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Solution Manual for Physical Mathematics, 2nd Edition

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Solution Manual for Physical Mathematics, 2nd Edition Solutions to the Exercises 1 Solutions to the Exercises on Linear Algebra 1. What is the most general function of three Grassmann numbers θ1, θ2, θ3? Solution: Grassmann numbers anticommute, that is {θi,θj }= θiθj +θj θi = 0 and θ 2 = θ 2 = θ 2 = 0. So the most general function of three Grassmann numbers is f (θ1, θ2, θ3) = a + b θ1 + c θ2 + d θ3 + e θ1θ2 + f θ1θ3 + g θ2θ3 + h θ1θ2θ3. 2. Derive the cyclicity (1.24) of the trace from Eq.(1.23). Solution: Since Tr(AB) = Tr(B A), it follows with B replaced by BC D that Tr(ABC D) = Tr(DABC) = Tr(CD AB) = Tr(BC DA). 3. Show that (AB) T = B T A T , which is Eq.(1.26). Solution: With a sum over the repeated index 4 understood, (AB) T ik = [(AB)]ki = Ak4 B4i = A T 4k B T i4 = B T i4 A T 4k = B T A T . 4. Show that a real hermitian matrix is symmetric. Solution: Aik = A † = A ∗ ki = Aki. 5. Show that (AB) † = B † A † , which is Eq.(1.29). Solution: With a sum over the repeated index 4 understood, (AB) † ik = (AB)∗T = (AB)∗ ki = Ak ∗ 4 B4 ∗ i = A ∗ 4 T Bi ∗T = Bi ∗T A4 ∗T = B †A † . 6. Show that the matrix (1.41) is positive on the space of all real 2-vectors but not on the space of all complex 2-vectors. 1 2 Solutions to the Exercises ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟θ = = = 0. ⎜0 0 0 −1 0⎟⎜ 0 0 0 0 0 0 −1⎟ ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎠ −1 1 b −1 1 b 0 0 0 0 0 0 0 0 i i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Solution: a b 1 1 a = a 2 + b 2 is positive if a and b are both real (and nonzero), but a¯ b¯ 1 1 a = a 2 + b 2 + a¯b − b¯a is complex when the imaginary part of a¯b ∗= 0.

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Solution Manual for Physical
Mathematics, 2nd Edition by Kevin
Cahill

,Solutions to the Exercises




1 Solutions to the Exercises on Linear Algebra

1. What is the most general function of three Grassmann numbers θ1, θ2, θ3?
Solution: Grassmann numbers anticommute, that is {θi , θ j } = θi θ j +θ j θi = 0
and θ12 = θ22 = θ3 2 = 0. So the most general function of three Grassmann
numbers is

f (θ1, θ2, θ3) = a + b θ1 + c θ2 + d θ3 + e θ1θ2 + f θ1θ3 + g θ2θ3 + h θ1θ2θ3.
2. Derive the cyclicity (1.24) of the trace from Eq.(1.23).
Solution: Since Tr( AB) = Tr(B A), it follows with B replaced by BC D that
Tr( ABC D) = Tr(DA BC) = Tr(CD AB) = Tr(BC DA).
3. Show that ( AB) T = BT AT, which is Eq.(1.26).
Solution: With a sum over the repeated index 4 understood,

( AB) T = [( AB)]ki = Ak4 B4i = AT4k B T i4 = B T i4 AT4k = BT AT .
ik ik
4. Show that a real hermitian matrix is symmetric.
Solution:

Aik = Aik = A∗ki = Aki .

5. Show that ( AB)† = B† A†, which is Eq.(1.29).
Solution: With a sum over the repeated index 4 understood,

( AB)† ik = ( A B)∗T = ( A B)∗ ki = A∗k 4 B4∗i
ik

= A∗4kT Bi∗T ∗T ∗T
4 = Bi 4 A 4 k = B A
† † .
ik
6. Show that the matrix (1.41) is positive on the space of all real 2-vectors but
not on the space of all complex 2-vectors.

1

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