‖ = 0. In general, the larger the value of ‖Ax − b‖,the more poorly x serves as an approximate solution of the system.

If W is a finite-dimensional subspace of an inner product space V , and if u isa vector in V , then projWu is the best approximation to u from W in thesense that

‖u− projWu‖ < ‖u−w‖for every vector w in W that is different form projWu.

Theorem 6.13 (Best Approximation Theorem).

Least Squares Problem: Given a linear system Ax = b of m equations in n un-knowns, find a vector x, if possible, that minimizes ‖Ax− b‖ with respect to Euclideaninner product on R

m. Such a vector is called a least squares solution of Ax = b.

Remark To understand the origin of the term least squares, let e = Ax−b, which we canview as the error vector the results from the approximation x. If e = (e1, e2, . . . , em), then

a least squares solution minimizes ‖e‖ = (e21 + e22 + · · ·+ e2m)1/2

; hence it also minimizes‖e‖2 = e21 + e22 + · · ·+ e2m. Hence the term least squares.

To solve the least squares problem, let W be the column space of A. For each n× 1matrix x, the product Ax is a linear combination of the column vectors of A. Thus, as xvaries over Rn, the vector Ax varies over all possible linear combinations of the columnvectors of A; that is, Ax varies over the entire column space W . Geometrically, solvingthe least squares problem amounts to finding a vector x in Rn such that Ax is the closestvector in W to b

(

Figure 6.1)

.


W = column space of AAx

b

b

b

Figure 6.1: A least squares solution x produces the vector Ax in W closest to b.

It follows from the Best Approximation Theorem (6.13) that the closest vector in Wto b is the orthogonal projection of b on W . Thus, for a vector x to be a least squaressolution of Ax = b, this vector must satisfy

Ax = projWb (6.15)

One could attempt to find least squares solutions of Ax = b by first calculating thevector projWb and then solving (6.15); however, there is a better approach. It followsfrom the Projection Theorem (6.11) and Formula (6.12) of Section 6.1 that

b− Ax = b− projWb

is orthogonal to W . But W is the column space of A, so it follows from Theorem 6.7that b − Ax lies in the nullspace of AT . Therefore, a least squares solution of Ax = b

must satisfyAT (b− Ax) = 0

or, equivalently,ATAx = ATb

This is called the normal system associated with Ax = b. Thus the problem of findinga least squares solution of Ax = b has been reduced to the problem of finding an exactsolution of the associated normal system.

For any linear system Ax = b, the associated normal system

ATAx = ATb

is consistent, and all solutions of the normal system are least squares solutionsof Ax = b. Moreover, if W is the column space of A, and x is any least squaressolution of Ax = b, then the orthogonal projection of b on W is

projWb = Ax

Theorem 6.14.

Uniqueness of Least Squares Solutions

Before we examine some numerical example, we shall establish conditions under whicha linear system is guaranteed to have a unique least squares solution. We shall need thefollowing theorem.


If A is an m× n matrix, then the following are equivalent.

(a) A has linearly independent column vectors.

(b) ATA is invertible.

Theorem 6.15.

If A is an m×n matrix with linearly independent column vectors, then for everym× 1 matrix b, the linear system Ax = b has a unique least squares solution.This solution is given by

x = (ATA)−1ATb (6.16)

Moreover, if W is the column space of A, then the orthogonal projection of bon W is

projWb = Ax = A(ATA)−1ATb (6.17)

Theorem 6.16.

Find the least squares solution of the linear system Ax = b given by

x1 − x2 = 4

3x1 + 2x2 = 1

2x1 + 4x2 = 3

and find the orthogonal projection of b on the column space of A.

Example 6.20.

Solution


Find the least squares solution of Ax = b, where

A =

1 2 −12 3 1−1 −1 −23 5 0

and b =

1010

Example 6.21.

Solution


Find the orthogonal projection of the vector u = (−3,−3, 8, 9) on the subspaceof R4 spanned by the vectors

u1 = (3, 1, 0, 1), u2 = (1, 2, 1, 1), u3 = (−1, 0, 2,−1)

Example 6.22.

Solution


Exercise 6.3

1. Find the normal system associated with the given linear system.

(a)

1 −12 34 5

[

x1

x2

]

=

2−15

(b)

2 −1 03 1 2−1 4 51 2 4

x1

x2

x3

=

−1012

2. Find the least squares solution of the linear system Ax = b, and find the orthogonalprojection of b onto the column space of A.

(a) A =

1 1−1 1−1 2

, b =

70−7

(b) A =

2 −21 13 1

, b =

2−11

(c) A =

1 0 −12 1 −21 1 01 1 −1

, b =

6093

(d) A =

2 0 −11 −2 22 −1 00 1 −1

, b =

0606

3. Find least squares solutions for each of the following linear systems:

(a)

1 10 21 1

[

x1

x2

]

=

210

(b)

2 3−1 32 4

[

x1

x2

]

=

321

(c)

1 2 4−2 −3 −71 3 5

x1

x2

x3

=

123

(d)

1 2 −13 5 5−1 −1 −23 5 0

x1

x2

x3

=

1110

4. Find the orthogonal projection of u onto the subspace of R4 spanned by the vectorsv1,v2, and v3.


(a) u = (6, 3, 9, 6); v1 = (2, 1, 1, 1),v2 = (1, 0, 1, 1),v3 = (−2,−1, 0,−1)

(b) u = (−2, 0, 2, 4); v1 = (1, 1, 3, 0),v2 = (−2,−1,−2, 1),v3 = (−3,−1, 1, 3)

Answer to Exercise 6.3

1. (a)

[

21 2525 35

] [

x1

x2

]

=

[

2020

]

(b)

15 −1 5−1 22 305 30 45

x1

x2

x3

=

−1913

2. (a) x1 = 5, x2 =1

2;

11

2

−9

2

−4

(b) x1 =3

7, x2 = −2

3;

46

21

− 5

2113

21

(c) x1 = 12, x2 = −3, x3 = 9;

3390

(d) x1 = 14, x2 = 30, x3 = 26;

26−24

3. (a) x =

[

1

2

1

2

]

(b) x =

[

− 1

37

21

37

]

(c) x =

−85

33

130

33

−40

33

(d) x =

−23

6

7

3

1

6

4. (a) (7, 2, 9, 5) (b)(

−12

5,−4

5, 12

5, 16

5

)

6.4 Complex Inner Product Space

An inner product on a complex vector space V is a function that asso-ciates a complex number 〈u,v〉 with each pair of vectors u and v in V in sucha way that following conditions are satisfied:

(i) 〈u,v〉 = 〈v,u〉 for all u and v in V .

(ii) 〈u+ v,w〉 = 〈u,w〉+ 〈v,w〉 for all u, v, w in V .

(iii) 〈αu,v〉 = α〈u,v〉 for all u, v in V and all scalars α.

(iv) 〈u,u〉 ≥ 0 with equality if and only if u = 0.

Definition 6.6.

A complex vector space with an inner product is called a complex inner product

space.The following additional properties follow immediately from the four inner product

conditions:

(i) 〈0,v〉 = 〈v, 0〉 = 0

(ii) 〈u,v +w〉 = 〈u,v〉+ 〈u,w〉


(iii) 〈u, αv〉 = α〈u,v〉

If u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are vectors in Cn, then we canshow that the inner product

〈u,v〉 = u · v = uTv = u1, v1 + u2v2 + · · ·unvn

satisfies the four inner product conditions. This inner product is known as theEuclidean inner product on Cn. ♣

Example 6.23.

If U =

[

u1 u2

u3 u4

]

and V =

[

v1 v2v3 v4

]

are any 2×2 matrices with complex entries,

then the following formula defines a complex inner product on complex M2×2:

〈U, V 〉 = u1v1 + u2v2 + u3v3 + u4v4

For example, if U =

[

0 i1 1 + i

]

and V =

[

1 −i0 2i

]

, then

〈U, V 〉 = (0)(1) + i(−i) + (1)(0) + (1 + i)(2i)

= (0)(1) + i(i) + (1)(0) + (1 + i)(−2i)

= 0 + i2 + 0− 2i− 2i2 = 1− 2i ♣

Example 6.24.

If f(x) = f1(x) + if2(x) is a complex-valued function of the real variable x, andif f1(x) and f2(x) are continuous on [a, b], then we define

∫ b

a

f(x)dx =

∫ b

a

[f1(x) + if2(x)]dx =

∫ b

a

f1(x)dx+ i

∫ b

a

f2(x)dx

We can show that if the functions f = f1(x) + if2(x) and g = g1(x) + ig2(x) arevectors in complex C[a, b], then the following formula defines an inner producton complex C[a, b]:

〈f , g〉 =

∫ b

a

[f1(x) + if2(x)][g1(x) + ig2(x)]dx

=

∫ b

a

[f1(x) + if2(x)][g1(x)− ig2(x)]dx

=

∫ b

a

[f1(x)g1(x) + f2(x)g2(x)]dx+ i

∫ b

a

[f2(x)g1(x)− f1(x)g2(x)]dx ♣

Example 6.25.


In complex inner product spaces, as in real inner product spaces, the norm (orlength) of a vector u is defined by

‖u‖ = 〈u,u〉1/2

and the distance between two vectors u and v is defined by

d(u,v) = ‖u− v‖

If u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are vectors in Cn with theEuclidean inner product, then

‖u‖ = 〈u,u〉1/2 =√

|u1|2 + |u2|2 + · · ·+ |un|2

and

d(u,v) = ‖u− v‖ = 〈u− v,u− v〉1/2

=√

|u1 − v1|2 + |u2 − v2|2 + · · ·+ |un − vn|2

Example 6.26.

Unitary, Normal, and Hermitian Matrices

If A is a matrix with complex entries, then the conjugate transpose of A, denoted by A?,is defined by

A? = AT

where A is the matrix whose entries are the complex conjugates of the corresponding

entries in A and ATis the transpose of A. The conjugate transpose is also called the

Hermitian transpose.

If

A =

[

1 + i −i 02 3− 2i i

]

, then A =

[

1− i i 02 3 + 2i −i

]

so

A? = AT=

1− i 2−i 3 + 2i0 −i

♣

Example 6.27.


If A and B are matrices with complex entries and k is any complex number,then

(a) (A?)? = A

(b) (A+B)? = A? +B?

(c) (kA)? = kA?

(d) (AB)? = B?A?

Theorem 6.17 (Properties of the Conjugate Transpose).

A square matrix A with complex entries is called unitary if

A−1 = A?.

Definition 6.7.

If A is an n× n matrix with complex entries, then the following are equivalent.

(a) A is unitary.

(b) The row vectors of A form an orthonormal set in Cn with the Euclideaninner product.

(c) The column vectors of A form an orthonormal set in Cn with the Euclideaninner product.

Theorem 6.18 (Equivalent Statements).


The matrix

A =

1 + i

2

1 + i

21− i

2

−1 + i

2

(6.18)

has row vectors

r1 =

(

1 + i

2,1 + i

2

)

and r2 =

(

1− i

2,−1 + i

2

)

.

Relative to the Euclidean inner product on Cn, we have

‖r1‖ =

√

∣

∣

∣

∣

1 + i

2

∣

∣

∣

∣

2

+

∣

∣

∣

∣

1 + i

2

∣

∣

∣

∣

2

=

√

1

2+

1

2= 1

‖r2‖ =

√

∣

∣

∣

∣

1− i

2

∣

∣

∣

∣

2

+

∣

∣

∣

∣

−1 + i

2

∣

∣

∣

∣

2

=

√

1

2+

1

2= 1

and

r1 · r2 =

(

1 + i

2

)(

1− i

2

)

+

(

1 + i

2

)(−1 + i

2

)

=

(

1 + i

2

)(

1 + i

2

)

+

(

1 + i

2

)(−1− i

2

)

=i

2− i

2= 0

so the row vectors form an orthonormal set in C2. Thus A is unitary and

A−1 = A? =

1− i

2

1 + i

21− i

2

−1− i

2

(6.19)

We can verify that matrix (6.19) is the inverse of matrix (6.18) by showing thatAA? = A?A = I. ♣

Example 6.28.

A square matrix A with complex entries is called Hermitian if

A = A?.

Definition 6.8.


If

A =

1 i 1 + i−i −5 2− i1− i 2 + i 3

then

A =

1 −i 1− ii −5 2 + i

1 + i 2− i 3

,

so

A? = AT=

1 i 1 + i−i −5 2− i1− i 2 + i 3

= A

which mean that A is Hermitian. ♣

Example 6.29.

A square matrix A with complex entries is called normal if

AA? = A?A

Definition 6.9.

Every Hermitian matrix A is normal since

AA? = AA = A?A,

and every unitary matrix A is normal since

AA? = I = A?A. ♣

Example 6.30.

The eigenvalues of a Hermitian matrix are real numbers.

Theorem 6.19.

Let A =

[

2 1 + i1− i 3

]

. Find the eigenvalues A.

Example 6.31.

Solution


Exercise 6.4

1. Let u = (u1, u2) and v = (v1, v2). Show that 〈u,v〉 = 3u1v1 + 2u2v2 defines aninner product on C2?

2. Compute 〈u,v〉 using the inner product in Exercise 1.

(a) u = (2i,−i), v = (−i, 3i)

(b) u = (0, 0), v = (1− i, 7− 5i)

(c) u = (1 + i, 1− i), v = (1− i, 1 + i)

(d) u = (3i,−1 + 2i), v = (3i,−1 + 2i)

3. Compute 〈u,v〉 using the inner product defined by

〈u,v〉 = u1v1 + (1 + i) u1v2 + (1− i) u2v1 + 3u2v2.

(a) u = (2i,−i), v = (−i, 3i)

(b) u = (0, 0), v = (1− i, 7− 5i)

(c) u = (1 + i, 1− i), v = (1− i, 1 + i)

(d) u = (3i,−1 + 2i), v = (3i,−1 + 2i)

4. Use the inner product of Example 6.24 to find 〈U, V 〉 if

U =

[

−i 1 + i1− i i

]

and V =

[

3 −2− 3i4i 1

]

5. Let V be the vector space of complex-valued functions of the real variable x, andlet f = f1(x) + if2(x) and g = g1(x) + ig2(x) be vectors in V . Does

〈f , g〉 = [f1(0) + if2(0)][g1(0) + ig2(0)]

define an inner product on V ? If not, list all conditions that fail to hold.

6. Use the Euclidean inner product to find ‖w‖ if

(a) w = (−i, 3i) (b) w = (1− i, 1 + i)

(c) w = (0, 2− i) (d) w = (0, 0)

7. Use the inner product of Example 6.24 to find ‖A‖ if

(a) A =

[

−i 7i6i 2i

]

(b) A =

[

−1 1 + i1− i 3

]

8. Use the Euclidean inner product on C2 to find d(x,y) if

(a) x = (1, 1), y = (i,−i)

(b) x = (1− i, 3 + 2i), y = (1 + i, 3)

9. Let complex M2×2 have the inner product of Example 6.24. Find d(A,B) if


(a) A =

[

i 5i8i 3i

]

and B =

[

−5i 07i −3i

]

(b) A =

[

−1 1− i1 + i 2

]

and B =

[

2i 2− 3ii 1

]

10. In each part, find A?.

(a) A =

2i 1− i4 3 + i

5 + i 0

(b) A =

2i 1− i −1 + i4 5− 7i −ii 3 1

(c) A =[

7i 0 −3i]

(d) A =

[

a11 a12 a13a21 a22 a23

]

11. Find k, `, and m to make A a Hermitian matrix if A =

−1 k −i3− 5i 0 m

` 2 + 4i 2

.

12. Use Theorem 6.18 to determine which of the following are unitary matrices.

(a)

[

i 00 i

]

(b)

i√2

1√2

− i√2

1√2

(c)

[

1 + i 1 + i1− i −1 + i

]

(d)

− i√2

i√6

i√3

0 − i√6

i√3

i√2

i√6

i√3

13. In each part, verify that the matrix is unitary and find its inverse.

(a)

3

5

4

5i

−4

5

3

5i

(b)

1√2

1√2

−1 + i

2

1 + i

2

14. Find the eigenvalues of the following matrices.

(a) A =

[

4 1− i1 + i 5

]

(b) A =

[

6 2 + 2i2− 2i 4

]

(c)

5 0 00 −1 −1 + i0 −1− i 0

15. (a) Find a 2× 2 matrix that is both Hermitian and unitary and whose entries arenot all real numbers.

(b) What can you say about the inverse of a matrix that is both Hermitian andunitary?


Answer to Selected Exercise 6.4

2. (a) −12 (b) 0 (c) 2i (d) 37 3. (a) 4 + 5i (b) 0 (c) 4− 4i (d) 42

4. 9− 5i 5. No. Condition 4 fails. 6. (a)√10 (b) 2 (c)

√5 (d) 0

7. (a) 3√10 (b)

√14 8. (a) 2 (b) 2

√2 9. (a) 7

√2 (b) 2

√3

10. (a)

[

−2i 4 5− i1 + i 3− i 0

]

(b)

−2i 4 −i1 + i 5 + 7i 3−1 − i i 1

(c)

−7i03i

(d)

a11 a21a12 a22a13 a23

11. k = 3 + 5i, l = i, m = 2− 4i 12. (a), (b)

13. (a) A−1 =

[

3

5−4

5

−4

5i −3

5i

]

(b) A−1 =

[

1√2

−1+i2

1√2

1−i2

]

14. (a) λ = 3, 6 (b) λ = 2, 8 (c) λ = −2, 1, 5

15. (a)

[

0 i−i 0

]

is one possibility.

Chapter 6 Inner Product Spacesmathstat.sci.tu.ac.th/~archara/MA131/MA131-116/stnote131-116-ch… ·...

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