Math 550 Notes Chapter 10 Jesse Crawford Department of Mathematics Tarleton State University Fall 2010 (Tarleton State University) Math 550 Chapter 10 Fall 2010 1 / 28
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.
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Outline
1 Change of Basis
2 Trace
3 Determinant of an Operator
4 Determinant of a Matrix
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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)).
(Tarleton State University) Math 550 Chapter 10 Fall 2010 4 / 28
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
).
(Tarleton State University) Math 550 Chapter 10 Fall 2010 5 / 28
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.
(Tarleton State University) Math 550 Chapter 10 Fall 2010 6 / 28
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.
(Tarleton State University) Math 550 Chapter 10 Fall 2010 8 / 28
Outline
1 Change of Basis
2 Trace
3 Determinant of an Operator
4 Determinant of a Matrix
(Tarleton State University) Math 550 Chapter 10 Fall 2010 9 / 28
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.
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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.
(Tarleton State University) Math 550 Chapter 10 Fall 2010 11 / 28
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 ).
(Tarleton State University) Math 550 Chapter 10 Fall 2010 12 / 28
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.
(Tarleton State University) Math 550 Chapter 10 Fall 2010 13 / 28
Outline
1 Change of Basis
2 Trace
3 Determinant of an Operator
4 Determinant of a Matrix
(Tarleton State University) Math 550 Chapter 10 Fall 2010 14 / 28
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.
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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.
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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 ).
(Tarleton State University) Math 550 Chapter 10 Fall 2010 17 / 28
Outline
1 Change of Basis
2 Trace
3 Determinant of an Operator
4 Determinant of a Matrix
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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.
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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.
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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.
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Lemma (10.23)Interchanging two entries in a permutation multiplies the sign by −1.
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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.
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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.
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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.
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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
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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).
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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 ).
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