Operators and Matrices Preprint
MULTIPLICITIES, BOUNDARY POINTS, AND JOINT NUMERICAL RANGES
WAI -S HUN C HEUNG , X UHUA L IU AND T IN -YAU TAM
(Communicated by C.-K. Li) Abstract. The multiplicity of a point in the joint numerical range W (A1 ,A2 ,A3 ) R3 is studied for n n Hermitian matrices A1 ,A2 ,A3. The relative interior points of W (A1 ,A2 ,A3 ) have multiplicity greater than or equal to n 2. The lower bound n 2 is best possible. Extreme points and sharp points are studied. Similar study is given to the convex set V (A) := {xT Ax : x Rn ,xT x = 1} C , where A Cnn is symmetric. Examples are given.
1. Introduction Let Cnn be the set of n n complex matrices. The classical numerical rang of A Cnn is W (A) := {x Ax : x Cn , x x = 1}. It is the image of the unit sphere
under the quadratic map x x Ax. Toeplitz-Hausdorff theorem asserts that W (A) is a compact convex set [9]. When n = 2 , W (A) is an elliptical disk (possibly degenerate) [9], known as the elliptical range theorem. A point W (A) is called an extreme point if is not in any open line segment that is contained in W (A). A point W (A) is a sharp point if is the intersection point of two distinct supporting lines of W (A) [9, p. 50]. We have the following inclusions for W (A), which are proper in general:
{sharp points} {extreme points} {boundary points}.

Sn1 = {x Cn : x x = 1}

Donoghue [7] showed that sharp points of W (A) are eigenvalues of A. Indeed the following is a characterization of the sharp points. T HEOREM 1.1. ([9, p. 5051]) Let A Cnn and W (A). Then is a sharp point if and only if A is unitarily similar to I B with W (B).
Mathematics subject classication (2010): Primary: 15A60. Keywords and phrases: Joint numerical range, multiplicity, extreme point, sharp point, boundary point.
c , Zagreb Paper OaM-0268
W.S. C HEUNG , X UHUA L IU AND T.Y. TAM
Given W (A), Embry [8] introduced M = M (A) := {x Cn : x Ax = x x}. In general M is not a subspace but it is homogeneous. Thus the span of M , denoted by M , satises M = M + M := {x + y : x, y M }. Stampi [14, Lemma 2] showed that M is a subspace of Cn if is an extreme point. Embry [8] established the converse and some related results. T HEOREM 1.2. (Embry) Let A Cnn and W (A). Then 1. is an extreme point if and only if M is a subspace of Cn. 2. if is a non-extreme boundary point, then M = wLW (A) Mw , where L is the supporting line of W (A), passing through . In this case M = Cn if and only if W (A) L. 3. if W (A) is nondegenerate, then is an interior point if and only if M = Cn.
i.e., wA ( ) is the maximal number of linearly independent vectors x Sn1 such that x Ax = . We call wA ( ) the multiplicity of . It is well known that W (A) is a line segment [ , ] if and only if A is essentially Hermitian, in which case wA ( ) = n for any ( , ). With this fact, one can deduce from Theorem 1.2 that wA ( ) = n for any relative interior point W (A), thus provides an afrmative answer to a question of Uhlig in [15, p. 18]. We will study multiplicities of relative interior points and some characterizations of extreme points and sharp points of two variations of the classical numerical range. 2. Joint numerical range of three Hermitian matrices
Let Hn be the set of n n Hermitian matrices. Let A = A1 + iA2 be the Hermitian decomposition of A Cnn , where A1 , A2 Hn. Since x Ax = x A1 x + ix A2 x
and x A1 x, x A2 x R, one may identify W (A) as the set W (A1 , A2 ) := {(x A1 x, x A2 x) : x Cn , x x = 1} R2.

wA ( ) := dim M ,

We remark that the results of Dongonhue, Stampi, and Embry are for any bounded linear operator on a complex Hilbert space. Now consider for each W (A)
M ULTIPLICITIES AND NUMERICAL RANGES
Given A1 ,. , Ak Hn , Au-Yeung and Poon [1] and other authors, for example, Binding and Li [4], Au-Yeung and Tsing [3], Li and Poon [10], considered the following generalization of W (A): W (A1 ,. , Ak ) := {(x A1 x,. , x Ak x) : x Cn , x x = 1}, which is a joint numerical range of A1 ,. , Ak. We remark that W (A) can be viewed as W (A1 , A2 , 0). Au-Yeung and Tsing [2] proved that W (A1 , A2 , A3 ) is convex when n 3. Moreover W (A1 , A2 , A3 ) is an ellipsoid (possibly degenerate) when n = 2 (see [6] for a conceptual reason). D EFINITION 2.1. For W (A1 , A2 , A3 ), where A1 , A2 , A3 Hn , the multiplicity, denoted by wA1 ,A2 ,A3 ( ), is the maximal number of linearly independent vectors x Sn1 such that (x A1 x, x A2 x, x A3 x) = . To simplify notation, given A1 , A2 , A3 Hn , we let f : Sn1 R3 be the map dened by f (x) := (x A1 x, x A2 x, x A3 x), x Sn1.

T HEOREM 2.2. Let A1 , A2 , A3 Hn and W (A1 , A2 , A3 ). 1. When n = 2 ,
(a) if W (A1 , A2 , A3 ) is a nondegenerate ellipsoid, then wA1 ,A2 ,A3 ( ) = 1.
(c) if W (A1 , A2 , A3 ) = { } (in this case A1 , A2 , A3 are scalar multiples of identity), then wA1 ,A2 ,A3 ( ) = 2. 2. When n 3 , if IntRW (A1 , A2 , A3 ), then wA1 ,A2 ,A3 ( )
and may not be a constant. The lower bound n 2 is best possible.
Proof. (1) Suppose n = 2. Notice that (c) is trivial.
(a) Suppose that W (A1 , A2 , A3 ) is a nondegenerate ellipsoid and W (A1 , A2 , A3 ). Then orthogonally project W (A1 , A2 , A3 ) onto the hyperplane P R3 with orthonormal basis {p := (p1 , p2 , p3 ), q := (q1 , q2 , q3 )} R3 so that the image is a nondegenerate elliptical disk E and the projection of is on the (relative) boundary of E. With respect to the basis {p, q} , E = {(pT (x A1 x, x A2 x, x A3 x), qT (x A1 x, x A2 x, x A3 x)) : x S1 } = {(x (p1 A1 + p2 A2 + p3 A3 )x, x (q1 A1 + q2A2 + q3 A3 )x : x S1 }
(b) if W (A1 , A2 , A3 ) is a nondegenerate elliptical disk or a nondegenerate line segment, then (i) wA1 ,A2 ,A3 ( ) = 2 if IntRW (A1 , A2 , A3 ), (ii) wA1 ,A2 ,A3 ( ) = 1 if IntRW (A1 , A2 , A3 ).
We denote by IntR S the relative interior of S Rk.
which is naturally identied as W (A), where A = (p1 A1 + p2 A2 + p3 A3 ) + i(q1 A1 + q2 A2 + q3 A3 ). By Theorem 1.2, wA ( ) = 1 and thus wA1 ,A2 ,A3 ( ) = 1. The proof of (b) is similar to that of (a). (2) The statement is trivial for n=3. Suppose that n 4 and let IntRW (A1 , A2 , A3 ). Suppose on the contrary that wA1 ,A2 ,A3 ( ) = k < n 2. Let {x1 ,. , xk } Sn1 be a (maximal) linearly independent set such that = f (xi ), i = 1,. , k. Since k < n 2 , there is u Sn1 such that x1 ,. , xk , u are linearly independent and thus = f (u). Because IntRW (A1 , A2 , A3 ) and W (A1 , A2 , A3 ) is convex in R3 [2], there is v Sn1 such that (a) f (v) = and f (v) = f (u), (b) the line segment L := [ f (u), f (v)] W (A1 , A2 , A3 ), and (c) L. In other words, ( f (u), f (v)) W (A1 , A2 , A3 ), where ( f (u), f (v)) denotes the open line segment. Since f (u) = f (v), u, v are linearly independent. Since k < n 2 , there would exist a unit vector w u, v, x1 ,. , xk. Let Ai , i = 1, 2, 3 , denote the compression of Ai onto the 3 -dimensional subspace u, v, w. So W (A1 , A2 , A3 ) and hence there would exist a unit vector y u, v, w and f (y) = . Write y = u + v + w for , , C. Notice that and cannot be both zero, otherwise f (y) = f (v) = . Then y, x1 ,. , xk would be linearly independent, a contradiction. The following example shows that the bound n 2 is best possible. E XAMPLE 2.3. Let n A1 = 3 and

It is known that W (A1 , A2 , A3 ) R3 is the convex hull of W (B1 , B2 , B3 ) and the origin, i where B1 = , B2 = and B3 =. Since W (B1 , B2 , B3 ) is the 10 i 0 unit sphere [1, 10], W (A1 , A2 , A3 ) is the unit ball in R3 and it is not hard to deduce (a) and (b), and (c) can be obtained by direct computation:
(a) wA1 ,A2 ,A3 (0) = n 2 ,
(b) wA1 ,A2 ,A3 ( ) = 1 for | | = 1 , (c) wA1 ,A2 ,A3 ( ) = n 1 for 0 < | | < 1. D EFINITION 2.4. Let A1 , A2 , A3 Hn. A point W (A1 , A2 , A3 ) is called
1. an extreme point if is not in an open line segment that is contained in W (A1 , A2 , A3 ). 2. a sharp point if is the intersection point of three distinct supporting planes of W (A1 , A2 , A3 ).

0n2, A2 = 0 1

01 0n2, A3 = 10

0 i 0n2. i 0

Evidently an extreme point W (A1 , A2 , A3 ) is a boundary point. However, a boundary point is not necessarily an extreme point. For example, W (B1 , B2 , B3 ) = conv{(2, 0, 0),W (A1 , A2 , A3 )} if B1 = Aand B2 = Aand B3 = A, where A1 = , A2 = 01 , A3 = i. i 0
An extreme point may not be a sharp point (see the boundary points in Example 2.3). Binding and Li [4, Denition 2.4] introduced conical point of any subset Rk. Since W (A1 , A2 , A3 ) is convex when n 3 , sharp points are conical points and the following result is an analog to Theorem 1.1 and can be deduced from [4, Proposition 2.5 or Theorem 2.7]. T HEOREM 2.5. (Binding and Li) Let A1 , A2 , A3 Hn with n 3 and W (A1 , A2 , A3 ). Then is a sharp point if and only if there is a unitary matrix U Cnn such that U AiU = i Im Bi , i = 1, 2, 3 (2.1) with = (1 , 2 , 3 ) convW (B1 , B2 , B3 ). W = W (A1 , A2 , A3 ) := {x Cn : (x A1 x, x A2 x, x A3 x) = x x}.
T HEOREM 2.6. Let A1 , A2 , A3 Hn with n
1. is an extreme point if and only if is a boundary point and W is a subspace of Cn. In particular, if W (A1 , A2 , A3 ) is degenerate, then is an extreme point if and only if W is a subspace of Cn. 2. if is a non-extreme boundary point and if P is a supporting plane of W (A1 , A2 , A3 ) at , then W zPW (A1 ,A2 ,A3 )Wz. (2.2) In addition, if P W (A1 , A2 , A3 ) is (i) a at convex set containing as a relative interior point, or (ii) a line segment, then W = zPW (A1 ,A2 ,A3 )Wz ; (iii) a at convex set S in which is not a relative interior point of S , then W = zLWz , where L is the longest line segment in W (A1 , A2 , A3 ) that contains . In case of (i) or (ii), W = Cn if and only if W (A1 , A2 , A3 ) P.

Notice that W is homogenous so that W = W + W. It is natural to ask whether Theorem 1.2 (1) can be extended to W (A1 , A2 , A3 ). Unfortunately, for Example 2.3 with n = 3 , W0 is a 1 -dimensional subspace, but 0 is clearly not an extreme point of the unit ball W (A1 , A2 , A3 ). But the problem can be resolved in the following theorem. It generalizes the rst two parts of Theorem 1.2. 3 and W (A1 , A2 , A3 ). Then
Similar to the W (A) case, for each W (A1 , A2 , A3 ) we dene
On the other hand, pick arbitrary z P W (A1 , A2 , A3 ) with z = . (i) If P W (A1 , A2 , A3 ) is a at convex set in which is an interior point, then there exists z = z and (z, z ) P W (A1 , A2 , A3 ). Pick x Wz and x Wz. Clearly x, x are linearly independent since z = z. Let A1 , A2 , A3 be the compressions of A1 , A2 , A3 onto x, x respectively. As an ellipsoid containing z, z , W (A1 , A2 , A3 ) W (A1 , A2 , A3 ) must be degenerate. Thus by Theorem 2.2(1b), there are two linearly independently vectors u, v x, x Sn1 such that f (u) = f (v) = . But x x, x = u, v W +W = W. So Wz W for all z P W (A1 , A2 , A3 ). So we have the other inclusion W zPW (A1 ,A2 ,A3 )Wz. (ii) If P W (A1 , A2 , A3 ) is a line segment, then there exists z = z and is in the open segment (z, z ). Similar to (i), W (A1 , A2 , A3 ) W (A1 , A2 , A3 ) is a line segment and apply Theorem 2.2(1b). (iii) Project W (A1 , A2 , A3 ) onto the hyperplane H (spanned by p = (p1 , p2 , p3 ) and (q1 , q2 , q3 )) that is orthogonal to the line L. So L is projected into a point H. Hence is an extreme point of W (A1 + iA2 ) where A1 = p1 A1 + p2 A2 + p3 A3 and A2 = q1 A1 + q2A2 + q3 A3. So M is a subspace by Theorem 1.2(1). Moreover W
Since P, W zPW (A1 ,A2 ,A3 )Wz we have (2.2)

W zPW (A1 ,A2 ,A3 )Wz.

Proof. (1) Evidently all extreme points of W (A1 , A2 , A3 ) are boundary points. Suppose that is a boundary point of W (A1 , A2 , A3 ). Since W (A1 , A2 , A3 ) is convex [2], there is a supporting plane P of W (A1 , A2 , A3 ) at . Let p = (p1 , p2 , p3 ) be a unit vector perpendicular to P. Project W (A1 , A2 , A3 ) onto p. So := pT = p+ p22 + p33 is an extreme point of the classical numerical range W (A) (a line segment), where A := p1 A1 + p2 A2 + p3 A3. So must be a maximal or minimal eigen value of the Hermitian matrix A. Thus M (A) is an eigenspace of A. In addition if is an extreme point of W (A1 , A2 , A3 ), we have W (A1 , A2 , A3 ) = M (A). Suppose that is a boundary point and W is a subspace of Cn. If were not an extreme point, there would exist distinct , W (A1 , A2 , A3 ) such that ( , ). Let u, v Sn1 such that = f (u), = f (v) and let Ai ( i = 1, 2, 3 ) denote the compression of Ai onto the 2 -dimensional subspace u, v. So f (u), f (v) W (A1 , A2 , A3 ) W (A1 , A2 , A3 ) and would be contained in the convex hull of the ellipsoid W (A1 , A2 , A3 ). Since is a boundary point of W (A1 , A2 , A3 ), this forces W (A1 , A2 , A3 ) to be an elliptical disk (possibly degenerate but not a point since = ). Hence W (A1 , A2 , A3 ) and by Theorem 2.2(1), there would exist two linearly independent unit vectors x, y u, v such that f (x) = f (y) = . But W is a subspace and clearly x, y W. So u u, v = x, y W and we would have = f (u) = which is absurd. (2) Suppose that is a non-extreme boundary point. Let P be a supporting plane of W (A1 , A2 , A3 ) at . Referring to the rst paragraph of the proof of (1), M (A) is an eigenspace of A and M (A) = zPW (A1 ,A2 ,A3 )Wz.

zLWz = M so that W zLWz. Then use the argument in (ii) to have W zLWz since L. C OROLLARY 2.7. Let A1 , A2 , A3 Hn with n wA1 ,A2 ,A3 ( ) = 1 , then is an extreme point. 4 and W (A1 , A2 , A3 ). If
E XAMPLE 2.8. With respect to Theorem 2.6(2)(iii) it is possible that there is only one supporting plane P at such that S := P W (A1 , A2 , A3 ) is a at convex set and is not in the relative interior of S R3. For example, if A1 = A2 = = B 0 B 0, 1 i 1 i = C 0 C 0, i 1 i 1
A3 = I3 (I3), then W (A1 , A2 , A3 ) = conv {W (B,C, I2 ),W (B,C, I2 ), L} , where L is the line segment joining (0, 0, 1) and (0, 0, 1). Notice that W (B,C, I2 ) is the circular disk {(x, y, 1) : (x 1)2 + (y 1)2 = 1} so that W (A1 , A2 , A3 ) is the convex hull of the cylinder K := {(x, y, z) : 1 z
but is not an extreme point. The vector x S5 such that = (x A1 x, x A2 x, x A3 x) must be of the form x = 1 (u, v), u, v S2 in view of the zero third coordinate. So 2 u = 1 (1, i, 0) and v = 2 (1, i, 0), |1 | = |2 | = 1. Thus x = 1 (1 , i2 , 0, 2 , i2 , 0). So 2 W = W = span {(1, i, 0, 0, 0, 0), (0, 0, 0, 1, i, 0)}.
For any point := (t, 0, 0) for 0 < t < 1 , W is not contained in W. So (2.3) does not hold though (2.2) is true. 3. Joint numerical range of two real symmetric matrices
Brickman [5] (also see [12]) studied the real analog of the numerical range of A Cnn : V (A) = {xT Ax : x Rn , xT x = 1} 3. In addition V (A) is an ellipse (possibly
and proved that V (A) is convex when n degenerate) when n = 2. Indeed [12]
S := {(x, 0, z) : 0 x 1, 1 z W (A) = V (A)
and L. The point := (1, 0, 0) K lies on the edge of the at portion 1} W (A1 , A2 , A3 )

1, (x 1)2 + (y 1)2 = 1}

8 where

A iA. iA A 2. Clearly

So convexity of W (A) follows from the convexity of V (A) when n V (A) = V ( A + AT ) 2
so that we can restrict our study to symmetric A. Moreover if A = AT , then W (A) = convV (A) [12] and in particular, if n 3 , then W (A) = V (A). See [13] for an interesting unied treatment for W (A), W (A1 , A2 , A3 ) and V (A); [11] for more general notions in the context of semisimple Lie algebras; and [10] for related results. The following is a list of some basic properties of V (A), similar to those of W (A). L EMMA 3.1. Let A Cnn be symmetric. 1. V ( A + In) = V (A) + , , C. 2. V (OT AO) = V (A) for any orthogonal matrix O O(n). 3. V (A) = { } if and only if A = In.

4. If B Cmm is a principal submatrix of A, then V (B) V (A). 5. (McIntosh [12]) convV (A) = W (A).
D EFINITION 3.2. Let A Cnn be symmetric and V (A). The multiplicity of , denoted by vA ( ), is the maximal number of linearly independent vectors x Sn1 := Sn1 Rn such that xT Ax = . R The following result and its proof are similar to Theorem 2.2. T HEOREM 3.3. Let A Cnn be symmetric. 1. Suppose n = 2 and write A = a11 a12 a21 a22
(a) V (A) = { } if and only if A = I2 for some C. In this case, vA ( ) = 2. (b) V (A) is a nondegenerate line segment [ , ] if and only if either a11 = a22 or a12 = 0 , but not both. Moreover vA ( ) = vA ( ) = 1 and vA ( ) = 2 if ( , ). (c) V (A) is a nondegenerate ellipse if and only if both a11 = a22 and a12 = 0. In this case, vA ( ) = 1 for all V (A).
2. When n 3 , if IntRV (A), then vA ( ) The lower bound n 2 is best possible.

with a12 = a21.

n 2 and may not be a constant.
Proof. (1) When n = 2 , V (A) = {a11 cos2 + (a12 + a21) cos sin + a22 sin2 : 0 < 2 } (a12 + a21) (a11 + a22) (a11 a22) ={ cos 2 + sin 2 + : 0 < 2 } 2 which is an ellipse with center (tr A)/2. We only need to consider the multiplicities for the following cases since the rest is straightforward computation. (b) V (A) is a nondegenerate line segment [ , ]. If is one of the endpoints, it corresponds to two values of [0, 2 ) of difference . So the corresponding x are negative to each other. Hence vA ( ) = 1. If ( , ), there are four desired values of which yield two linearly independent vectors, and hence vA ( ) = 2. (c) V (A) is a nondegenerate ellipse. Each V (A) corresponds to two values of [0, 2 ) of difference . Hence vA ( ) = 1. (2) The statement is trivial for n = 3. Suppose n 4 and IntRV (A). Suppose on the contrary that vA ( ) = k < n 2. Let g : Sn1 C be the map dened by R

(a) vA (0) = n 2 ,

Let {x1 ,. , xk } Sn1 be a (maximal) linearly independent set such that = g(xi ), R i = 1,. , k. Choose u Sn1 such that x1 ,. , xk , u are linearly independent and clearly R = g(u). Because IntRV (A) and V (A) is convex [12], there is v Sn1 such that R (a) g(v) = and g(v) = g(u), (b) the line segment L = [g(u), g(v)] V (A), and (c) L. Since g(u) = g(v), u, v are linearly independent. Since k < n 2 , there is w Sn1 and w u, v, x1 ,. , xk. Let A denote the compression of A onto the three R dimensional subspace u, v, w. Since V (A) is convex, there would exist a unit vector y u, v, w and g(y) = . Write y = u + v + w for , , R. Notice that and cannot be both zero, otherwise g(y) = g(v) = . But then y, x1 ,. , xk would be linearly independent, a contradiction. The following example shows that the lower bound n 2 is best possible and vA ( ) may not be a constant. E XAMPLE 3.4. Let A =

1 i i 1

0 Cnn , where n
1 i. Then (a) i 1 and (b) of the following can be deduced immediately and (c) can be computed directly: unit disk since it is the direct sum of V (B) and the origin where B =
(b) vA ( ) = n 1 if 0 < | | < 1 , and (c) vA ( ) = 1 if | | = 1.

g(x) := xT Ax,

x Sn1. R

3. Then V (A) is the

D EFINITION 3.5. Let A Cnn be symmetric and V (A). We dene V = V (A) := {x Rn : xT Ax = xT x}. The ideas of conical points in [4] can be applied to V (A) since V (A) can be identied as the joint numerical range of A1 and A2 : {(xT A1 x, xT A2 x) : xT x = 1, x Rn } R2 , where A = A1 +iA2 and A1 and A2 are real symmetric matrices. Adapting the approach in [4] yields the following result. We now provide a different proof and remark that the approach applies to Theorem 2.5. T HEOREM 3.6. (Binding and Li) Let A Cnn be symmetric and V (A). Then is a sharp point if and only if there is an orthogonal matrix O O(n) such that OT AO = Im B, with convV (B). Proof. Suppose that there is an orthogonal matrix O O(n) such that OT AO = Im B, with convV (B) and B C(nm)(nm) is symmetric. By Lemma 3.1 Since is not contained in the compact convex set convV (B), is a sharp point of V (A). Conversely suppose that is a sharp point of V (A). Without loss of generality, we may assume that = a11 otherwise we perform an orthogonal similarity on A. a11 a1i of A satises For each i = 2,. , n , the principal submatrix Ai := ai1 aii V (Ai ) V (A). Since = a11 is a sharp point, V (Ai ) must be a line segment (possibly degenerate). By Lemma 3.3, a1i = ai1 must be zero. Thus A = A, where A C(n1)(n1) is symmetric, and in particular is an eigenvalue of A. So V (A) = conv{ ,V (A)}. Since is a sharp point, is not contained in the relative interior of So is either in V (A) or not (even when A C22 with V (A) an ellipse). convV (A). it forces that convV (A) since is a sharp point and we are done. If V (A), Otherwise, repeat the argument on A to arrive at the desired conclusion. Similar to W (A1 , A2 , A3 ), Theorem 1.2(1) cannot be extended to V (A) by observing Example 3.4 with n = 3 : V0 is a 1 -dimensional subspace, but 0 is not an extreme point of the unit disk V (A).

V = zLVz ,

T HEOREM 3.7. Let A Cnn be symmetric with n
1. is an extreme point if and only if is a boundary point and V is a subspace of Rn.

2. if is a non-extreme boundary point, then where L is the supporting plane of V (A), passing through . In this case V = Rn if and only if V (A) L.
V (A) = V (OT AO) = conv { ,V (B)}.

3 and V (A). Then

V zPV (A)Vz.
On the other hand, suppose z L V (A) and z = . Then there exists z = z and is in the open segment (z, z ). Pick x Vz and x Vz. Clearly x, x are linearly in dependent since z = z. Let A be the compressions of A onto x, x respectively. Then V (A) must degenerate since [z, z ] W (A) and is a bound the ellipse V (A) ary point of V (A). Thus by Theorem 3.3(1b), there are two linearly independently vectors u, v x, x Sn1 such that f (u) = f (v) = . But x x, x = u, v = R V + V = V. So Vz V for all z L V (A). So we have the other inclusion W zLV (A)Vz. We remark that the above technique can be used to prove Theorem 1.2 and is different from the approach of Embry in [8]. C OROLLARY 3.8. Let V Cnn with n is an extreme point.
Acknowledgement. Thanks are given to C.-K. Li for helpful comments and references. In section 3, W (A) = W ( 1 (A + AT )) for any A Cnn. The referee asked the 2 following question: Let A Cnn with n 3. Does there exist a symmetric complex matrix S Cnn satisfying W (A) = W (S)?
Proof. (1) All extreme points of V (A) are boundary points. Suppose that = 1 + i2 ( 1 , 2 R) is an extreme point of V (A). Since V (A) is convex [5, 12], there is a supporting line L of V (A) at . Let p := p1 + ip2 C ( p1 , p2 R) be a unit vector perpendicular to L. Project V (A) onto p. If A = A1 + iA2 is the Hermitian decomposition, then := p+ pis an extreme point of V (p1 A1 + p2 A2 ) (a line segment), where p1 A1 + p2 A2 is clearly real symmetric. By the spectral theorem for real symmetric matrices, V (p1 A1 + p2 A2 ) is the eigenspace of p1 A1 + p2 A2. Since is an extreme point of V (A), V (A) = V (p1 A1 + p2 A2 ) and is a subspace of Rn. Suppose that V (A) is a boundary point and V is a subspace of Rn. If were not an extreme point, there would exist distinct , V (A) such that ( , ). Let u, v Sn1 such that = g(u), = g(v) and let A denote the compression of A R onto the 2 -dimensional subspace u, v. So g(u), g(v) V (A) V (A) and would be Since is a boundary point of V (A), contained in the convex hull of the ellipse V (A). this forces V (A) to be a line segment (but not a point since = ). Hence V (A) and thus there would exist two linearly independent unit vectors x, y u, v such that g(x) = g(y) = . Thus u, v = x, y. But V is a subspace and clearly x, y V. So u x, y V and we would have = g(u) = which is absurd. (2) Suppose that is a non-extreme boundary point of V (A). Let L be the supporting line of V (A) at . Referring to the rst paragraph of the proof of (1), V (p1 A1 + p2 A2 ) is the eigenspace of p1 A1 + p2 A2 and V (p1 A1 + p2 A2 ) = zLV (A)Vz. Since L , clearly V zPV (A)Vz and we have

4 and V (A). If vA ( ) = 1 , then
W.S. C HEUNG , X UHUA L IU AND T.Y. TAM REFERENCES
(Received November 20, 2009)

EP R PR

Operators and Matrices
www.ele-math.com oam@ele-math.com
[1] Y. H. AU -Y EUNG AND Y. T. P OON, A remark on the convexity and positive deniteness concerning Hermitian matrices, Southeast Asian Bull. Math., 3 (1979), 8592. [2] Y. H. AU -Y EUNG AND N. K. T SING, An extension of the Hausdorff-Toeplitz theorem, Proc. Amer. Math. Soc., 89 (1983), 215218. [3] Y. H. AU -Y EUNG AND N. K. T SING, Some theorems on the numerical range, Linear and Multilinear Algebra, 15 (1984), 215218. [4] P. L. B INDING AND C. K. L I , Joint ranges of Hermitian matrices and simultaneous diagonalization, Linear Algebra Appl., 151 (1991), 157168. [5] L. B RICKMAN, On the eld of values of a matrix, Proc. Amer. Math. Soc., 12 (1961), 6166. [6] C. D AVIS , The Toeplitz-Hausdorff theorem explained, Canad. Math. Bull., 14 (1971), 245246. [7] W. F. D ONOGHUE , J R., On the numerical range of a bounded operator, Michigan Math. J., 4 (1957), 261263. [8] M. R. E MBRY, The numerical range of an operator, Pacic J. Math., 32 (1970), 647650. [9] R. A. H ORN AND C. R. J OHNSON, Topics in Matrix Analysis, Cambridge University Press, 1991. [10] C. K. L I AND Y. T. P OON, Convexity of the joint numerical range, SIAM J. Matrix Anal. Appl., 21 (1999), 668678. [11] C. K. L I AND T. Y. TAM, Numerical ranges arising from simple Lie algebras, J. Canad. Math. Soc., 52 (2000), 141171. [12] A. M C I NTOSH, The Toeplitz-Hausdorff theorem and ellipticity conditions, Amer. Math. Monthly, 85 (1978), 475477. [13] Y. T. P OON, Generalized numerical ranges, joint positive deniteness and multiple eigenvalues, Proc. Amer. Math. Soc., 125 (1997), 16251634. [14] J. G. S TAMPFLI , Extreme points of the numerical range of hyponormal operators, Michigan Math. J., 13 (1966), 8789. [15] F. U HLIG, An inverse eld of values problem, Inverse Problems, 24 (2008), 055019, 19 pages.
Xuhua Liu Department of Mathematics and Statistics Auburn University AL 368495310, USA e-mail: xzl0002@auburn.edu Tin-Yau Tam Department of Mathematics and Statistics Auburn University AL 368495310, USA e-mail: tamtiny@auburn.edu
Wai-Shun Cheung Department of Mathematics University of Hong Kong Hong Kong
e-mail: cheungwaishun@gmail.com

The Inhomogeneous system Ax = y, y = 0
Definition: The system Ax = y is inhomogeneous if its not homogeneous. Mathematicians love denitions like this! It means of course that the vector y is not the zero vector. And this means that at least one of the equations has a non-zero right hand side. As an example, we can use the same system as in the previous lecture, except well change the right hand side to something non-zero: x1 + 2x2 x4 = 1 2x1 3x2 + 4x3 + 5x4 = 2. 2x1 + 4x2 2x4 = 3 Those of you with sharp eyes should be able to tell at a glance that this system is inconsistent that is, there are no solutions. Why? Were going to proceed anyway because this is hardly an exceptional situation. The augmented matrix is 2. (A|y) = We cant discard the 5th column here augmented matrix is 0 And the reduced echelon form is 3 0. 1 The third equation, from either of these, now reads 0x1 + 0x2 + 0x3 + 0x4 = 1, or 0 = 1. This is false! How can we wind up with a false statement? The actual reasoning that led us here is this: If the original system has a solution, then performing elementary row operations will give us an equivalent system with the same solution. But this equivalent system of equations is inconsistent. It has no solutions; that is no choice of x1 ,. , x4 satises the equation. So the original system is also inconsistent. 1 since its not zero. The row echelon form of the 3 4.
In general: If the echelon form of (A|y) has a leading 1 in any position of the last column, the system of equations is inconsistent. Now its not true that any inhomogenous system with the same matrix A is inconsistent. It depends completely on the particular y which sits on the right hand side. For instance, if , y= 2 then (work this out!) the echelon form of (A|y) is 0 and the reduced echelon form is 3 4. 0 Since this is consistent, we have, as in the homogeneous case, the leading variables x1 and x2 , and the free variables x3 and x4. Renaming the free variables by s and t, and writing out the equations solved for the leading variables gives us x1 x2 x3 x4 = = = = 8s + 7t 7 4s 3t + 4. s t
This looks like the solution to the homogeneous equation found in the previous section except for the additional scalars 7 and + 4 in the rst two equations. If we rewrite this using vector notation, we get 8 x1 3 4 4 x2 : s, t R + +t =s xI = 0 0 1 x3 0 x4 Compare this with the general solution xH to the homogenous equation found before. Once again, we have a 2-parameter family (or set) of solutions. We can get a particular solution 2
by making some specic choice for s and t. For example, taking s = t = 0, we get the particular solution xp = 0. 0 We can get other particular solutions by making other choices. Observe that the general solution to the inhomogeneous system worked out here can be written in the form xI = xH + xp. In fact, this is true in general: Theorem: Let xp and xp be two solutions to Ax = y. Then their dierence xp xp is a solution to the homogeneous equation Ax = 0. The general solution to Ax = y can be written as xI = xp + xH where xH denotes the general solution to the homogeneous system. Proof: Since xp and xp are solutions, we have A(xp xp ) = Axp A p = y y = 0. x So their dierence solves the homogeneous equation. Conversely, given a particular solution xp , then the entire set xp + xH consists of solutions to Ax = y: if z belongs to xH , then A(xp + z) = Axp + Az = y + 0 = y and so xp + z is a solution to Ax = y. Going back to the example, suppose we write the general solution to Ax = y in the vector form xI = {sv1 + tv2 + xp , s, t R} , where 8 4 , v2 = 3 , and xp = 4 v1 = 0 0 Any dierent choice of s, t, for example taking s = 1, t = 1, gives another solution: xp = 1. 1 We can rewrite the general solution as xI = (s 1 + 1)v1 + (t 1 + 1)v2 + xp = (s 1)v1 + (t 1)v2 + xp. v2 + xp = sv1 + t 3

As s and t run over all possible pairs of real numbers we get exactly the same set of solutions as before. So the general solution can be written as xp + xH as well as xp + xH ! This is a bit confusing unless you recall that these are sets of solutions, rather than single solutions; (, t) s and (s, t) are just dierent sets of coordinates. But running through either set of coordinates (or parameters) produces the same set. Remarks Those of you taking a course in dierential equations will encounter a similar situation: the general solution to a linear dierential equation has the form y = yp + yh , where yp is any particular solution to the DE, and yh denotes the set of all solutions to the homogeneous DE.

xp + z xH 0 z

Figure 1: The lower plane (the one passing through 0) represents xH ; the upper is xI. Given the particular solution xp and a z in xH , we get another solution to the inhomogeneous equation. As z varies in xH , we get all the solutions to Ax = y. We can visualize the general solutions to the homogeneous and inhomogeneous equations weve worked out in detail as follows. The set xH is a 2-plane in R4 which goes through the origin since x = 0 is a solution. The general solution to Ax = y is obtained by adding the vector xp to every point in this 2-plane. Geometrically, this gives another 2-plane parallel to the rst, but not containing the origin (since x = 0 is not a solution to Ax = y unless y = 0). Now pick any point in this parallel 2-plane and add to it all the vectors in the 2-plane corresponding to xH. What do you get? You get the same parallel 2-plane! This is why xp + xH = xp + xH. Exercise: Using the same example above, let xp be the solution obtained by taking s = 2, t = 1. Verify that both xp xp and xp xp are solutions to the homogeneous equation.