Chapter 10
INEQUALITIES IN MATRIX THEORY
The Volume of an m-Dimensional Parallelepiped
For an m-dimensional rectangular parallelepiped in R m , the lengths of whose mutually perpendicular sides are l1, l2, ... , lm, the volume (for m = 3, area for m = 2 and the so-called hyper-volume for m > 3) is defined to be P 1£ i£ m li. Here the length is the Euclidean length of a vector side. If the vectors defining concurrent sides of a rectangular parallelepiped are the unit vectors ei , i = 1, 2, ... , m, obviously the volume equals 1 = |det (e1 | e2 | ... | em)| and if the vectors are ciei , it equals |c1c2…cm| | = |det (c1e1 | c2e2 | ... | cmem)|. This formula remains valid even if the vectors are not orthogonal, i.e., if the concurrent sides are v1, v2, ... , vm, the volume of the parallelepiped is given by V = |det (v1 | v2 | ... | vm)|, for, if Pi is the orthogonal projection on the orthogonal complement of span {v1, v2, ... , vm}, then det (v1, v2, ... , vm) = det (v1, P1v2, P2v3, ... , Pm-1vm) and the column vectors on the r.h.s. are concurrent sides of a rectangular parallelepiped of the same volume as that of the original parallelepiped. (Two volumes corresponding to the same base and the same height are equal!)
Another useful formula for the volume which may be derived from the above is
V2 = det [(vi, vj)]1
£ i,j£ m = det [(v1 | v2 | ... | vm)*( v1 | v2 | ... | vm)],where (vi, vj) is the inner product of vi and vj. The formula
V2 = det [(ai, aj)]1
£ i,j£ m,where a1, a2, ... , ak are the concurrent side vectors of a parallelepiped in any inner product space (real or complex) and V is the volume of this k-dimensional parallelepiped and where the dimension of the space need not be k or even finite, is a generalized version of the above formula. The proof is analogous to that of the first formula for
det [(ai, aj)]1
£ i,j£ k = det [(Pi-1ai, Pj-1aj)]1£ i,j£ k = det [diag [(Pi-1ai, Pi-1ai)]1£ i£ k],and the last expression is the product of squares of sides of a rectangular parallelepiped of the same volume as the original parallelepiped.
Change of Variables in Multiple Integrals
Let x, y Î K m. Consider ò Dy f(y) dVy where y ranges over the volume of an n-dimensional region Dy. Let the vector variable y be related with x according to y = y(x), where y is a continuously differentiable vector function of x. Let the Jacobian of the transformation be denoted by
.
As y ranges over Dy, let x run over Dx. An infinitesimal parallelepiped in the region Dx with concurrent sides (du1, du2, ... , dun) corresponds to an infinitesimal parallelepiped in the region Dy with concurrent sides given by (dv1, dv2, ... , dvn) = ((¶ y/¶ x)du1, (¶ y/¶ x)du1, ... , (¶ y/¶ x)dun) = (¶ y/¶ x)(du1, du2, ... , dun). Thus ò Dy f(y) dVy = ò Dy f(y) |det (dv1, dv2, ... ,dvn)| = ò Dx f(y(x)) |det (¶ y/¶ x)| |det (du1, du2, ... ,dun)|. Hence
ò
Dy f(y) dVy = ò Dy f(y(x)) |det(¶ y/¶ x)| dVx is the required change of variable formula in multiple integrals.In the sequel, while considering surfaces and volumes, unless otherwise stated the matrices under consideration are real.
Surface and Volume of a Hyper-Sphere
Let |A| ¹ 0 and consider the multiple integral . If we put y = Ax, then, x = A-1y, and so
.
But as , it follows that
I =
p n/2/|det A| = p n/2/|l 1(A) l 2(A) ... l n(A)|,where l i(A), i = 1, 2, ... , n are the n eigenvalues of A. Hence
.
Corollary 1. Let B be positive definite. Then
.
Proof: Let A = Ö B. Then, (Bx, x) = (Ax, Ax) and l i(A) = Ö l i(B), i = 1 (1) n. #
Corollary 2. Let An(r ) denote the surface area of the n-dimensional sphere Sn(r ) of radius r . Then
An(
r ) = 2p n/2r n-1/G (n/2).Proof: Taking B = I, the identity matrix
.
Thus, An(1) = 2p n/2/G (n/2). Since An(r ) = r n-1An(1), the result follows. #
Remark: Note that A2(r) = 2p r, the circumference of a circle with radius r, A3(r) = 2p 3/2r2 /[(1/2)p 1/2] = 4p r2, the area of the surface of sphere of radius r, which are the familiar formulae. Note that A1(r) = 2, the number of the end points of the line segment [-r, r]!
Corollary 3. The volume Vn(r ) of the n-sphere of radius r is given by
.
Proof: Vn(r) = ò 0¥ An(r)dr = ò 0¥ [2p n/2rn-1/G (n/2)] dr = 2p n/2r n/[nG (n/2)] = [p n/2/G (n/2 + 1)] r n. #
Remark: V1(r) = (Ö p /(Ö p /2))r = 2r, the volume of 1-sphere of radius one (i.e., the length of the interval {x : |x| < r}), V2(r) = (p /G (2))r2 = p r2, area inside a circle of radius r, and V3(r) = p Ö p /((3/2)(1/2)Ö p )r3 = (4/3)p r3, the familiar formula for the volume of a sphere of radius r.
Corollary 4. Let A and B be positive definite and 0 < q < 1. Then, |q A + (1-q )B| > |A|q |B|1-q .
Proof: By Corollary 1, and using Holder's inequality with 1/p = l , 1/q = 1-l , we have
from which the desired inequality follows. #
The Volume of an n-Dimensional Ellipsoid
Let the ellipsoid be (Ax, x) < 1, where A is n´ n positive definite. Let A = UDU*. Let dx stand for the
infinitesimal volume element in R n. Then the volume of the ellipsoid is
VA =
ò (Ax,x)<1 dx = ò (Dy,y)<1 dy = ò (Ö Dy,Ö Dy)<1 dy = |D|-1/2ò (z,z)<1 dz = p n/2|D|-1/2/G + 1) = p n/2|A|-1/2/G (n/2+ 1) = p n/2|l 1(A)l 2(A) … l n(A)|-1/2/G (n/2 + 1) = [G (n/2 + 1)]-1ò R n e-(Ax,x)dx.The restriction of the ellipsoid (Ax, x) < 1, to the orthogonal complement of a set of k-linearly independent vectors u1, u2, ... , uk is given by the conditions: (Ax, x) < 1, (x, ui) = 0, 1 £ i £ k and constitutes an n-k dimensional ellipsoid. For, without loss of generality we may assume ui, 1 £ i £ k, orthogonal, and complete these to an orthonormal basis ui, 1 £ i £ n. Let U = (u1 | u2 | ... | un), a column partitioned form. Then x = UU*x and with V = (uk+1 | uk+2 | ... | un) for y Î < u1, u2, ... , uk > ^ , y = VV*y and so the above restriction conditions are equivalent to y Î < u1, u2, ... , uk > ^ , 1 > (AVV*y, VV*y) = ((V*AV)V*y, V*y) and V being of rank n-k, z = V*y runs over an n-k dimensional space as y has form y = Vt, t running over R n-k and moreover V*AV is an (n-k)´ (n-k) positive definite matrix.
The volume of the largest m-dimensional section of an ellipse is to be obtained in the sequel.
Principal Semi-Axes of a Section of an Ellipsoid
The lengths of principal semi-axes of the ellipsoid E n : (Ax, x) < 1 are 1/Ö l n(A), 1/Ö l n-1(A), ... , 1/Ö l 1(A), written in non-increasing order of magnitude. For, letting A = Udiag(l n(A), l n-1(A), ... , 1/l 1(A))U*, the unitary transformation y = U*x transforms the ellipsoid to (diag (l n(A), l n-1(A), ... , 1/l 1(A)) y, y) < 1, i.e.,
.
Moreover, the volume of the ellipsoid is given by
.
Let Vn-m be an n-m dimensional subspace of R n and let E m denote the restriction of E n to the orthogonal complement of Vn-m . Then E m is described by {(Ax, x) < 1 : x ^ Vn-m}. Let the lengths of the principal semi-axes of E m be denoted, in non-decreasing order of magnitude, by 1/Ö m m(A), 1/Ö m m-1(A), ... , 1/Ö m 1(A). Let Xm = Vn-m^ . By the Courant-Fischer min-max characterization:
l
k = minWn-k+1max0¹ xÎ Wn-k+1x*Ax/x*x £ minWn-k+1Ì Xmmax0¹ xÎ Wn-k+1x*Ax/x*x = m m+1-(n-k+1) = m m-n+k.Hence m m-n+k ³ l k, so that 1/Ö m m £ 1/Ö l n, 1/Ö m m-1 £ 1/Ö l n-1, … , 1/Ö m 1 £ 1/Ö l n-m+1, implying thereby that the lengths of semi-axles of restriction of an ellipse are never larger than the corresponding lengths of the original ellipse. The max-min characterization, on the other hand gives:
m
k(A) = maxWkÌ Xm min0¹ xÎ Wk x*Ax/x*x £ maxWk min0¹ xÎ Wk x*Ax/x*x = l k(A),i.e., 1/Ö m k(A) ³ 1/Ö l k(A). These inequalities lead us to:
Vm
£ p m/2 /[Ö (l n-m+1 …l n-1l n)G (m/2 + 1)],Vm
³ p m/2 /[Ö (l ml m-1…l 1)G (m/2 + 1)].Corollary: The largest and the smallest volumes of m-dimensional sections of an ellipse (Ax,x) < 1 are, respectively, given by:
max
E m Vm = p m/2 /[Ö (l n-m+1 …l n-1l n)G (m/2 + 1)],min
E m Vm = p m/2 /[Ö (l ml m-1…l 1)G (m/2 + 1)].Proof: The foregoing implies that the r.h.s.’s in the above give, respectively, an upper bound and a lower bound. But these bounds are attained for the m-dimensional sections, respectively, given by:
º { x ^ um+1, um+2, ... , un | x*Ax < 1}
& º { x ^ u1, u2, ... , un-m | x*Ax < 1}. #
For A positive definite, let us define |A|k = l nl n-1 ... l n-k+1, which is the product of the k smallest eigenvalues of A.
Theorem. Let Vk denote a k-dimensional subspace of R n. Then for a real positive definite matrix A,
,
where ‘dx’ stands for the infinitesimal volume element in Vk (i.e., is the restricted Lebesgue measure) and the ‘max’ is over the class of all k-dimensional subspaces.
Proof: Let the ellipsoid Ek(r ; Vk) = {x : (Ax, x) < r , x Î Vk} have volume Vk(r ) = r k/2Vk(1). Then, as
,
the result follows from the expressions for the maximum and the minimum volumes of a k-dimensional section of an ellipse in the previous corollary. #
Corollary (Ky Fan). If A, B are real p.d. matrices, |q A+(1-q )B|k > (|A|k)q (|B|k)1-q , 0 £ q £ 1, 1 £ k £ m.
Proof: Using Holder's inequality with 1/p = q , 1/q = 1-q , the result follows from previous theorem. #
Corollary (Ky Fan). For a real symmetric (hermitian) matrix A, let Sk(A) denote the sum of the smallest n-k+1 eigen values of A: Sk(A) = l n(A) + l n-1 (A) + ... + l k(A). Then
Sk(
q A + (1-q )B) ³ q Sk(A) + (1-q )Sk(B), 1 £ k £ n, 0 £ q £ 1.Proof: For all e > 0 sufficiently small, I + e A and I + e B are p.d. and l i(I + e A) = 1 + e l i(A), l i(I + e B) = 1 + e l i(B) and l i(q (I + e A) + (1-q )(I + e B)) = 1 + e l i(q A + (1-q )B), as e being positive the non-increasing order of the eigenvalues is preserved. By the previous corollary applied to I + e A and I + e B, we have: P i=n(-1)k (1 + e l i(q A + (1-q )B)) ³ (P i=n(-1)k (1 + e l i(A)))q (P i=n(-1)k (1 + e l i (B)))1-q . It follows that: 1 + e S i=n(-1)k l i(q A + (1-q )B)) ³ 1 + e q S i=n(-1)k l i(A) + e (1-q )S i=n(-1)k l i (B) + O(e 2). Canceling 1, dividing by e (>0) and taking limit as e ® 0, the result follows. #
Corollary (Ky Fan). If Tk(A) = l 1(A) + l 2(A) + ... + l k(A) and A, B are real symmetric (Hermitian)
Tk(
q A + (1-q )B) £ q Tk(A) + (1-q ) Tk(B), 1 £ k £ n, 0 £ q £ 1.Proof: Sk(-A) = -(l 1(A) + l 2(A) + ... + l n-k+1(A) = -Tn-k+1(A), and the result follows from the previous corollary. #
Aliter: Tr(A) is the sum of all the eigenvalues of A and Tr(q A + (1-q )B) = q Tr(A) + (1-q )Tr(B). The result follows after subtracting the inequality in the previous corollary from this equality. #
Theorem (Ky Fan). Let A be p.d. and R = {(z1, z2, ... , zn-k+1)} : (zi, zj) = d ij , 1 £ i, j £ n-k+1}, the collection of all (n-k+1)-ples of orthonormal vectors. Then: l 1l 2 ... l n-k+1 = maxR det ((zi, Azj))1£ i,j£ n-k+1, and l nl n-1 ... l k = minR det ((zi, Azj))1£ i,j£ n-k+1. (The proof is valid for complex p.d. matrices.)
Proof: First note that the max on the right hand side is attained when zi is taken as an eigen vector corresponding to the eigen value l i , 1 £ i £ n-k+1, in a unitary diagonalization of A. Similarly, the min on the right hand side is attained when zi is taken as the eigen vector corresponding to the eigen value l n-i+1, 1 £ i £ n-k+1. Next, ((zi, Azj))' = ((Azj, zi)) = (z1 | z2 | ... | zn-k+1)*A(z1 | z2 | ... | zn-k+1) = B*AB, say, where the column partitioned matrix B = (z1 | z2 | ... | zn-k+1) is sub-unitary. Hence by the Poincaré separation theorem l i(B*AB) £ l i(A), 1 £ i £ n-k+1, and l n-k+1-j(B*AB) ³ l n-j(A), 0 £ j £ n-k. Hence, with D = det((zi, Azj)) = det(B*AB) = l 1(B*AB)l 2(B*AB) ... l n-k+1(B*AB), as the eigenvalues of B*AB and A are positive, we have D £ l 1(A)l 2(A) … l n-k+1(A) and D ³ l n(A)l n-1(A) … l k(A) proving thereby that maxR D £ l 1(A)l 2(A) … l n-k+1(A) and minR D ³ l n(A)l n-1(A) … l k(A), which, since the right hand expressions are attained, proves the theorem. #
Theorem. If A is positive definite, then |A| £ a11 a22 ... ann. (The result is valid also for the complex case, as the second proof will show.)
Proof: p n/2/Ö |A| = ò R n e-(Ax,x) dx = ò R n exp [-S 1£ i,j£ n aijxjxi] dx = ò R n exp [-S 1£ i,j£ n s is jaijxjxi] dx, by changing the variable x to -x , where s k = 1 if k ¹ 1 and s k = -1 if k = 1. Taking the arithmetic mean (A.M.)of the last two integrals and using A.M. ³ G.M., the geometric mean, we have
p
n/2/Ö |A| ³ ò R n exp [-a11x12 -S 2£ i,j£ n aijxjxi] dx.Similarly, replacing x2 by –x2, and so on upto xn-1 by –xn-1 and at each stage proceeding analogously we end up with p n/2/Ö |A| ³ ò R n exp [-S 1£ i£ n aiixi2] dx = p n/2/Ö ( a11 a22 ... ann), from which the result follows. #
Aliter: We can write: , say. Then B is positive definite and so B-1 is positive definite and hence the matrix (Bij) of co-factors of B is also positive definite. If Mji denotes the (j, i)-th minor of A, expanding by the 1st column
,
as (Bij) is positive definite. Hence
,
and the result follows by induction over n. #
Note: This proof is valid also when A is complex p.d. matrix (or even p.s.d. when, of course, the result is trivial).
Corollary. Let B be a real square matrix. Then: |B| < P 1£ i£ n (S 1£ k£ n bik2). (Hadamard's inequality)
Proof: |B| = |BB'| and (BB')ii = S 1£ k£ n bik2 , 1 £ i £ n. #
Application of Poincare Separation Theorem to Ellipsoids
An m-dimensional section of an n-dimensional ellipsoid
E n = {x Î R n : (Ax, x) = 1}, obtained by restricting its hyper surface to an m-dimensional subspace Wm is given by E m = {x Î Wm : (Ax, x) = 1}. Let {u1, u2, ... ,um} be an orthonormal basis of Wm. Then if x Î Wm, x = a 1u1 + ... + a mum for some vector a = (a 1, ... , a m) Î R m. Thus, x = (u1 | u2 | ... | um)a = Ua , say. Note that for the n´ m U we have U*U = I, i.e., U is sub-unitary. Thus E m = {a Î R m : (AUa , Ua ) = 1} = {a Î R m : (U*AUa , a ) = 1}. Hence the k-th principal semi-axis of E m is given by ak(m) = (l m-k+1(U*AU))-1/2. By the Poincaré separation theorem, therefore ak(m) £ ak(n) and am-k+1(m) £ an-k+1(n) i.e., an-k+1(m) £ am-k+1(m) £ am-k+1(n), providing an easier approach than before. #