Chapter 25

Integration and Matrices

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Problem 25.1.1

Consider the 2×22 \times 2 matrix over R\mathbb {R}

F(s)=(cosh⁡(s)sinh⁡(s)sinh⁡(s)cosh⁡(s)). F(s) = \begin{pmatrix} \cosh (s) & \sinh (s) \\ \sinh (s) & \cosh (s) \end{pmatrix}.

Find

G(t)=∫0tF(s) ds. G(t) = \int _{0}^{t} F(s) \, ds.

Then calculate the commutator [F(s),G(t)][F(s), G(t)].

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Problem 25.1.2

Consider the analytic function f:C→Cf : \mathbb {C} \rightarrow \mathbb {C}, f(z)=z2f(z) = z^{2} and the 2×22 \times 2 matrix

A=(0110). A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Calculate f(A)f(A) applying the Cauchy integral theorem.

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Problem 25.1.3

Let AA be an n×nn \times n matrix over C\mathbb {C}. If λ\lambda is not an eigenvalue of AA, then the matrix (A−λIn)(A - \lambda I_{n}) has an inverse, namely the resolvent

Rλ=(A−λIn)−1. R_{\lambda } = (A - \lambda I_{n})^{-1}.

Let λj\lambda_{j} be the eigenvalues of AA. For ∣λ∣≥a\left|\lambda \right| \geq a, where aa is any positive constant greater than all the ∣λj∣\left|\lambda_{j}\right| the resolvent can be expanded as

Rλ=−1λ(In+1λA+1λ2A2+⋯ ). R_{\lambda } = -\frac{1}{\lambda }\left(I_{n} + \frac{1}{\lambda }A + \frac{1}{\lambda ^{2}}A^{2} + \cdots \right).

Calculate

−12πi∮∣λ∣=aλmRλ dλ,m=0,1,2,…. -\frac{1}{2\pi i}\oint _{\left|\lambda \right|=a} \lambda ^{m} R_{\lambda } \, d\lambda , \qquad m = 0, 1, 2, \ldots .
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Problem 25.1.4

Let Cn×N\mathbb {C}^{n \times N} be the vector space of all n×Nn \times N complex matrices. Let Z∈Cn×NZ \in \mathbb {C}^{n \times N}. Then Z∗≡Z‾TZ^{*} \equiv \overline{Z}^{T}, where TT denotes transpose. One defines a Gaussian measure μ\mu on Cn×N\mathbb {C}^{n \times N} by

dμ(Z):=1πnNexp⁡(−tr⁡(ZZ∗)) dZ d\mu (Z) := \frac{1}{\pi ^{nN}} \exp (-\operatorname {tr}(ZZ^{*})) \, dZ

where dZdZ denotes the Lebesgue measure on Cn×N\mathbb {C}^{n \times N}. The Fock space F(Cn×N)\mathcal{F}(\mathbb {C}^{n \times N}) consists of all entire functions on Cn×N\mathbb {C}^{n \times N} which are square integrable with respect to the Gaussian measure dμ(Z)d\mu (Z). With the scalar product

⟨f,g⟩:=∫Cn×Nf(Z)g(Z)‾ dμ(Z),f,g∈F(Cn×N) \left\langle f, g \right\rangle := \int _{\mathbb {C}^{n \times N}} f(Z)\overline{g(Z)} \, d\mu (Z), \qquad f, g \in \mathcal{F}(\mathbb {C}^{n \times N})

one has a Hilbert space. Show that this Hilbert space has a reproducing kernel KK. This means a continuous function K(Z,Z′):Cn×N×Cn×N→CK(Z, Z') : \mathbb {C}^{n \times N} \times \mathbb {C}^{n \times N} \rightarrow \mathbb {C} such that

f(Z)=∫Cn×NK(Z,Z′)f(Z′) dμ(Z′) f(Z) = \int _{\mathbb {C}^{n \times N}} K(Z, Z') f(Z') \, d\mu (Z')

for all Z∈Cn×NZ \in \mathbb {C}^{n \times N} and f∈F(Cn×N)f \in \mathcal{F}(\mathbb {C}^{n \times N}).

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Problem 25.1.5

Let AA be an n×nn \times n positive definite matrix over R\mathbb {R}, i.e.x→TAx→>0\overrightarrow {x}^{T}A\overrightarrow {x} > 0 for all x→∈Rn\overrightarrow {x} \in \mathbb {R}^{n}. Calculate

∫Rnexp⁡(−x→TAx→) dx→. \int _{\mathbb {R}^{n}} \exp (-\overrightarrow {x}^{T}A\overrightarrow {x}) \, d\overrightarrow {x}.
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Problem 25.1.6

Let VV be an N×NN \times N unitary matrix, i.e.VV∗=INVV^{*} = I_{N}. The eigenvalues of VV lie on the unit circle; that is, they may be expressed in the form exp⁡(iθn)\exp (i\theta_{n}), θn∈R\theta_{n} \in \mathbb {R}. A function f(V)=f(θ1,…,θN)f(V) = f(\theta_{1}, \ldots , \theta_{N}) is called a class function if ff is symmetric in all its variables. Weyl gave an explicit formula for averaging class functions over the circular unitary ensemble

∫U(N)f(V) dV \int _{U(N)} f(V) \, dV =1(2π)NN!∫02π⋯∫02πf(θ1,…,θN)∏1≤j<k≤N∣eiθj−eiθk∣2 dθ1⋯dθN. = \frac{1}{(2\pi )^{N}N!} \int _{0}^{2\pi } \cdots \int _{0}^{2\pi } f(\theta _{1}, \ldots , \theta _{N}) \prod _{1 \leq j < k \leq N} \left|e^{i\theta _{j}}-e^{i\theta _{k}}\right|^{2} \, d\theta _{1} \cdots d\theta _{N}.

Thus we integrate the function f(V)f(V) over U(N)U(N) by parametrizing the group by the θi\theta_{i} and using Weyl’s formula to convert the integral into an NN-fold integral over the θi\theta_{i}. By definition the Haar measure dVdV is invariant under V→U~VU~∗V \rightarrow \widetilde{U}V\widetilde{U}^{*}, where U~\widetilde{U} is any N×NN \times N unitary matrix. The matrix VV can always be diagonalized by a unitary matrix, i.e.

V=W(eiθ10⋮⋱⋮0⋯eiθN)W∗ V = W \begin{pmatrix} e^{i\theta _{1}} & & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & e^{i\theta _{N}} \end{pmatrix} W^{*}

where WW is an N×NN \times N unitary matrix. Thus the integral over VV can be written as an integral over the matrix elements of WW and the eigenphases θn\theta_{n}. Since the measure is invariant under unitary transformations, the integral over the matrix elements of UU can be evaluated straightforwardly, leaving the integral over the eigenphases. Show that for ff a class function we have

∫U(N)f(V) dV=1(2π)N∫02π⋯∫02πf(θ1,…,θN)det⁡((eiθn(n−m))) dθ1⋯dθN. \int _{U(N)} f(V) \, dV = \frac{1}{(2\pi )^{N}} \int _{0}^{2\pi } \cdots \int _{0}^{2\pi } f(\theta _{1}, \ldots , \theta _{N}) \operatorname {det}\left((e^{i\theta _{n}(n-m)})\right) \, d\theta _{1} \cdots d\theta _{N}.
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Problem 25.1.7

Let AA be an n×nn \times n positive definite matrix over R\mathbb {R}. Let q→\overrightarrow {q} and J→\overrightarrow {J} be column vectors in Rn\mathbb {R}^{n}. Calculate

Z(J→)=∫−∞∞⋯∫−∞∞dq1⋯dqnexp⁡(−12q→TAq→+J→Tq→). Z(\overrightarrow {J}) = \int _{-\infty }^{\infty } \cdots \int _{-\infty }^{\infty } dq_{1} \cdots dq_{n} \exp \left(-\frac{1}{2}\overrightarrow {q}^{T}A\overrightarrow {q} + \overrightarrow {J}^{T}\overrightarrow {q}\right).

Note that

∫−∞∞dq e−(aq2+bq+c)=πae(b2−4ac)/(4a).(1) \int _{-\infty }^{\infty } dq \, e^{-(aq^{2}+bq+c)} = \sqrt{\frac{\pi }{a}} e^{(b^{2}-4ac)/(4a)}. \tag {1}
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Problem 25.1.8

Let AA be an n×nn \times n positive definite matrix over R\mathbb {R}, i.e.all the eigenvalues, which are real, are positive. We also have AT=AA^{T} = A. Consider the analytic function f:Rn→Rf : \mathbb {R}^{n} \rightarrow \mathbb {R}

f(x→)=exp⁡(−12x→TA−1x→). f(\overrightarrow {x}) = \exp \left(-\frac{1}{2}\overrightarrow {x}^{T}A^{-1}\overrightarrow {x}\right).

Calculate the Fourier transform of ff. The Fourier transform is defined by

f^(k→):=∫Rnf(x→)eik→⋅x→ dx→ \hat{f}(\overrightarrow {k}) := \int _{\mathbb {R}^{n}} f(\overrightarrow {x}) e^{i\overrightarrow {k}\cdot \overrightarrow {x}} \, d\overrightarrow {x}

where k→⋅x→≡k→Tx→≡k1x1+⋯+knxn\overrightarrow {k}\cdot \overrightarrow {x} \equiv \overrightarrow {k}^{T}\overrightarrow {x} \equiv k_{1}x_{1} + \cdots + k_{n}x_{n} and dx→=dx1…dxnd\overrightarrow {x} = dx_{1}\ldots dx_{n}. The inverse Fourier transform is given by

f(x→)=1(2π)n∫Rnf^(k→)e−ik→⋅x→ dk→ f(\overrightarrow {x}) = \frac{1}{(2\pi )^{n}} \int _{\mathbb {R}^{n}} \hat{f}(\overrightarrow {k}) e^{-i\overrightarrow {k}\cdot \overrightarrow {x}} \, d\overrightarrow {k}

where dk→=dk1…dknd\overrightarrow {k} = dk_{1}\ldots dk_{n}. With a>0a > 0 we have

∫Re−(ax2+bx+c) dx=πae(b2−4ac)/(4a). \int _{\mathbb {R}} e^{-(ax^{2}+bx+c)} \, dx = \sqrt{\frac{\pi }{a}} e^{(b^{2}-4ac)/(4a)}.
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Problem 25.1.9

Let AA be an n×nn \times n matrix. Let ω,μ∈R\omega , \mu \in \mathbb {R}. Assume that

∥etA∥≤Meωt,t≥0 \left\| e^{tA}\right\| \leq M e^{\omega t}, \qquad t \geq 0

and μ>ω\mu > \omega. Then we have

(μIn−A)−1≡∫0∞e−μtetA dt.(1) (\mu I_{n} - A)^{-1} \equiv \int _{0}^{\infty } e^{-\mu t} e^{tA} \, dt. \tag {1}

Calculate the left- and right-hand side of (1) for the matrix A=σ1A = \sigma_{1}.

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Problem 25.1.10

Consider the differentiable manifold

S3={(x1,x2,x3,x4):x12+x22+x32+x42=1}. S^{3} = \left\{ (x_{1}, x_{2}, x_{3}, x_{4}) : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = 1\right\} .
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(i)

Show that the matrix

U(x1,x2,x3,x4)=−i(x3+ix4x1−ix2x1+ix2−x3+ix3) U(x_{1}, x_{2}, x_{3}, x_{4}) = -i \begin{pmatrix} x_{3}+ix_{4} & x_{1}-ix_{2} \\ x_{1}+ix_{2} & -x_{3}+ix_{3} \end{pmatrix}

is unitary. Show that the matrix is an element of SU(2)SU(2).

(ii)

Consider the parameters (θ,ψ,ϕ)(\theta , \psi , \phi ) with 0≤θ<π0 \leq \theta < \pi, 0≤ψ<4π0 \leq \psi < 4\pi, 0≤ϕ<2π0 \leq \phi < 2\pi. Show that

x1(θ,ψ,ϕ)+ix2(θ,ψ,ϕ)=cos⁡(θ/2)ei(ψ+ϕ)/2 x_{1}(\theta , \psi , \phi ) + ix_{2}(\theta , \psi , \phi ) = \cos (\theta /2) e^{i(\psi +\phi )/2} x3(θ,ψ,ϕ)+ix4(θ,ψ,ϕ)=sin⁡(θ/2)ei(ψ−ϕ)/2 x_{3}(\theta , \psi , \phi ) + ix_{4}(\theta , \psi , \phi ) = \sin (\theta /2) e^{i(\psi -\phi )/2}

is a parametrization. Thus the matrix given in (i) takes the form

−i(sin⁡(θ/2)ei(ψ−ϕ)/2cos⁡(θ/2)e−i(ψ+ϕ)/2cos⁡(θ/2)ei(ψ+ϕ)/2−sin⁡(θ/2)e−i(ψ−ϕ)/2) -i \begin{pmatrix} \sin (\theta /2) e^{i(\psi -\phi )/2} & \cos (\theta /2) e^{-i(\psi +\phi )/2} \\ \cos (\theta /2) e^{i(\psi +\phi )/2} & -\sin (\theta /2) e^{-i(\psi -\phi )/2} \end{pmatrix}

and

U∗=U−1=(ie−i(ψ−ϕ)/2sin⁡(θ/2)e−i(ψ+ϕ)/2cos⁡(θ/2)ei(ψ+ϕ)/2cos⁡(θ/2)−iei(ψ−ϕ)/2sin⁡(θ/2)). U^{*} = U^{-1} = \begin{pmatrix} ie^{-i(\psi -\phi )/2}\sin (\theta /2) & e^{-i(\psi +\phi )/2}\cos (\theta /2) \\ e^{i(\psi +\phi )/2}\cos (\theta /2) & -ie^{i(\psi -\phi )/2}\sin (\theta /2) \end{pmatrix}.
(iii)

Let (ξ1,ξ2,ξ3)=(θ,ψ,ϕ)(\xi_{1}, \xi_{2}, \xi_{3}) = (\theta , \psi , \phi ) with 0≤θ<π0 \leq \theta < \pi, 0≤ψ<4π0 \leq \psi < 4\pi, 0≤ϕ<2π0 \leq \phi < 2\pi. Show that

124π2∫0πdθ∫04πdψ∫02πdϕ∑j,k,ℓ=13ϵjkℓtr⁡(U−1∂U∂ξjU−1∂U∂ξkU−1∂U∂ξℓ)=1 \frac{1}{24\pi ^{2}} \int _{0}^{\pi } d\theta \int _{0}^{4\pi } d\psi \int _{0}^{2\pi } d\phi \sum _{j,k,\ell =1}^{3} \epsilon _{jk\ell } \operatorname {tr}\left(U^{-1}\frac{\partial U}{\partial \xi _{j}} U^{-1}\frac{\partial U}{\partial \xi _{k}} U^{-1}\frac{\partial U}{\partial \xi _{\ell }}\right) = 1

where ϵ123=ϵ321=ϵ132=+1\epsilon_{123} = \epsilon_{321} = \epsilon_{132} = +1, ϵ213=ϵ321=ϵ132=−1\epsilon_{213} = \epsilon_{321} = \epsilon_{132} = -1 and 00 otherwise.

(iv)

Consider the metric tensor field

g=dx1⊗dx1+dx2⊗dx2+dx3⊗dx3+dx4⊗dx4. g = dx_{1} \otimes dx_{1} + dx_{2} \otimes dx_{2} + dx_{3} \otimes dx_{3} + dx_{4} \otimes dx_{4}.

Using the parametrization show that

gS3=14(dθ⊗dθ+dψ⊗dψ+dϕ⊗dϕ+cos⁡(θ)dψ⊗dϕ+cos⁡(θ)dϕ⊗dψ). g_{S^{3}} = \frac{1}{4}\left(d\theta \otimes d\theta + d\psi \otimes d\psi + d\phi \otimes d\phi + \cos (\theta ) d\psi \otimes d\phi + \cos (\theta ) d\phi \otimes d\psi \right).
(v)

Consider the three differential one forms e1e_{1}, e2e_{2}, e3e_{3} defined by

(e1e2e3)=(−x4−x3x2x1x3−x4−x1x2−x2x1−x4x3)(dx1dx2dx3dx4). \begin{pmatrix} e_{1} \\ e_{2} \\ e_{3} \end{pmatrix} = \begin{pmatrix} -x_{4} & -x_{3} & x_{2} & x_{1} \\ x_{3} & -x_{4} & -x_{1} & x_{2} \\ -x_{2} & x_{1} & -x_{4} & x_{3} \end{pmatrix} \begin{pmatrix} dx_{1} \\ dx_{2} \\ dx_{3} \\ dx_{4} \end{pmatrix}.

Show that gS3=de1⊗de1+de2⊗de2+de3⊗de3g_{S^{3}} = de_{1} \otimes de_{1} + de_{2} \otimes de_{2} + de_{3} \otimes de_{3}.

(vi)

Show that

dej=∑k,ℓ=13ϵjkℓek∧eℓ de_{j} = \sum _{k,\ell =1}^{3} \epsilon _{jk\ell } e_{k} \wedge e_{\ell }

i.e.de1=2e2∧e3de_{1} = 2e_{2} \wedge e_{3}, de2=2e3∧e1de_{2} = 2e_{3} \wedge e_{1}, de3=2e1∧e2de_{3} = 2e_{1} \wedge e_{2}.

Problem 25.2.1

Let AA, BB be n×nn \times n matrices over C\mathbb {C}. Let β,ϵ∈R\beta , \epsilon \in \mathbb {R}. Show that

exp⁡(β(A+B))≡exp⁡(βA)(In+∫0βdϵ e−ϵABeϵ(A+B)). \exp (\beta (A+B)) \equiv \exp (\beta A) \left(I_{n} + \int _{0}^{\beta } d\epsilon \, e^{-\epsilon A} B e^{\epsilon (A+B)}\right).

Multiply both sides from the left with exp⁡(−βA)\exp (-\beta A) and differentiate with respect to β\beta.

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Problem 25.2.2

Let α∈R\alpha \in \mathbb {R}. Consider an n×nn \times n matrix A(α)A(\alpha ), where the entries of AA depends smoothly on α\alpha. Then one has the identity

ddαeA(α)≡∫01e(1−s)A(α)dA(α)dαesA(α) ds≡∫01esA(α)dA(α)dαe(1−s)A(α) ds. \frac{d}{d\alpha } e^{A(\alpha )} \equiv \int _{0}^{1} e^{(1-s)A(\alpha )} \frac{dA(\alpha )}{d\alpha } e^{sA(\alpha )} \, ds \equiv \int _{0}^{1} e^{sA(\alpha )} \frac{dA(\alpha )}{d\alpha } e^{(1-s)A(\alpha )} \, ds.

Let n=2n = 2 and

A(α)=(cos⁡(α)−sin⁡(α)sin⁡(α)cos⁡(α))⇒dA(α)dα=(−sin⁡(α)−cos⁡(α)cos⁡(α)−sin⁡(α)). A(\alpha ) = \begin{pmatrix} \cos (\alpha ) & -\sin (\alpha ) \\ \sin (\alpha ) & \cos (\alpha ) \end{pmatrix} \Rightarrow \frac{dA(\alpha )}{d\alpha } = \begin{pmatrix} -\sin (\alpha ) & -\cos (\alpha ) \\ \cos (\alpha ) & -\sin (\alpha ) \end{pmatrix}.

Calculate the right-hand sides of the identities.

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Problem 25.2.3

Let AA, BB be positive definite matrices. Then we have the integral representation (x≥0x \geq 0)

ln⁡(A+xB)−ln⁡(A)≡∫0∞(A+uIn)−1xB(A+xB+uIn)−1 du. \ln (A+xB) - \ln (A) \equiv \int _{0}^{\infty } (A+uI_{n})^{-1} xB (A+xB+uI_{n})^{-1} \, du.

Let

A=(2−1−11),B=(1112). A = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.

Calculate the left- and right-hand side of the integral representation.

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Problem 25.2.4

Consider the 2×22 \times 2 matrices

A(t)=(cos⁡(t)−sin⁡(t)sin⁡(t)cos⁡(t)),B(t)=∫0tA(s) ds. A(t) = \begin{pmatrix} \cos (t) & -\sin (t) \\ \sin (t) & \cos (t) \end{pmatrix}, \qquad B(t) = \int _{0}^{t} A(s) \, ds.

Find the commutator [A(t),B(t)][A(t), B(t)]. Discuss. What is the condition such that [A(t),B(t)]=02[A(t), B(t)] = 0_{2}?

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Problem 25.2.5

Let PjP_{j} (j=0,1,2,…j = 0, 1, 2, \ldots) be the Legendre polynomials

P0(x)=1,P1(x)=x,P2(x)=12(3x2−1),…. P_{0}(x) = 1, \qquad P_{1}(x) = x, \qquad P_{2}(x) = \frac{1}{2}(3x^{2}-1), \qquad \ldots .

Show that the infinite dimensional matrix A=(ajk)A = (a_{jk})

ajk:=∫−1+1Pj(x)dPk(x)dx dx a_{jk} := \int _{-1}^{+1} P_{j}(x) \frac{dP_{k}(x)}{dx} \, dx

where j,k=0,1,…j, k = 0, 1, \ldots is given by

A=(0000000⋯2000000⋯0200000⋯2020000⋯0202000⋯2020200⋯0202020⋯⋮⋮⋮⋮⋮⋮⋮⋱). A = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 0 & 2 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 2 & 0 & 2 & 0 & 0 & 0 & \cdots \\ 2 & 0 & 2 & 0 & 2 & 0 & 0 & \cdots \\ 0 & 2 & 0 & 2 & 0 & 2 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.

Note that the matrix is not symmetric. The matrix AA can be considered as a linear operator in the Hilbert space ℓ2(N0)\ell_{2}(\mathbb {N}^{0}). Is ∥A∥<∞\left\| A\right\| < \infty?

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Problem 25.2.6

Let AA, BB, CC be positive definite n×nn \times n matrices. Then (Lieb inequality)

tr⁡(eln⁡(A)−ln⁡(B)+ln⁡(C))≤tr⁡∫0∞A(B+uIn)−1C(B+uIn)−1 du. \operatorname {tr}(e^{\ln (A)-\ln (B)+\ln (C)}) \leq \operatorname {tr}\int _{0}^{\infty } A(B+uI_{n})^{-1} C(B+uI_{n})^{-1} \, du.
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(i)

Let

A=(2−1−11),B=(1112),C=(4225). A = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \qquad C = \begin{pmatrix} 4 & 2 \\ 2 & 5 \end{pmatrix}.

Calculate the left-hand side and right-hand side of the inequality.

(ii)

Show that a sufficient condition such that one has an equality is that AA, BB, CC commute. However this condition is not necessary.