Preservers of eigenvalue inclusion sets

Preservers of eigenvalue inclusion sets Jim Hartman, Aaron Herman, and Chi-Kwong Li Abstract For a square matrix A, let S(A) be an eigenvalue inclusion set such as the Gershgorin region, the Brauer region in terms of Cassini ovals, and the Ostrowski region. Characterization is obtained for maps Φ on n n matrices satisfying S(Φ(A) Φ(B)) = S(A B) for all matrices A and B. From these results, one can deduce the structure of additive or (real) linear maps satisfying S(A) = S(Φ(A)) for every matrix A. Keywords: Preservers, eigenvalue inclusion sets, Gershgorin region, Brauer region, Cassini ovals, Ostrowski region. AMS Subject Classification: 15A86, 15A18. 1 Introduction Motivated by pure and applied problems, researchers often need to understand the eigenvalues of matrices. For example, in numerical analysis or population dynamics, a square matrix A satisfies lim m A m = 0 if and only if each eigenvalue has modulus less than 1; in stability theory of differential equations, the solution of the system of differential equations x = Ax is stable if and only if all the eigenvalues of A lie in the left half plane; in the study of quadratic forms a Hermitian matrix is positive definite if and only if all the eigenvalues are positive real; see [2]. However, it is sometimes difficult to compute the eigenvalues efficiently and accurately, due to reasons such as the dimension of the matrix is too high, there are numerical or measuring errors in the entries, etc. So, researchers consider eigenvalue inclusion sets; see [2, 5]. For instance, the well known Gershgorin theorem asserts that the eigenvalues of a matrix lie in the union of circular disks centered at the diagonal entries with radii determined by the off diagonal entries (see Section 2). These eigenvalue inclusion sets give effective ways to estimate the location of the eigenvalues for matrices. To further improve the estimate, one may apply simple transformations such as diagonal similarities to a matrix to get better or easier estimates of the location of the eigenvalues of a given matrix. Department of Mathematics and Computer Science, The College of Wooster, Wooster, OH 44691 (hartman@wooster.edu). Research was done while this author was spending his research leave at the College of William and Mary in the academic year 2008-2009. Department of Mathematics, College of William and Mary, Williamsburg, VA 23185 (apherm@wm.edu). Research was partially supported by an NSF CSUMS grant. Department of Mathematics, College of William and Mary, Williamsburg, VA 23185 (ckli@math.wm.edu). Research was partially supported by an NSF grant and the William and Mary Plumeri award. Li is an honorary professor of the University of Hong Kong and an honorary professor of Taiyuan University of Technology. 1

In this paper, we study maps on matrices leaving invariant the eigenvalue inclusion sets such as the Gershgorin region, the Brauer region in terms of Cassini ovals, and the Ostrowski region (see Sections 2, 3, and 4, respectively). The study of maps on matrix spaces leaving invariant some properties, functions or subsets is known as preserver problems. Early study on the subject focused on linear preservers; i.e., linear maps having the preserving properties; see [3] and its references. Recently, researchers work on more general preservers; see [4] and its references. In our study, let S(A) be an eigenvalue inclusion set for the matrix A. We characterize maps Φ on square matrices satisfying S(Φ(A) Φ(B)) = S(A B) for any two matrices A and B. It is shown that such Φ are real affine maps. From our results or proofs, one can deduce the structure of additive or (real) linear maps satisfying the following: S(A) = S(Φ(A)). (1.1) Note that if one just assumes that a map Φ satisfies (1.1) for every matrix A, the structure of Φ can be quite arbitrary. For instance, one can partition the set of matrices into equivalence classes so that two matrices A and B belong to the same class if S(A) = S(B). If Φ sends each of these classes back to itself, then Φ satisfies (1.1) for every matrix A. We will always assume that n 2 to avoid trivial consideration. The following notation and definitions will be used in our discussion. M n : the set of n n complex matrices. {E 11, E 12,..., E nn }: the standard basis for M n. P n : the group of permutation matrices in M n. DU n : the group of diagonal unitary matrices in M n. GP n = {DP : D DU n, P P n }: the group of generalized permutation matrices in M n. 2 Gershgorin Regions In this section, we consider preservers of the Gershgorin region of A = (a ij ) M n defined by G(A) = n k=1 G k(a), where for k = 1,..., n, G k (A) = {µ C : µ a kk R k } with R k = a kj. It is known that G(A) contains all the eigenvalues of A; one may see [2, Chapter 6] for general properties of G(A). Theorem 2.1 A map Φ : M n M n satisfies G(Φ(A) Φ(B)) = G(A B) for all A, B M n (2.1) if and only if Φ is a composition of maps of the following forms: 2

(1) A A 1 Q 1. for A M n with rows A 1,..., A n, where Q 1,..., Q n GP n are such that A n Q n the (j, j) entry of Q j is 1 for j = 1,..., n. (2) A P AP t + S, where P P n and S M n. (3) A = (a rs ) (ψ rs (a rs )), where ψ rs is either the complex conjugation map z z or identity map z z, and the latter always occurs if r = s. It is easy to verify that maps of the form (1), (2), and (3) in the theorem indeed satisfy (2.1). Hence the sufficiency of the theorem is clear. To prove the necessity part of the theorem, we need some lemmas. Lemma 2.2 Let w = e i2π/n and { } 1 w k Γ = : j {1,..., n 1}, k {1,..., n}. (2.2) 1 wj Suppose µ C \ Γ. Then R P n is such that the set of entries of the vector µ(w, w 2,..., w n ) (w, w 2,..., w n )R equals {(µ 1)w j : 1 j n} if and only if R = I n. Proof. The sufficiency is clear. To verify the necessity part, assume that the kth entry of the vector µ(w, w 2,..., w n ) (w, w 2,..., w n )R equals µw k w i with k i. Then µw k w i = µw j w j for some j {1,..., n} so that µ(w j w k ) = w j w i. If j = i, then j k and hence µ = 0 = (1 w n )/(1 w); if j i, then µ = (1 w k j )/(1 w i j ). This contradicts the fact that µ / Γ. The next lemma should be known to researchers on distance preserving maps. We include a proof for completeness. Lemma 2.3 Let V = C 1 m. A map f : V V satisfies l 1 (f(x) f(y)) = l 1 (x y) for all x, y V if and only if there is z V and Q GP n such that f has the form (x 1,..., x m ) (f 1 (x 1 ),..., f m (x m ))Q + z, where f j is either the identity map µ µ or the complex conjugation µ µ. Proof. The sufficiency is clear. We consider the converse. We divide the proof into two assertions. Assertion 1 Let B = {x V : l 1 (x) 1} and E = {x B : x (u + v)/2 with different u, v B} be the set of extreme points of B. Then x E if and only if x has only one nonzero entry with modulus 1. Proof. Let x = (x 1,..., x m ) B. We consider three possible cases. 3

Case 1 Suppose x has a single nonzero entry at the jth position with x j = 1. If u = (u 1,..., u n ), v = (v 1,..., v n ) B satisfy x = (u + v)/2, then x j = (u j + v j )/2. Since x j = 1 u j, v j, we see that u j = v j = x j. Because u, v B, all other entries of u and v must be zero. Thus, x = u = v. Hence x E. Case 2 Suppose x has a single nonzero entry at the jth position such that x j < 1. Then there exists a δ 0 such that x j + δ, x j δ 1. We can then let u = (0,..., 0, x j + δ, 0,..., 0), v = (0,..., 0, x j δ, 0,..., 0) B such that x = (u + v)/2. Thus, x / E. Case 3 Suppose x B has at least two nonzero entries, say, at the jth and kth positions such that x j = ρ j e it j, x k = ρ k e it k and ρ j, ρ k (0, 1), where ρ j + ρ k 1. Then there exists u = (u 1,..., u m ) with two nonzero entries, namely, u j = δe it j and u k = δe it k with min{ρ j, ρ k } > δ > 0 so that x + u, x u B are different vectors and x = [(x + u) + (x u)]/2 / E. Combining these three cases we see that x E if and only if x has only one nonzero entry with modulus 1. Assertion 2 The map g(x) = f(x) f(0) is real linear and satisfies g(e) = E, and f has the asserted form with f(0) = z. Proof. By the result in [1], we know that g(x) = f(x) f(0) is real linear. Note that g(x) = 0 if and only if 0 = l 1 (g(x)) = l 1 (x), i.e., x = 0. Hence g is a real linear injective map, and therefore is bijective. As a result, g(b) = B. Clearly, x = (u + v)/2 for distinct u, v B if and only if g(x) = (g(u) + g(v))/2 with distinct g(u), g(v) B. Thus, g(e) = E. Let {e 1, e 2,..., e m } be the standard basis for C 1 m. Because g(e) = E, we see that g(e j ) = µ j e rj for some r j {1,..., m} and µ j C with µ j = 1. Since l 1 (e j + γe k ) = l 1 (g(e j ) + γg(e k )) for all γ R and j k, we see that (r 1,..., r m ) is a permutation of (1,..., m). Now, g(ie j ) = ν j e sj for some s j {1,..., m} and ν j C with ν j = 1. Since l 1 (e j + γie j ) = l 1 (g(e j ) + γg(ie j )) for all γ R, we see that s j = r j and ν j {iµ j, iµ j }. Thus, g has the form (x 1,..., x m ) (f 1 (x 1 ),..., f m (x m ))Q, where f 1,..., f m and Q satisfy the conclusion of the lemma. Thus, f has the asserted form. Proof of Theorem 2.1. The sufficiency is clear, as remarked before. We consider the necessity. Replacing Φ by the map X Φ(X) Φ(0), we may assume that Φ(0) = 0 and G(Φ(X)) = Φ(X) for all X M n in addition to the assumption that G(Φ(A) Φ(B)) = G(A B) for all A, B M n. Let Γ be defined as in Lemma 2.2. Assertion 1 Let D = diag (w, w 2,..., w n 1, 1) for w = e i2π/n. There is a permutation matrix P such that P Φ(D)P t = D. Furthermore, if µ C \ Γ, then P Φ(µD)P t = µd. To verify the assertion, note that G(Φ(µD)) = G(µD) = {µw j : 1 j n}. Hence, Φ(µD) has digonal entries µw j with j = 1,..., n, and all off-diagonal entries equal to zero. Thus, there is a permutation matrix P such that P Φ(D)P t = D. We may assume that Φ(D) = D. Otherwise, replace Φ by the map A P t Φ(A)P. Suppose µ C \ Γ. Then G(Φ(µD) Φ(D)) = G(µD D) = {(µ 1)w j : j = 1,..., n}. 4

Thus, the vector of diagonal entries of Φ(µD) Φ(D) equals µ(w, w 2,..., w n ) (w, w 2,..., w n )R for some R P n, and the entries constitute the set {(µ 1)w j : 1 j n}. By Lemma 2.2, the (j, j) entry of Φ(µD) is µw j for each j = 1,..., n, i.e., Φ(µD) = µd. Assertion 2 Assume that P in the conclusion in Assertion 1 is I n. Then Φ(W k ) W k for k = 1,..., n, where W k = a j E kj : a 1,..., a k 1, a k+1,..., a n C. Moreover, define Φ k : C 1 (n 1) C 1 (n 1) by Φ k (a 1,..., a k 1, a k+1,..., a n ) = (b 1,..., b k 1, b k+1,..., b n ) if Φ( a je kj ) = b je kj. Then Φ k satisfies the conclusion of Lemma 2.3 with z = 0. To prove the assertion, let A = a je kj W k and Φ(A) = B = (b ij ). Let µ (0, ) \ Γ satisfy µ 1 w > a kj = R k. Then Φ(µD) = µd by Assertion 1. Moreover, since G(B) = G(A) = {z C : z R k }, Thus, min{µ w i w j : 1 i < j n} = µ 1 w > R k max{ b jj : 1 j n}. (2.3) G(µD B) = G(Φ(µD) Φ(A)) = G(µD A) = {µw j : j k} {γ : γ µw k R k }. For i = 1,..., n, the (i, i) entry of µd B equals µw i b ii is the center of one of the disks in G(µD A). Thus, µw i b ii = µw j implies that µ w i w j = b ii < µ 1 w by (2.3). It follows that i = j and b ii = 0. Thus, µd B has diagonal entries µw, µw 2,..., µw n 1, µ. Since G(µD B) = G(µD A), we see that only the kth row of B can have nonzero off diagonal entries, and b kj = R k. Now, define Φ k as in Assertion 2. Since G(Φ(A 1 ) Φ(A 2 )) = G(A 1 A 2 ) for any A 1, A 2 W k and Φ(0) = 0, we see that l 1 (Φ k (u) Φ k (v)) = l 1 (u v) for all u, v C 1 (n 1). Thus, the last statement of Assertion 2 follows. Assertion 3 The map Φ has the asserted form. By Assertion 2, we may compose the map Φ with maps of the form (1) (3) described in the theorem and assume that (I) Φ(µD) = µd whenever µ C \ Γ, and (II) Φ(X) = X whenever X n k=1 W k. Under conditions (I) and (II), we will show that Φ(A) = A for each A M n. First, we consider the special case when A = µd + B with µ C \ Γ and B W k where G(A) consists of n disjoint disks with at most one of them having positive radius. Suppose Φ(A) = C = (c ij ). Let ν > 0 be 5

such that ν, νµ / Γ and G(νµD A) = G(µ(ν 1)D B) consists of n connected components. Then Φ(νµD) = νµd by (I), and G(νµD C) = G(Φ(νµD) Φ(A)) = G(νµD A) = G(µ(ν 1)D B) consists of n circular disks with only one of them having positive radius. Considering the centers of the n disks, we conclude that the diagonal entries of νµd C equal µ(ν 1)w j for j = 1,..., n. Similarly, since G(C) = G(A) consists of n disks, we see that C has diagonal entries lying in {µw j : j = 1,..., n}. Suppose the c jj = µw k such that k j. Then the vector of diagonal entries of Φ(νµD) Φ(A) equals νµ(w, w 2,..., w n ) µ(w, w 2,..., w n )R for some R P n, and has entries which constitute the set {µ(ν 1)w j : 1 j n}. It follows that the vector ν(w, w 2,..., w n ) (w, w 2,..., w n )R has entries in {(ν 1)w j : 1 j n}. By Lemma 2.2, we see that c jj = µw j. Since G(C) = G(A), we see that only the kth row of C can have off-diagonal entries. Since G(Φ(A) Φ(B)) = G(A B) = {µw j : 1 j n}, we see that Φ(A) Φ(B) is a diagonal matrix, which is µd. It follows that Φ(A) = µd + Φ(B) = µd + B. Next, we consider a general matrix A = (a ij ) M n. Suppose Φ(A) = C = (c ij ). Let µ (0, ) \ Γ be such that G(µD A) consists of n disjoint disks. Moreover, we can choose µ > 0 such that µ 1 w > n j=1 ( a jj + c jj ) so that µw j c jj µw k a kk for any j k. Since Φ(µD) = µd, G(µD C) = G(Φ(µD) Φ(A)) = G(µD A). By our choice of µ, we see that µw j c jj = µw j a jj so that c jj = a jj for j = 1,..., n. Furthermore, since G µd + a kj E kj C = G Φ µd + a kj E kj Φ(A) = G µd + a kj E kj A, we see that G k µd + a kj E kj C = {µw k c kk }. Thus, a kj = c kj for all j k. Since k is arbitrary, we see that C = A as asserted. Corollary 2.4 Let Φ : M n M n. The following conditions are equivalent. (a) The map Φ is real linear and satisfies G(Φ(A)) = G(A) for all A M n. (b) The map Φ is additive and satisfies G(Φ(A)) = G(A) for all A M n. (c) The map Φ has the form in Theorem 2.1 with S = 0. 6

Proof. The implications (c) (a) (b) are clear. We focus on (b) (c). Clearly, X M n satisfies G(X) = {0} if and only if X = 0 and thus Φ(X) = 0 if and only if X = 0 since G(Φ(A)) = G(A) for all A M n. For any B M n, since G(Φ(B)+Φ( B)) = G(Φ(B B)) = {0}, it follows that Φ( B) = Φ(B). Therefore, G(Φ(A B)) = G(Φ(A) + Φ( B)) = G(Φ(A) Φ(B)) for all A, B M n. Thus Φ has the asserted form in Theorem 2.1. Since Φ(0) = {0}, we see that S = 0 in case condition (2) in Theorem 2.1 holds. The conclusion follows. Corollary 2.5 A (complex) linear map Φ : M n M n satisfies G(Φ(A)) = G(A) for all A M n if and only if it is a compositions of maps of the forms in (1) or (2) described in Theorem 2.1 with S = 0. It is clear that one can have the column vector version of Gershgorin regions. This is the same as considering G(A t ), and we can prove analogous results on preservers. 3 Brauer Regions In this section, we consider preservers of the Brauer region of a matrix A = (a ij ) M n defined as C(A) = 1 i<j n C ij (A), with C ij (A) = {µ C : (µ a ii )(µ a jj ) R i R j } is a Cassini oval, with R i = j i a ij as defined in Section 2. The values a ii and a jj are the foci of the oval. One may see the discussion of the Cassini oval in standard references; for example, [2, Chapter 6] and Wikipedia. The following facts can be easily verified and will be used in our discussion. Lemma 3.1 Let A = (a ij ) M n. (a) The set C(A) consists of a collection of isolated points if and only if A has at most one row with nonzero off diagonal entries. In such case, C(A) coincides with the spectrum of A. (b) If a ii a jj 2 > 4R i R j, then C ij (A) consists of two closed convex regions, each of which contains a focus. Consequently, if a ii a jj 2 > 4 max{ R k 2 : 1 k n} whenever i j, then C(A) consists of n disjoint connected regions, each of which contains a diagonal entry of A. (c) Suppose a kk is an isolated point of C(A). Then either C(A) is a collection of isolated points so that (a) holds or the kth row of A has only zero off diagonal entries. Theorem 3.2 A map Φ : M n M n satisfies C(Φ(A) Φ(B)) = C(A B) for all A, B M n 7

if and only if one of the following holds. (a) n = 2 and Φ has the form ( ) ( ) a11 uτ A P 1 (a 12 ) P t a11 uτ + S or A P 1 (a 21 ) P t + S, vτ 2 (a 21 ) a 22 vτ 2 (a 12 ) a 22 where S M 2, P P 2, u, v C satisfy uv = 1, and τ j is the identity map or the conjugation map for j = 1, 2. (b) Φ has the form described in Theorem 2.1. Proof. The sufficiency is clear. We consider the necessity. We may replace Φ by the map A Φ(A) Φ(0) and assume that C(A) = C(Φ(A)) for any A M n. Case 1 Suppose n = 2. For any µ C, we have C(Φ(µE 12 )) = C(µE 12 ) = {0}. It follows that Φ(µE 12 ) = f 1 (µ)e 12 + f 2 (µ)e 21 with f 1 (µ)f 2 (µ) = 0. Similarly, for any ν C, C(Φ(νE 21 )) = C(νE 21 ) = {0}, we see that Φ(νE 21 ) = g 1 (ν)e 12 +g 2 (ν)e 21 with g 1 (ν)g 2 (ν) = 0. Since C(Φ(µE 12 ) Φ(νE 21 )) = C(µE 12 νe 21 ) is a circular disk centered at the origin with radius µν, Φ(µE 12 ) 0 and Φ(νE 21 ) 0 whenever µν 0. Now for any µ 1, µ 2 C, We see that C(Φ(µ 1 E 12 ) Φ(µ 2 E 12 )) = C(µ 1 E 12 µ 2 E 12 ) = {0}. (1) Φ(µE 12 ) = f(µ)e 12 for all µ C, or (2) Φ(µE 12 ) = f(µ)e 21 for all µ C. We may assume that (1) holds. Otherwise, replace Φ by the map A Φ(A) t. It will then follow that Φ(νE 21 ) = g(ν)e 21 for all ν C. Note that C(Φ(µE 12 ) Φ(νE 21 )) = C(µE 12 νe 21 ) = {z C : z 2 µν }. Thus, we see that f(µ)g(ν) = µν. In particular, f(µ) = 0 if and only if µ = 0, and g(ν) = 0 if and only if ν = 0. Since C(Φ(E 11 )) = C(E 11 ) = {1, 0}, we see that Φ(E 11 ) has diagonal entries 1, 0 and at most one nonzero off diagonal entry. Since C(Φ(E 11 ) Φ(X)) = Φ(E 11 X) = {1, 0} for X {E 12, E 21 }, we see that Φ(E 11 ) = E 11 or Φ(E 11 ) = E 22. We may assume the former case holds. Otherwise, we may replace Φ by the map A P Φ(A t )P t with P = E 12 + E 21. Then we have Φ(E 11 ) = E 11, Φ(µE 12 ) = f(µ)e 12 and Φ(νE 21 ) = g(ν)e 21. Now one may use the facts that C(Φ(µE 11 ) Φ(E 11 )) = {µ 1, 0}, C(Φ(µE 11 ) Φ(E 12 )) = {µ, 0}, and C(Φ(µE 11 ) Φ(E 21 )) = {µ, 0} to conclude that Φ(µE 11 ) = µe 11. Similarly, we can argue that Φ(µE 22 ) = µe 22. Up to this point, we may assume that for j {1, 2} and µ C, Φ(µE jj ) = µe jj and there are two functions f, g C C such that Φ(µE 12 ) = f(µ)e 12 and Φ(νE 21 ) = g(ν)e 21, with ν C. Now suppose Φ(A) = (b ij ) for A = (a ij ). Since C(Φ(A) Φ(a ij E ij )) = C(A a ij E ij ) = {a 11, a 22 } for (i, j) {(1, 2), (2, 1)}, we see that b 12 = f(a 12 ) and b 21 = g(a 21 ). For i {1, 2}, since C(Φ(A) Φ(µE ii )) = C(A µe ii ) for all µ C, we see that b ii = a ii. Thus, ( ) a11 f(a Φ(A) = 12 ) g(a 21 ) a 22 8

such that f(a 12 )g(a 21 ) = a 12 a 21. Now, for any µ 1, µ 2 C, since C (Φ(E 21 + µ 1 E 12 ) Φ(µ 2 E 12 )) = C(E 21 + µ 1 E 12 µ 2 E 12 ), we see that uf(µ 1 ) uf(µ 2 ) = µ 1 µ 2 if u = g(1). By Lemma 2.3, we see that uf = τ 1 is the identity map or the conjugation map. Similarly, using the fact that C(Φ(E 12 + ν 1 E 21 ) Φ(ν 2 E 21 )) = C(E 12 + ν 1 E 21 ν 2 E 21 ), we can show that vg = τ 2 is the identity map or the conjugation map, where v = f(1). Combining the above arguments, we get condition (a). Case 2 Suppose n > 2. We prove condition (b) holds by establishing several assertions. We will assume Γ is defined as in Lemma 2.2. Assertion 1 Let D = diag (w, w 2,..., w n 1, 1) for w = e i2π/n. Then there is a permutation matrix P such that P Φ(µD)P t has (j, j) entry equal to µw j for j = 1,..., n, for any µ C \ Γ. The verification to this assertion is identical to that of Assertion 1 in the proof of Theorem 2.1. In the following, we will always assume that D is defined as in Assertion 1, and the matrix P in the conclusion is the identity matrix. Assertion 2 For k {1,..., n}, if then Φ(W k ) W k. W k = a j E kj : a 1,..., a k 1, a k+1,..., a n C, For any µ > 0 and k {2,..., n}, since C(Φ(µD + E k1 )) = C(µD + E k1 ) = {µw j : 1 j n}, we see that Φ(µD + E k1 ) has diagonal entries µw,..., µw n 1, µ, and there is at most one row of Φ(µD + E k1 ) with nonzero off diagonal entries by Lemma 3.1(a). Note that C(Φ(µD + E k1 ) Φ(D)) = C(µD + E k1 D) = {(µ 1)w j : 1 j n}. Suppose A = n j=2 a je 1j 0. Since C(Φ(A)) = C(A) = {0}, we see that all diagonal entries of Φ(A) equal zero and there is at most one row of Φ(A) with nonzero off diagonal entries. Let µ > 0 be sufficiently large so that the (j, j) entry of Φ(µD + E k1 ) is µw j for sufficiently large µ > 0. Then C(Φ(µD + E k1 ) Φ(A)) = C(µD + E k1 A) is the union of {µw j : 1 < j n, j k} and the Cassini oval with two connected regions containing the foci µw and µw k, which is two disconnected regions. Thus, Φ(µD + E k1 ) Φ(A) has nonzero off diagonal entries in row 1 and row k. Since this is true for all k {2,..., n}, we see that the only nonzero row of Φ(A) is row 1. Furthermore, we see that for sufficiently large µ, Φ(µD + E k1 ) has nonzero off diagonal entries at row k for k {2,..., n}. 9

Now, for k {2,..., n}, let A = a je kj 0. Since C(Φ(A)) = C(A) = {0}, we see that Φ(A) has zero diagonal entries and has at most one nonzero row. Since C(Φ(µD + E m1 ) Φ(A)) = C(µD + E m1 A) for m {1,..., n} \ {k} and sufficiently large µ, we see that the kth row of Φ(A) is nonzero. Combining the above arguments, we see that Assertion 2 holds. Assertion 3 Suppose A = (a ij ) and Φ(A) = (b ij ). Then for k {1,..., n}, we have (i) b kk = a kk and (ii) b kj E kj = Φ a kj E kj. Step 1 We prove (i). Since C(µD A) = C(Φ(µD) Φ(A)) = C(µD Φ(A)) for all sufficiently large µ > 0, we see that Φ(A) has (j, j) entry equal to a jj. See Assertion 3 in the proof of Theorem 2.1. Step 2 We prove (ii) for the special case when A = (a ij ) has two rows with nonzero diagonal entries such that C(A) consists of n 2 distinct points and a Cassini oval consisting of two disconnected components. Without loss of generality, we assume that the first two rows of A have nonzero off diagonal entries. Since Φ(A) = (b ij ) satisfies b jj = a jj for j {1,..., n}, C(A) = C(Φ(A)) and C(Φ(µD + E m1 ) Φ(A)) = C(µD + E m1 A) for m {1,..., n} and sufficiently large µ, we see that only the first two rows of Φ(A) have nonzero off diagonal entries. Moreover, for k = 1, 2, C(Φ(A) Φ( a kje kj )) = C(A a kje kj ). Thus by Assertion 2, the kth row of Φ(A) Φ( a kje kj ) equals zero, i.e., b kje kj = Φ( a kje kj ). Step 3 We prove (ii) for the case when A = (a ij ) M n has a single row with nonzero off diagonal entries. Without loss of generality, we may assume the first row of A has nonzero off diagonal entries. Let X M n have two nonzero rows such that X = µd + j 1 a 1jE 1j + E kj, µ C\Γ, k 1, and Φ(X) = (y ij ). By Steps 1 and 2, we see that y ii = x ii for all i {1,..., n}, only rows 1 and k of Φ(X) have nonzero off diagonal entries, and that j 1 y 1jE 1j = Φ( j 1 a 1jE 1j ). Because C (Φ(A) Φ(X)) = C (b ij ) µd + y 1j E 1j + y kj E kj = C (A X) j 1 consists of n distinct points for sufficiently large µ > 0, we can conclude that Φ(A) Φ(X) has only 1 row with nonzero off diagonal entries. Since C(Φ(A)) = C(A) implies that Φ(A) has only 1 row with nonzero off diagonal entries, it follows that this row must be row 1 or row k. As this result holds for any k {2,..., n}, we see that only row 1 of Φ(A) has nonzero off diagonal entries, and hence only row k of Φ(A) Φ(X) has nonzero off diagonal entries. Therefore the first row of Φ(A) Φ(X) equals zero, i.e., b 1j E 1j = y 1j E 1j = Φ a 1j E 1j j 1 j 1 j 1 10

. Step 4 We prove (ii) for a diagonal matrix A = (a ij ). If Φ(A) = B = (b ij ), then we can show a jj = b jj for all j {1,..., n} as in the previous two cases. Since C(B) = C(A) does not contain a disk with positive radius, we see that B has at most one row with nonzero off diagonal entries. If B has such a row, then we can find X W k so that B Φ(X) has two rows with nonzero off diagonal entries. But then C(A X) = C(B Φ(X)) contains a disk with positive radius, which is a contradiction. Hence, we see that Φ(A) = A if A is a diagonal matrix. Step 5 We prove (ii) for a matrix A = (a ij ) M n with at least three rows having nonzero off diagonal entries. Let Φ(A) = B = (b ij ) and k {1,..., n}. We are going to show that b kj = Φ( a kje kj ). To this end, let X = (x ij ) M n be such that X has exactly two nonzero rows indexed by k and k, where the off diagonal entries in these two rows are the same as those of A = (a ij ). Moreover, the diagonal entries x 11,..., x nn are chosen so that (1) C(X) consists of n 2 distinct points and a Cassini oval consisting of two connected components, (2) C(Φ(X) Φ(A)) = C(X A) consists of n 2 distinct points and a Cassini oval consisting of two connected components which include {x kk a kk } and {x k k a k k }. then Since we have proved the result for the special cases covering the case for X, if Φ(X) = Y = (y ij ), y kj E kj = Φ x kj E kj = Φ a kj E kj. By (2), we see that It follows that y kj E kj = b kj E kj. b kj E kj = Φ a kj E kj. k j Since this is true for any k {1,..., n}, the conclusion of the Assertion 3 holds. Assertion 4 Condition (b) in the theorem holds. By Assertion 2, we may define Φ k : C 1 (n 1) C 1 (n 1) by Φ k (a 1,..., a k 1, a k+1,..., a n ) = (b 1,..., b k 1, b k+1,..., b n ) if Φ( a je kj ) = b je kj. We claim that l 1 (Φ 1 (a) Φ 1 (a )) = l 1 (a a ) for any a = (a 2,..., a n ), a = (a 2,..., a n) C 1 (n 1). To see this, consider A k = n j=2 a je 1j + E k1 and A = n j=2 a j E 1j. By Assertion 3 and the fact that C(Φ(A k )) = C(A k ), R 1 (Φ(A k ))R k (Φ(A k )) = 11

R 1 (A k )R k (A k ) = l 1 (a). Also, since C(Φ(A k ) Φ(A )) = C(A k A ), by Assertion 3 we have l 1 (a a ) = R 1 (A k A )R k (A k ) = R 1 (Φ(A k ) Φ(A ))R k (Φ(A k )) = l 1 (Φ 1 (a) Φ 1 (a ))l 1 (Φ k (1, 0,..., 0)) By Assertion 3, R k (Φ(A k )) = R k (Φ(E k1 )). Thus, for k {2,..., n}, l 1 (Φ k (1, 0,..., 0)) = R k (Φ(E k1 )) = l 1 (a a )/l 1 (Φ 1 (a) Φ 1 (a )) = ν > 0. Since C(Φ(E 21 ) Φ(E k1 )) = C(E 21 E k1 ) = {z C : z 1}, we have 1 = R 2 (Φ(E 21 ))R k (Φ(E k1 ) for k {3,..., n}. (Here we use the fact that n 3.) Thus, ν = 1, and l 1 (a a ) = l 1 (Φ 1 (a) Φ 1 (a )). So, Φ 1 satisfies the conclusion of Lemma 2.3. Similarly, we can show that Φ k satisfies the conclusion of Lemma 2.3 for k {2,..., n}. Consequently, Φ has the asserted form. As with the Gershgorin discs, it is clear that one can have the column version of Cassini ovals by considering C(A t ). However, there is no extension for C ijk (A) = {µ C : (µ a ii )(µ a jj )(µ a kk ) R i R j R k } or higher dimensions. Note that we cannot deduce the structure of additive preservers of C(A) using Theorem 3.2 as in Corollary 2.4 because C(A) = {0} does not imply that A = 0. Nevertheless, we can apply a similar proof to characterize additive preservers of C(A) and then deduce the results on (real or complex) linear preservers of C(A). 4 Ostrowski Regions In this section, we consider the Ostrowski region of A M n defined by O ξ (A) = n k=1 O ξ,k(a), for a given ξ (0, 1), where O ξ,k (A) = {µ C : µ a kk R k (A) ξ R k (A t ) 1 ξ }, k {1,..., n}. One may see the discussion of the Ostrowski region in [2, Chapter 6]. Theorem 4.1 Let ξ (0, 1). A map Φ : M n M n satisfies O ξ (Φ(A) Φ(B)) = O ξ (A B) for all A, B M n 12

if and only if there exist P P n and maps ψ ij : C C, where ψ jj is the identity map and for i j, ψ ij has the form z ν ij z or z ν ij z for a norm one complex number ν ij, such that one of the following holds. (a) (ξ, n) = (1/2, 2) and the map Φ has the form in Theorem 3.2 (a). (b) There is S M n such that the map Φ has the form A = (a ij ) P (ψ ij (a ij ))P t + S. (c) ξ = 1/2 and there is S M n such that Φ has the form A = (a ij ) P (ψ ij (a ij )) t P t + S. Proof. The sufficiency is clear. We consider the proof of the necessity. The proof for the case when n = 2 is similar to that of Theorem 3.2. So, we assume that n > 2 and O ξ (Φ(A) Φ(B)) = O ξ (A B) for all A, B M n. Replacing Φ by the map X Φ(X) Φ(0), we may assume that O ξ (Φ(A)) = O ξ (A) for all A M n. Let D = diag (w, w 2,..., w n 1, 1) with w = e i2π/n. Furthermore, let Γ be defined as in Lemma 2.2. Assertion 1 There is a permutation matrix P such that one of the following holds for any µ C\Γ. (i) P Φ(E ij )P t = u ij E ij and P Φ(µD + E ij )P t = µd + v ij E ij with u ij, v ij C satisfying u ij = v ij = 1 whenever 1 i, j n and i j. (ii) P Φ(E ij )P t = u ij E ji and P Φ(µD + E ij )P t = µd + v ij E ji with u ij, v ij C satisfying u ij = v ij = 1 whenever 1 i, j n and i j. To prove the assertion, let µ C\Γ. Suppose ν C is such that ν, νµ / Γ. Since O ξ (Φ(νµD)) = O ξ (νµd) = {νµw j : 1 j n}, there is a permutation P (depending on µ and ν) such that P Φ(νµD)P t = νµd + F, where F has zero diagonal entries and R j (F )R j (F t ) = 0 for all j = 1,..., n. We will show that condition (i) or (ii) holds. Once this is done, we conclude that P is independent of the choice of µ and ν by examining Φ(E ij ) for 1 i, j n. For simplicity, we assume that P = I n. Otherwise, replace Φ by the map X P t Φ(X)P. For pairs (i, j) with i j consider V ij = Φ(µD + E ij ). Since O ξ (V ij ) = O ξ (µd + E ij ), we see that V ij has diagonal entries µw,..., µw n 1, µ. Since Φ(νµD) and νµd have the same diagonal entries, and O ξ (Φ(νµD) V ij ) = O ξ ((ν 1)µD E ij ) = {(ν 1)µw j : 1 j n}, the vector of the diagonal of the matrix (νµd V ij )/µ equals ν(w, w 2,..., w n ) (w, w 2,..., w n )R with R P n, and has entries in {(ν 1)w j : 1 j n}. Since ν C \ Γ, it follows from Lemma 2.2 that V ij = µd + F ij such that F ij has zero diagonal and R k (F ij )R k (Fij t ) = 0 for k = 1,..., n. For pairs (i, j) with i j let U ij = Φ(E ij ). Since O ξ (U ij ) = O ξ (E ij ) = {0} we see that for k = 1,..., n, R k (U ij )R k (U t ij ) = 0 and U ij has zero diagonal entries. Moreover, O ξ (V ij U ji ) = O ξ (µd + E ij E ji ) contains non-degenerate circular disks centered at µw i and µw j. Considering the disk with center µw i, we see that either R i (U ij ) 0 = R i (U t ij ) and R i(v t ji ) 0 = R i(v ji ) or R i (U ij ) = 0 R i (U t ij ) and R i(v t ji ) = 0 R i(v ji ). Suppose R 1 (U 12 ) 0 = R 1 (U t 12 ). If U 12 has a nonzero (k, j) entry with j > 2, then the jth row of U 12 is zero as R j (U 12 )R j (U t 12 ) = 0. Since O ξ(v j1 U 1j ) = O ξ (µd + E j1 E 1j ) contains a unit 13

disk centered at µw j, either R j (V t j1 )R j(u 1j ) 0 = R j (V j1 ) = R j (U t 1j ) or R j(v j1 )R j (U t 1j ) 0 = R j (V t j1 ) = R j(u 1j ). In the former case, O ξ (E 1j E 12 ) = O ξ (U 1j U 12 ) contains a non-degenerate circular disk; in the latter case, O ξ (µd + E j1 E 12 ) = O ξ (V j1 U 12 ) contains a non-degenerate circular disk centered at µw j. In both cases, we have a contradiction. By the above paragraph, only the second column of U 12 can be nonzero. Now, suppose U 12 has a nonzero (k, 2) entry with k > 2. Then R k (U 12 ) 0 = R k (U t 12 ). Since O ξ(v k1 U 1k ) = O ξ (µd + E k1 E 1k ) contains a unit disk centered at µw k, either R k (V t k1 )R k(u 1k ) 0 = R k (V k1 ) = R k (U t 1k ) or R k (V k1 )R k (U t 1k ) 0 = R k(v t k1 ) = R k(u 1k ). In the former case, O ξ (µd+e k1 E 12 ) = O ξ (V k1 U 12 ) contains a non-degenerate circular disk centered at µw k ; in the latter case, O ξ (E 1k E 12 ) = O ξ (U 1k U 12 ) contains a non-degenerate circular disk. In both cases, we have a contradiction. Thus, we conclude that U 12 = u 12 E 12 for some nonzero u 12. If R 1 (U 12 ) = 0 R 1 (U t 12 ), we can use a similar argument to show that U 12 = u 12 E 21 for some nonzero u 12. i j. Applying the above argument to U ij, we see that U ij = u ij E ij or u ij E ji for pairs (i, j) with Interchanging the roles of U ij and V ij in the above proof, we see that V ij = µd + v ij E ij or µd + v ij E ji for pairs (i, j) with i j. Now, suppose U 12 = u 12 E 12. Since O ξ (V j1 U 12 ) = O ξ (µd+e j1 E 12 ), we see that V j1 = µd+ v j1 E j1 with u 12 ξ v j1 (1 ξ) = 1 for all j = 2,..., n. Next, by the fact that O ξ (V j1 U kj ) = O ξ (µd+ E j1 E kj ), we see that U kj = u kj E jk such that u kj (1 ξ) v j1 ξ = 1 for all (j, k) with j k, j > 1 and k {1,..., n}. In particular, there is u j such that u kj = u j for all k j. Since O ξ (U 1j U j1 ) = O ξ (E 1j E j1 ), we see that U j1 = u j1 E j1 with max{ u 1j ξ u j1 (1 ξ), u 1j (1 ξ) u j1 ξ } = 1 for j = 2,..., n; since O ξ (V ij U ji ) = O ξ (µd + E ij E ji ), we see that v ij ξ u ji (1 ξ) = 1 = v ij (1 ξ) u ji ξ for all pairs (i, j) with i j. It follows that u ij = v ij = 1 for pairs (i, j) with i j. Thus, condition (i) holds. Suppose U 12 = u 12 E 21. We can use a similar argument to show that condition (ii) holds. Assertion 2 There exist functions ψ ij as described in the theorem such that the following holds. (I) If conclusion (i) of Assertion 1 holds, then P Φ(νE ij )P t = ψ ij (ν)e ij for all i j and ν C. (II) If conclusion (ii) of Assertion 1 holds, then ξ = 1/2 and P Φ(νE ij )P t = ψ ij (ν)e ji for all i j and ν C. To prove the assertion, let P satisfy the conclusion of Assertion 1. For simplicity, assume that P = I n. (I) Suppose condition (i) in Assertion 1 holds. Consider Φ(νE ij ). Since O ξ (Φ(νE ij ) µd v rs E rs ) = O ξ (Φ(νE ij ) Φ(µD + E rs )) = O ξ (νe ij µd E rs ) for any pairs (r, s) with r s, we see that Φ(νE ij ) = γe ij. Let ψ ij : C C be such that Φ(νE ij ) = ψ ij (ν)e ij. We claim that ψ ij (ν 1 ) ψ ij (ν 2 ) = ν 1 ν 2 for any ν 1, ν 2 C. Note that O ξ (Φ(ν 1 E ij +E ji ) (µd+v rs E rs )) = O ξ (Φ(ν 1 E ij +E ji ) Φ(µD+E rs )) = O ξ (ν 1 E ij +E ji µd E rs ) 14

for any pairs (r, s) with r s. It follows that Φ(ν 1 E ij + E ji ) = γe ij + v ji E ji. By the fact that O ξ (Φ(ν 1 E ij + E ji ) Φ(ν 1 E ij )) = {0}, we see that γ = ψ ij (ν 1 ). Now, O ξ (ψ ij (ν 1 )E ij + v ji E ji ψ ij (ν 2 )E ij ) = O ξ (Φ(ν 1 E ij + E ji ) Φ(ν 2 E ij )) = O ξ ((ν 1 ν 2 )E ij + E ji ). We get the desired conclusion. By Lemma 2.3, ψ ij has the asserted form. (II) Suppose condition (ii) in Assertion 1 holds. We can use a similar argument to that in the proof of (I) to conclude that Φ(νE ij ) = ψ ij (ν)e ji for all (i, j) with i j. Now, O ξ (ψ 12 (2)E 21 ψ 31 (1)E 13 )) = O ξ (Φ(2E 12 ) Φ(E 31 )) = O ξ (2E 12 E 31 ). Thus, 2 (1 ξ) = 2 ξ and hence ξ = 1/2. Assertion 3 The map has the asserted form. To prove the assertion, we may assume that condition (I) in Assertion 2 holds. Otherwise, replace Φ by the map A Φ(A t ). For simplicity, we assume that P = I n and ψ ij is the identity map for all pairs (i, j). We will show that Φ(A) = A for all A M n. Note that if µ C \ Γ with sufficiently large magnitude, we will have O ξ (Φ(A) µd v rs E rs ) = O ξ (Φ(A) Φ(µD + E rs )) = O ξ (A µd E rs ) = O ξ (νe ij + E ki µd E rs ) (4.1) We consider 5 special cases before the general case. Case 1 Suppose A = n j=1 d je jj is a diagonal matrix. Assume Φ(A) has a nonzero entry at the (p, q) position for some p q. Since O ξ (Φ(A)) = O ξ (A) = {d 1,..., d n }, we see that the pth column of Φ(A) is zero. But then we can choose a suitable µ C \ Γ so that O ξ (Φ(A) µd v qp E qp ) contains a non-degenerate disk centered at the p diagonal entry of Φ(A) µd v qp E qp, where as O ξ (A µd E qp ) = {d j µw j : 1 j n}, which is a contradiction. Thus, Φ(A) is a diagonal matrix. Since O ξ (φ(a) µd v 12 E 12 ) = O ξ (A µd E 12 ) = {d j µw j : 1 j n} for any µ C \ Γ, we see that Φ(A) = A. Case 2 Suppose A = νe ij + E ki with ν 0 and i / {j, k}. For notational simplicity, we assume that (i, j) = (1, 2) so that A = νe 12 + E 21 or (i, j, k) = (1, 2, 3) so that A = νe 12 +E 31. It is easy to adapt the arguments to the general case. Assume Φ(A) has a nonzero (p, q) entry with p q and (p, q) {1,..., n} {3,..., n}. Taking (r, s) = (q, p) in (4.1), we see that O ξ (Φ(A) µd v qp E qp ) = O ξ (A µd E qp ) has µw q as an isolated point. Since Φ(A) µd v qp E qp has nonzero (p, q) position, the qth column of Φ(A) has only one nonzero entry at the (q, p) position equal to v qp. But then O ξ (µd Φ(A)) contains a non-degenerate circular disk centered at µw q, whereas µw q is an isolated point of O ξ (µd A), which is a contradiction. Similarly, if Φ(A) has a nonzero (p, q) entry with (p, q) {k,..., n} {1, 2} with k = 3 or 4 depending on A = νe 12 + E 21 or A = νe 12 + E 31. Taking (r, s) = (q, p) in (4.1), we see that O ξ (Φ(A) µd v qp E qp ) = O ξ (A µd E qp ) has µw p as an isolated point. Thus, the pth row of 15

Φ(A) has only one nonzero entry at the (q, p) position equal to v qp. But then using Φ(µD) = µd, we see that O ξ (µd Φ(A)) will contain a non-degenerate circular disk centered at µw p, whereas O ξ (µd A) does not, which is a contradiction. Thus, Φ(A) only has a nonzero entry at the (p, q) positions with (p, q) K, where K = {(1, 2), (2, 1)} or K = {(1, 2), (2, 1), (3, 1), (3, 2)} depending on A = νe 12 +E 21 or A = νe 13 +E 31. If A = νe 12 +E 21, then O ξ (Φ(A) νe 12 ) = O ξ (A νe 12 ) = {0} and O ξ (Φ(A) E 21 ) = O ξ (A E 21 ) = {0}. We conclude that Φ(A) = νe 12 + E 21. Suppose A = νe 12 + E 31. Since O ξ (Φ(A) X) = O ξ (A X) for X {0, νe 12, E 23, µd + E 12 } for a suitable µ C \ Γ, we see that Φ(A) = A as asserted. Case 3 Suppose A = µd + R, where R has nonzero off diagonal entries in at most one row and µ satisfies µ 1 w > 2 or µ = 0. For simplicity, assume that the nonzero off diagonal entries of A = µd + n j=2 a 1jE 1j lie in the first row. For µ 1 w > 2 since (I) holds, we see that Φ(µD + E ij ) = µd + v ij E ij with v ij = 1 for pairs (i, j) with i j. Suppose Φ(A) = (y ij ). Since O ξ (Φ(A) Φ(µD + E ij )) = O ξ (A µd E ij ) for all pairs (i, j) with i j, we see that y ij = 0 for i > 1 and i j. Moreover, since O ξ (Φ(A) Φ(νE 1j + E 21 )) = O ξ (A νe 1j E 21 ) for all ν C by Case 2, we see that y 1j = a 1j for j = 2,..., n. Now, suppose µ = 0. Since O ξ (Φ(A) Φ(E ij )) = O ξ (A E ij ) for all pairs (i, j) with i j, we see that y ij = 0 for i > 1 and i j. Moreover, since O ξ (Φ(A) Φ(νE 1j + E 21 )) = O ξ (A νe 1j E 21 ) for all ν C by Case 2, we see that y 1j = a 1j for j = 2,..., n. Case 4 Suppose A = a ji E ji + k i a ike ik with a ji 0 and R i (A) 0. We may assume without loss of generality that (i, j) = (1, 2), a 21 0, and R 1 (A) 0. Let Φ(A) = Y = (y rs ). There is a choice of µ for which O ξ (Y µd) = O ξ (A µd) yields y kk = a kk = 0 for k = 1,..., n. We must have y rs = 0 for r > 2 because O ξ (Y (µd + νe 1r )) = O ξ (A (µd + νe 1r )) which shows that the latter set will have a degenerate circular disk at µw r whereas, if y rs 0, the former will have a non-degenerate circular disk at µw r for ν y 1r. Now if y 21 0 then O ξ Y a 1k E 1k = O ξ A a 1k E 1k = O ξ (a 21 E 21 ) = {0} k 1 k 1 which implies y 1k = a 1k for all k 1. Similarly, if y 12 0 then y 2k = a 2k for all k 2. Now if y 21 = 0, y 12 0 then y 21 = a 21 0 will be a contradiction. Thus y 21 = 0 implies y 12 = 0, which implies O ξ (Y ) = {0} O ξ (Φ(A)), again a contradiction. Hence we have Φ(A) = A. Case 5 Suppose A = (a ij ) has exactly two indices i and j with R i (A) 0 and R j (A) 0. For simplicity, assume that i = 1 and j = 2. Let Φ(A) = Y = (y ij ). By an appropriate choice of µ and using O ξ (Y µd) = O ξ (A µd), we get that y kk = a kk for k = 1,..., n. As before, if 16

y rs 0 for r > 2 and r s, then O ξ (Y µd) = O ξ (A µd) implies R r (Y t ) = 0. Now O ξ (Y (µd + E 1r )) = O ξ (A (µd + E 1r )). The latter set has a degenerate circular disk at a rr µw r whereas the former set has a non-degenerate circular disk centered at that point. Thus we must have y rs = 0 for r > 2 and r s. Now, using essentially the same arguments as in Case 4 yields Φ(A) = A. Again, generalizing to any i and j with i j, we have Φ(A) = A for all A with both R i (A) 0 and R j (A) 0. General Case We complete the proof by now taking A = (a ij ) to be arbitrary. Let Φ(A) = Y = (y ij ). As before, we can get y kk = a kk by using µd with an appropriate choice of µ. If R k (Y t ) 0 then using O ξ Y µd + a kj E kj = O ξ A µd + a kj E kj j i j i we get that the y kj = a kj for all j k since the latter set has a degenerate circular disk centered at a kk µw k. If R k (Y t ) = 0 then using O ξ Y µd + νe ik + a kj E kj = O ξ A µd + νe ik + a kj E kj j i j i where i j and ν a ik, we conclude that y kj degenerate circular disk centered at a kk µw k. = a kj for all j k since the latter set has a One can use a similar proof to obtain the structure of additive preservers of O ξ (A), and then deduce the results on linear preservers. References [1] Z. Charzyński, Sur les transformations isométriques des espaces du type (F), Studia Math. 13 (1953), 94-121. [2] R.A. Horn and C.R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985. [3] C.K. Li and S. Pierce, Linear preserver problems, Amer. Math. Monthly 108 (2001), 591 605. [4] P. Šemrl, Maps on matrix spaces, Linear Algebra Appl. 413 (2006), 364 393. [5] R.S. Varga, Geršgorin and his circles, Springer Series in Computational Mathematics, 36, Springer-Verlag, Berlin, 2004. 17