Hassani S. Mathematical Methods: For Students of Physics and Related Fields

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224 Finite-Dimensional Vector Spaces

7.4 Eigenvectors and Eigenvalues

One of the most important applications of the determinant is in ﬁnding cer-

tain vectors that are not aﬀected by transformations. As an example, consider

rotation which is a linear transformation of space onto itself (or a transfor-

mation from R

to R

). A general rotation in space is very complicated (see

Example 6.2.5 and the discussion immediately preceding it), but if we can

ﬁnd an axis which is unaﬀected by the operation, then the process becomes a

simple rotation about this axis.

When we say that a vector is unaﬀected, we mean that its direction (and

not necessarily its magnitude) is unchanged. We use n ×n matrices to repre-

sent transformations of R

.Ifx is a (column) vector in R

whose direction is

not aﬀected by the transformation T,thenwecanwrite

Tx= λx or (T −λ1)x = 0, (7.17)

where λ is a real number and we introduced the unit matrix to give meaning to

the subtraction of λ from T. In Equation (7.17), x is called the eigenvector

eigenvector,

eigenvalue, and

eigenvalue

equation

and λ the eigenvalue of the linear transformation. Since the zero vector triv-

ially satisﬁes (7.17), we demand that eigenvectors always be nonzero.Equation

(7.17) itself is called an eigenvalue equation; its solution involves calculat-

ing both the eigenvalues and the eigenvectors. It is clear from (7.17) that a

multiple of an eigenvector is also an eigenvector (see Problem 7.6). There-

fore, an eigenvalue equation (7.17) has no unique solution. By convention, we

normalize eigenvectors so that their length is unity.

To ﬁnd the solution to (7.17), we note that the matrix (T −λ1)musthave

no inverse, because if it did, then we could multiply both sides of the equation

by (T − λ1)

−1

and obtain

(T − λ1)

−1

(T −λ1)



 

x =(T −λ1)

−1



 

⇒ x = 0

which is not an acceptable solution. So, we must demand that the matrix

a necessary

condition for an

eigenvalue

equation to have

nontrivial

solutions is that

the determinant of

T − λ1bezero.

(T − λ1) have no inverse. This will happen only if the determinant of this

matrix vanishes. So, the problem is reduced to ﬁnding those λ’s which make

the determinant of the matrix vanish. In other words, the eigenvalues are the

solutions of the equation

det(T − λ1)=0. (7.18)

Once the eigenvalues are determined, we substitute them one by one in the

matrix equation (7.17) and ﬁnd the corresponding eigenvectors by solving the

resulting n linear equations in n unknowns. The best way to explain this is

through an example.

Example 7.4.1.

Let T be a linear transformation of space (or R

) represented by

the matrix

T =

⎛

⎝

100

012

021

⎞

⎠

7.4 Eigenvectors and Eigenvalues 225

The eigenvalue equation is

(T − λ1)x = 0 or

⎡

⎣

⎛

⎝

100

012

021

⎞

⎠

− λ

⎛

⎝

100

010

001

⎞

⎠

⎤

⎦

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

This can also be written as

⎛

⎝

1 − λ 00

01−λ 2

021− λ

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

(7.19)

whose nontrivial solution is obtained by setting the determinant of the matrix equal

to zero:

det

⎛

⎝

1 − λ 00

01−λ 2

021− λ

⎞

⎠

(1 − λ)det



1 − λ 2

21−λ



=(1− λ)

(1 − λ)

− 4

=0.

This equation has the solutions

1 − λ =0 or (1− λ)

=4 ⇒ 1 − λ = ±2.

It follows that there are three eigenvalues: λ

=1,λ

= −1, and λ

=3. Wenow

ﬁnd the eigenvectors corresponding to each eigenvalue.

Substituting λ

=1forλ in Equation (7.19) yields

⎛

⎝

000

002

020

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

It follows that x

=0=x

. Therefore, the ﬁrst eigenvector is

⎛

⎝

⎞

⎠

= x

⎛

⎝

⎞

⎠

with x

an arbitrary real number. This arbitrariness comes from the fact that a

multiple of an eigenvector is also an eigenvector. We choose x

= 1 to normalize the

eigenvector to unit length. Denoting this eigenvector by e

,wehave

⎛

⎝

⎞

⎠

To ﬁnd the second eigenvector, we substitute λ

= −1forλ in Equation (7.19).

This gives

⎛

⎝

200

022

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

+2x

⎞

⎠

⎛

⎝

⎞

⎠

226 Finite-Dimensional Vector Spaces

It follows that x

=0,and2x

+2x

=0orx

= −x

. Therefore, the second

eigenvector is

⎛

⎝

−x

⎞

⎠

= x

⎛

⎝

−1

⎞

⎠

with x

arbitrary. To normalize the eigenvector, we divide it by its length.

This

amounts to choosing x

=1/

√

2 (see Problem 7.7). We thus have

√

⎛

⎝

−1

⎞

⎠

For the third eigenvector, we substitute λ

= 3 in Equation (7.19) to obtain

⎛

⎝

−20 0

0 −22

02−2

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

−2x

+2x

− 2x

⎞

⎠

⎛

⎝

⎞

⎠

or x

=0,andx

= x

. Therefore, the third eigenvector is

⎛

⎝

⎞

⎠

= x

⎛

⎝

⎞

⎠

with x

arbitrary. To normalize the eigenvector, we divide it by its length and get

√

⎛

⎝

⎞

⎠



The unit eigenvectors

,and

of the preceding example are mutually

perpendicular as the reader may easily verify. This is no accident! The matrix

of that example happens to be symmetric, and for such matrices, we have the

following general property:

Box 7.4.1. Eigenvectors of a symmetric matrix corresponding to diﬀerent

eigenvalues are orthogonal.

To show this, let x and y be eigenvectors of a symmetric matrix T correspond-

ing to eigenvalues λ and λ



, respectively:

Tx = λx, Ty = λ



Multiply both sides of the ﬁrst equation by

y and the second by

x to get

yTx= λ

yx,

xTy = λ



xy. (7.20)

Here we are assuming that the inner product for the calculation of length is the usual

Euclidean one.

7.5 Orthogonal Polynomials 227

Now take the transpose of both sides of the ﬁrst equation in (7.20). This

gives

(

yTx)

= λ(

yx)

⇒

y = λ

But double transposing y gives back y.Furthermore,

T = T, because T is

symmetric. So,

xTy = λ

xy.

Subtracting both sides of this equation from those of the second equation in

(7.20), we obtain

0=(λ −λ



)

xy.

By assumption, λ = λ



;so,wemusthave

xy = 0, i.e., that x and y are

orthogonal.

7.5 Orthogonal Polynomials

The last section generalized the two- and three-dimensional “arrows” and

polynomials to higher dimensions in which many of the original properties of

vectors—such as the inner product—were retained. In this section, we want to

make two more generalizations which are necessary for many physical applica-

tions. The ﬁrst is the introduction of a weight function in the deﬁnition of

weight function

inner product. A weight function is a function that is positive deﬁnite

in the

interval (a, b) of integration of the inner product. More speciﬁcally, let p =

p(t)andq = q(t) be polynomials in P

[t]. We deﬁne their inner product as

p · q =

p(t)q(t)w(t) dt, (7.21)

where w(t) is a function that is never zero or negative for a<t<b,andits

form is usually dictated by the physical application. The reader may verify

that Equation (7.21) deﬁnes a positive deﬁnite inner product.

The second generalization is to consider the collection of all polynomials

of arbitrary degree. In other words, instead of conﬁning ourselves to P

[t]for

some ﬁxed n, we shall allow all polynomials without any restriction on their

degree. Clearly, such a collection is indeed a vector space; however it does

not have a ﬁnite basis. We denote this inﬁnite-dimensional space by P

(a,b)

[t],

in which notation both the weight function and the interval of integration are

included.

Given any basis for P

(a,b)

[t], we can apply the Gram–Schmidt process on it

to turn it into an orthonormal basis. Due to historical reasons, the normality

is not a desirable property for the basis vectors. So, one seeks polynomials

that are orthogonal, but not necessarily of unit length. Instead of normalizing

tradition and

history

“standardize”

polynomials

instead of “nor-

malizing” them.

the vectors, one standardizes them. Standardization is a rule—dictated by

tradition—that ﬁxes some of the coeﬃcients of the polynomials. The proce-

dure for ﬁnding these orthogonal polynomials is to start from the constant

Recall that8and

mean the same thing.

This just means that the function is positive and never zero.

228 Finite-Dimensional Vector Spaces

polynomial (of degree zero) and standardize it to get the ﬁrst polynomial.

Next apply the standardization to the polynomial of degree one (with two

unknown coeﬃcients), and make sure that it is perpendicular to the ﬁrst

polynomial, where the inner product is deﬁned by (7.21). These two require-

ments (standardization and perpendicularity) provide two equations and two

unknowns which can be solved to ﬁnd the coeﬃcients of the second polyno-

mial. The next polynomial has degree two with three unknown coeﬃcients.

Standardization and orthogonality to the ﬁrst two polynomials provide three

equations in three unknowns, the solution of which equations determines the

third polynomial. This process can be continued indeﬁnitely determining the

coeﬃcients of orthogonal polynomials up to any desired degree.

Example 7.5.1.

The procedure above is best illustrated by a concrete example.

The Legendre polynomial of degree n,denotedbyP

(t), is characterized byLegendre

polynomial the standardization P

(1) = 1. We denote the collection of these polynomials by

(−1,1)

[t], indicating that the interval of integration for them is from −1to+1

and that the weight function is unity. Because of standardization, we must choose

(t) = 1. The ﬁrst degree polynomial is generally written as P

(t)=α

+ α

Standardization gives α

+ α

= 1. Orthogonality to P

(t)gives

−1

(t)P

(t)w(t) dt =

−1

1 · (α

+ α

t) · 1 dt =2α

So, α

=0andα

= 1. Therefore, P

(t)=t.

For P

(t)=α

+ α

t + α

we have (reader please verify!)

+ α

= 1 (by standardization),

2α

+0·α

= 0 (by orthogonality to P

0 · α

· α

+0·α

= 0 (by orthogonality to P

The solution to these equations is α

= −

, α

=0,andα

= −

,sothatP

(t)=

(3t

− 1). Other Legendre polynomials can be found analogously. 

By their very construction, orthogonal polynomials, which are denoted by

(t), satisfy the following orthogonality condition:orthogonal

polynomials

deﬁned

(t)F

(t)w(t) dt =

0ifm = n,

if m = n,

(7.22)

where h

is just a positive number (depending on n,ofcourse)whichis

diﬀerent for diﬀerent types of F

As before, let us treat these polynomials

as vectors and write F

for F

(t). Then using the Kronecker delta of (7.9),

Equation (7.22) can be written as

·F

0ifm = n

if m = n

= h

There are many diﬀerent types of orthogonal polynomials, distinguished from each other

by diﬀerent intervals, and diﬀerent w(t). Diﬀerent symbols—such as P

(t), H

(t), T

(t),

etc., are used for diﬀerent types. We have used F

(t) to represent any one of these types

in our general discussion.

7.5 Orthogonal Polynomials 229

In particular, F

·F

= h

or |F

= h

. So, the “length” of F

√

Now consider the set of all functions deﬁned in the interval (a, b)anytwo

the set of all

functions (not just

polynomials) is

also a vector

space.

of which give a ﬁnite result when integrated as in Equation (7.22). The reader

may easily verify that this set is indeed a vector space.Iff = f(t)andg = g(t)

are two vectors in this space, then we deﬁne their inner product as

f ·g =

f(t)g(t)w(t) dt. (7.23)

It is clear that the F

belong to this space. Furthermore, it can be shown

that they form a convenient basis for the vector space. In fact, any function of

the space can be written as a (inﬁnite) linear combination of the orthogonal

polynomials

f =

∞



n=0

whose coeﬃcients can be determined by taking the inner product of both sides

with F

f · F



∞



n=0



·F

∞



n=0

·F

= a

·F

= h

because in the last inﬁnite sum all the terms are zero except one. We can

solve this equation for a

to obtain a

= f · F

.Thus,

f =

∞



n=0

where a

f ·F

. (7.24)

In terms of functions and polynomials, we have the important result:

expansion of

functions in terms

of orthogonal

polynomials

Theorem 7.5.2. Afunctionf(t), deﬁned in the interval (a, b), can be repre-

sented as an inﬁnite sum in orthogonal polynomials given by

f(t)=

∞



n=0

(t), where a

f(t)F

(t)w(t) dt. (7.25)

There are a number of so-called classical orthogonal polynomials used

classical

orthogonal

polynomials

in mathematical physics a number of whose properties we simply cite here.

We have already mentioned Legendre polynomials for which the interval is

(−1, +1) and w(x)=1.

For Legendre polynomials, h

=2/(2n + 2), i.e.,

−1

(t)P

(t) dt =

⎧

⎪

⎨

⎪

⎩

0ifm = n

2n +1

if m = n

2n +1

. (7.26)

If the interval is (−∞, ∞)andw(t)=e

−t

, then the resulting polynomials,

denoted by H

(t), are called Hermite polynomials. For Hermite polynomi- Hermite

polynomials

als, we have

A detailed discussion of Legendre polynomials and their origin can be found in Chapter

26.

230 Finite-Dimensional Vector Spaces

∞

−∞

(t)H

(t)e

−t

dt =

⎧

⎪

⎨

⎪

⎩

0ifm = n

√

π 2

n!ifm = n

√

π 2

n! δ

. (7.27)

If the interval is (0, ∞)andw(t)=t

−t

with m a positive integer,

then the resulting polynomials, denoted by L

(t), are called Laguerre poly-

nomials. For Laguerre polynomials, we have

Laguerre

polynomials

∞

(t)L

(t)t

−t

dt =

0ifk = n

√

π (n + m)!/n!ifk = n

√

(n + m)!

. (7.28)

There are other (classical) orthogonal polynomials which we shall not inves-

tigate here.

7.6 Systems of Linear Equations

Our discussion of determinants in Section 6.3 started with a system of two

linear equations in two unknowns and led to the result that if the determinant

of the matrix of coeﬃcients is nonzero, then the inverse of this matrix exists,

and the unknowns can be found conveniently using this inverse [see Equation

(6.52) and Theorem 6.3.1]. This was further generalized to the case of three

linear equations in three unknowns and stated in Equation (6.59). A system

of n linear equations in n unknowns can be handled in the same way. We

write such a system as

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

... a

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⇒ x = Ab (7.29)

and note that, if det A = 0, we can calculate A

−1

according to Box 7.3.1,

and multiply both sides of (7.29) by this inverse and obtain x = A

−1

b.The

case of the vanishing determinant is best treated in the context of a system

of equations for which the number of unknowns is not equal to the number of

equations.

The process that led to Equations (6.49) and (6.51) is called elimination,

and can be extended to m linear equations in n unknowns of the form

m linear equations

in n unknowns

Actually m need not be an integer. However, the space and scope of this book does

not permit us to consider the general case.

The interested reader may ﬁnd Hassani, S. Mathematical Physics: A Modern Intro-

duction to Its Foundations, Springer-Verlag, 1999, Chapter 7, a useful reference for all

orthogonal polynomials including many derivations and proofs that we have skipped here.

7.6 Systems of Linear Equations 231

+ a

+ ···+ a

= b

+ a

+ ···+ a

= b

. (7.30)

+ a

+ ···+ a

= b

We will now describe a general process known as Gauss elimination,

Gauss elimination

for ﬁnding all solutions of the given system of linear equations. The idea

is to replace the given system by a simpler system, which is equivalent to

the original system in the sense that it has precisely the same solutions. For

example, the degenerate equation

0 · x

+0·x

+ ···+0· x

= b

is equivalent to 0 = b

, which cannot be satisﬁed unless b

is zero.

In a more compact notation, we write only the ith equation, indicating

its form by a sample term a

and the statement that the equation is to be

summed over j from 1 to n by writing



j=1

= b

for i =1, 2,...,m. (7.31)

We distinguish two cases:

1. Every a

= 0, i.e., all coeﬃcients of the unknown x

vanish. Then, triv-

ially, the system (7.31) is equivalent to a smaller system of m equations

in the n − 1 unknowns x

,...,x

with x

arbitrary for any solution of

the smaller system.

2. Some a

= 0. By interchanging the ﬁrst equation with another if nec-

essary, we get an equivalent system with a

= 0. Dividing the ﬁrst

equation by a

, we then get an equivalent system in which a

=1.

Then subtracting a

times the new ﬁrst equation from each ith equa-

tion for i =2,...,m, we get an equivalent system of the form

+ a



+ a



+ ···+ a



= b



+ a



+ ···+ a



= b



. (7.32)



+ a



+ ···+ a



= b



Now we apply the same procedure to the system of equations in (7.32) in-

volving only x

through x

so that x

will appear only in the ﬁrst of these

equations. If case 2 always arises, the given system is said to be compatible.

compatible and

incompatible

systems of linear

equations

If case 1 arises once in a while, then we may get degenerate equations of the

form 0 = d

.Ifalld

turn out to be zero, these can be ignored; if one d

=0,

the original system (7.30) is incompatible (has no solutions). We summarize

these ﬁndings as

The reader may ﬁnd an adequate discussion of summations and “dummy” indices in

Section 9.2.

232 Finite-Dimensional Vector Spaces

Theorem 7.6.1. Any system (7.30) of m linear equations in n unknowns can

be reduced to an equivalent system of r linear equations whose ith equation has

the form

+ c

i,i+1

i+1

+ c

i,i+2

i+2

+ ···+ c

= d

(7.33)

plus m −r equations of the form 0=d

Written out in full, Equation (7.33) looks like

+ c

+ ···+ c

= d

+ c

···+ c

= d

+ c

···+ c

= d

, (7.34)

+ ···+ c

= d

(r ≤ m),

whichissaidtobeinechelon form.

echelon form of a

system of linear

equations

Solutions of any system of the echelon form (7.34) are easily described.

Consider the succession of the unknowns starting with x

and going down to

.Ifagivenx

appears as the ﬁrst variable in an equation of (7.34), then it

can be written in terms of all preceding unknowns:

= d

− c

i,i+1

i+1

− c

i,i+2

i+2

−···−c

. (7.35)

If x

does not appear as the ﬁrst variable in an equation of (7.34), then it can

be chosen arbitrarily. We thus have

Box 7.6.1. In the compatible case of Theorem 7.6.1, the set of all solu-

tions of Equation (7.30) are determined as follows. The m −r unknowns

not occurring in (7.34) can be chosen arbitrarily (they are free param-

eters). For any choice of these x

’s, the remaining x

can be computed by

substituting in (7.35).

Example 7.6.2. Consider the following four linear equations in three unknowns

(so m =4andn =3):

−x

+2x

=1,

+ x

− 3x

=0,

−x

+ x

= −2, (7.36)

+2x

− x

= −1.

The coeﬃcient of x

in the ﬁrst equation is zero. So, we switch this equation with

one of the other equations, say the second. Then we multiply the new ﬁrst equation

by the negative of the coeﬃcient of x

in each remaining equation and add the result

If r = n, then the last equation of (7.34) will be x

= d

, and (if the set of equations

is compatible) all unknowns will be determined.

7.6 Systems of Linear Equations 233

to that equation to eliminate x

. Thus, we add the new ﬁrst equation to the third

equation of (7.36), and subtract the new ﬁrst equation from the last equation of

(7.36). The result is

+ x

− 3x

=0,

−x

+2x

=1,

− 2x

= −2, (7.37)

+2x

= −1.

To eliminate x

from the last two equations, multiply the second equation of (7.37)

by 2 (or 1 for the last) and add it to the third (or last) equation. This will yield

+ x

− 3x

=0,

−x

+2x

=1,

=0, (7.38)

=0.

Multiply the second equation by −1, divide the third equation in (7.38) by 4, and

ﬁnally subtract the result from the last equation. The ﬁnal result is the following

echelon form:

+ x

− 3x

=0,

− 2x

= −1,

=0, (7.39)

0=0,

which corresponds to Equation (7.34) with r = n = 3. Thus, we have one equation

of the form 0 = d

for which d

is zero. So, the system has a solution. To ﬁnd

this solution, start with the third equation of (7.39) which gives x

= 0. Substitute

in the equation above it to get x

= −1, and these values in the ﬁrst equation to

obtain x

=1.



Example 7.6.3. As another example, consider the following:

+ x

=0,

− x

+ x

= −2,

−x

+2x

+ x

= −1, (7.40)

− 2x

+ x

=2.

Multiply the ﬁrst equation successively by −2, 1, and −1 and add it to the second,

third and fourth equations. The result will be

+ x

=0,

−3x

− x

= −2,

+2x

= −1,

−3x

+0· x

=2.