math 550 notes - chapter 10 chapter 10.pdf · title: math 550 notes - chapter 10 author: jesse...

Math 550 NotesChapter 10

Jesse Crawford

Department of MathematicsTarleton State University

Fall 2010

(Tarleton State University) Math 550 Chapter 10 Fall 2010 1 / 28

Notation

F denotes R or C.V is a finite-dimensional, nonzero vector space over F.I denotes all identity operators and all identity matrices.

Chapter 10 covers the trace and determinant and places a greateremphasis on matrices than the rest of the book.


Outline

1 Change of Basis

2 Trace

3 Determinant of an Operator

4 Determinant of a Matrix


Reminder: The Matrix of a Linear Map

ReminderSuppose U is a vector space with basis u1, . . . ,up, andV is a vector space with basis v1, . . . , vn.If T ∈ L(U,V ), the matrix of T wrt. these bases is denoted

M(T , (u1, . . . ,up), (v1, . . . , vn)).

If W is a third space with basis (w1, . . . ,wm), andS ∈ L(V ,W ), then ST ∈ L(U,W ), and

M(ST , (u1, . . . ,up), (w1, . . . ,wm))

=M(S, (v1, . . . , vn), (w1, . . . ,wm))M(T , (u1, . . . ,up), (v1, . . . , vn)).


Example

The matrix of the identity operator on F2

wrt. the bases ((4,2), (5,3)) and ((1,0), (0,1)) is

M(I, ((4,2), (5,3)), ((1,0), (0,1))) =(

4 52 3

)Proposition (10.2)

If (u1, . . . ,un) and (v1, . . . , vn) are bases of V ,thenM(I, (u1, . . . ,un), (v1, . . . , vn)) is invertible, and

M(I, (u1, . . . ,un), (v1, . . . , vn))−1 =M(I, (v1, . . . , vn), (u1, . . . ,un)).

Example

M(I, ((1,0), (0,1)), ((4,2), (5,3))) =(

4 52 3

)−1

=

( 32 −5

2−1 2

).


Change of Basis

Theorem (10.3)Suppose T ∈ L(V ), andlet (u1, . . . ,un) and (v1, . . . , vn) be bases of V .Let A =M(I, (u1, . . . ,un), (v1, . . . , vn)).Then,

M(T , (u1, . . . ,un)) = A−1M(T , (v1, . . . , vn))A.


Suppose the matrix of T ∈ L(F2) wrt. the basis ((1,0), (0,1)) is

M(T , ((1,0), (0,1))) =(

1 23 4

).

Then the matrix of T wrt. ((4,2), (5,3)) is

M(T , ((4,2), (5,3))) =(

4 52 3

)−1( 1 23 4

)(4 52 3

)

=

(−38 −5132 43

).

(1 23 4

)(10

)= 1

(10

)+ 3

(01

)(−38 −5132 43

)( 32−1

)= 1

( 32−1

)+ 3

(−5

22

)(Tarleton State University) Math 550 Chapter 10 Fall 2010 7 / 28

Rewriting Theorems in Terms of Matrices

Real Spectral Theorem (7.13)Suppose that V is a real inner-product space, andT ∈ L(V ).Then V has an orthonormal basis consisting of evec.’s of T iffT is self-adjoint.

Matrix VersionSuppose M is an n × n real matrix.Then there exists an n × n orthogonal matrix Rand an n × n diagonal matrix D, such that

D = RtMR,

if and only if M is symmetric.


Outline

1 Change of Basis

2 Trace




Revisiting the Characteristic Polynomial

Let V be a complex vector space.Recall that the characteristic polynomial of T ∈ L(V ) is

(z − λ1) · · · (z − λn),

where the λj ’s are the eval.’s of T , each repeated a number oftimes equal to its multiplicity.Expanding the characteristic polynomial yields

zn − (λ1 + · · ·+ λn)zn−1 + · · ·+ (−1)n(λ1 · · ·λn).

The negative of the coefficient of zn−1 in the characteristicpolynomial is called the trace of T , denoted trace T .trace T = λ1 + · · ·+ λn.


DefinitionSuppose V is a complex vector space, andlet T ∈ L(V ).Then trace T is the sum of the eigenvalues of T includingmultiplicities.

Observation: If a basis is chosen for V such thatM(T ) isupper-triangular, then trace T is the sum of the diagonal entries ofM(T ).Conjecture: this is always true regardless of the basis for V .

DefinitionSuppose A is a square matrix.Then trace A is the sum of the diagonal entries of A.


Proving the Conjecture

Proposition (10.9)If A and B are square matrices of the same size, then

trace (AB) = trace (BA).

Corollary (10.10)Suppose T ∈ L(V ).If (u1, . . . ,un) and (v1, . . . , vn) are bases of V , then

traceM(T , (u1, . . . ,un)) = traceM(T , (v1, . . . , vn)).

Theorem (10.11)If T ∈ L(V ), then trace T = traceM(T ).


Additional Facts About the Trace

Corollary (10.12)If S,T ∈ L(V ), then

trace (ST ) = trace (TS), and

trace (S + T ) = trace S + trace T .

Corollary (10.13)There do not exist operators S,T ∈ L(V ) such that

ST − TS = I.


Outline

1 Change of Basis

2 Trace




Determinant of an Operator

Let V be a complex vector space, T ∈ L(V ),and let λ1, . . . , λn be the eval.’s of T ,each repeated a number of times equal to its multiplicity.Expanding the characteristic polynomial yields

zn − (λ1 + · · ·+ λn)zn−1 + · · ·+ (−1)n(λ1 · · ·λn),

where the λj ’s are the eval.’s of T , each repeated a number oftimes equal to its multiplicity.

DefinitionIf T ∈ L(V ), the determinant of T , denoted det Tis (−1)dim V times the constant term in the characteristicpolynomial of T .If V is a complex vector space, det T = λ1 · · ·λn,the product of the eval.’s of T , including multiplicities.


Example

Let T ∈ L(C3) be the operator whose matrix is 3 −1 −23 2 −31 2 0

.

The eval.’s of this operator are 1, 2 + 3i , and 2− 3i , sodet T = 1(2 + 3i)(2− 3i) = 13.

PropositionSuppose V is a complex vector space, and T ∈ L(V ).If a basis is chosen for V wrt. which T has an upper triangularmatrix,then det T is the product of the diagonal entries of this matrix.


Proposition (10.14)An operator is invertible if and only if its determinant is nonzero.

PropositionIf T ∈ L(V ), and λ, z ∈ F, thenλ is an eigenvalue of T if and only ifz − λ is an eigenvalue of zI − T .The multiplicities are the same.

Theorem (10.17)Suppose T ∈ L(V ).The characteristic polynomial of T is det(zI − T ).


Outline

1 Change of Basis

2 Trace




Goal: define determinants for matrices so thatdet T = detM(T ), regardless of the basis chosen.Would be nice if the determinant were just the product of thediagonal entries, but this is not true.


Permutations

DefinitionA permutation of the list (1, . . . ,n) is a list(m1, . . . ,mn) that contains each of the numbers 1, . . . ,n exactlyonce.perm n = the set of all such permutations.

DefinitionSuppose (m1, . . . ,mn) ∈ perm n.Count the number of pairs (j , k), such that

I 1 ≤ j < k ≤ n, andI j appears after k in the list (m1, . . . ,mn).

The sign of (m1, . . . ,mn) is defined to be 1 if this number is evenand −1 if it is odd.


Example(2,6,1,3,4,5,7) ∈ perm 7.The pairs (j , k), such that

I 1 ≤ j < k ≤ n, andI j appears after k in the list (m1, . . . ,mn).

are(2,1), (6,1), (6,3), (6,4), (6,5).

The sign of this permutation is (−1)5 = −1.Alternatively, we can count the number of 2-elementtranspositions that transform (1, . . . ,7) into (2,6,1,3,4,5,7).

(1,2,3,4,5,6,7) → (2,1,3,4,5,6,7) → (2,1,3,4,6,5,7)→ (2,1,3,6,4,5,7) → (2,1,6,3,4,5,7)

(2,6,1,3,4,5,7) is an odd permutation.


Lemma (10.23)Interchanging two entries in a permutation multiplies the sign by −1.


The Determinant of a Matrix

DefinitionSuppose A is an n × n matrix whose entries are denoted by ai,j .Then the determinant of A, denoted det A, is

det A =∑

(m1,...,mn)∈perm n

sign(m1, . . . ,mn)am1,1 · · · amn,n.

ExampleThe determinant of a 2× 2 matrix A is

a11a22 − a21a12.


PropositionIf A is an upper-triangular square matrix, the determinant of A is theproduct of the diagonal entries of A.

Lemma (10.28)Suppose A is a square matrix, andB is obtained from A by interchanging two columns.Then det A = −det B.

Lemma (10.29)If A is a square matrix that has two equal columns, then det A = 0.


det is an Alternating Form

Notation:If A is an n × n square matrix,A = [a1 · · · an] means that the columns of A are a1, . . . ,an.

Lemma (10.30)Suppose A = [a1 · · · an] is an n × n matrix, and(m1, . . . ,mn) ∈ perm n.Then

det[am1 · · · amn ] = (sign(m1, . . . ,mn))det A.


det is a Multilinear Form

PropositionThe mapping

(Fn)n → F(a1, . . . ,an) 7→ det[a1 · · · an]

is multilinear, that is, linear in each component.

ExampleNote that (

24

)= 2

(10

)+ 4

(01

).

Therefore,

det(

7 23 4

)= 2 det

(7 13 0

)+ 4 det

(7 03 1

)(Tarleton State University) Math 550 Chapter 10 Fall 2010 26 / 28

The Key Result

Theorem (10.31)If A and B are square matrices of the same size, then

det(AB) = det(BA) = (det A)(det B).


Smooth Sailing

Corollary (10.32)Suppose T ∈ L(V ), and(u1, . . . ,un) and (v1, . . . , vn) are bases of V .Then

detM(T , (u1, . . . ,un)) = detM(T , (v1, . . . , vn)).

Theorem (10.33)If T ∈ L(V ), then

det T = detM(T ).

Corollary (10.34)If S,T ∈ L(V ), then det(ST ) = det(TS) = (det S)(det T ).


math 550 notes - chapter 10 chapter 10.pdf · title: math 550 notes - chapter 10 author: jesse...

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