- Email: [email protected]
On separation of eigenvalues by certain matrix subgroups
Linear Algebra and its Applications 440 (2014) 213–217
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On separation of eigenvalues by certain matrix subgroups ✩ Grega Cigler ∗ , Marjan Jerman Department of Mathematics, Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia
a r t i c l e
i n f o
Article history: Received 3 July 2013 Accepted 22 October 2013 Available online 14 November 2013 Submitted by C.-K. Li MSC: 15A18
a b s t r a c t Let A be an n × n complex matrix with rank r. It is shown that there are a monomial matrix M and a unitary matrix U such that each of the matrices M A and U A has r distinct non-zero eigenvalues. If H is an irreducible subgroup of GLn (C) and A = 0, it is shown that there is an X ∈ H such that X A has at least two distinct eigenvalues. © 2013 Elsevier Inc. All rights reserved.
Keywords: Distinct eigenvalues Irreducible group Monomial group Unitary group
1. Introduction In their article [3] the authors showed that every invertible complex 2 × 2 or 3 × 3 matrix is diagonally equivalent to a matrix with distinct eigenvalues. In [2] it was proved that the same statement holds for all square complex matrices regardless of their dimension. That is, for every invertible n × n complex matrix A there is an n × n diagonal invertible D such that D A has distinct eigenvalues. Note that the products P A Q and ( Q P ) A have the same spectrum since they are similar for any pair of invertible matrices P and Q . Here some similar questions are explored. Suppose that H is a subgroup of the general linear group GLn (C) and that A is an n × n complex matrix. We shall say that a matrix A ∈ Cn×n with rank r n is H-separable if there is a matrix X ∈ H such that X A has r distinct non-zero eigenvalues. ✩
*
The authors were supported by the Slovenian Research Agency. Corresponding author. E-mail addresses: [email protected] (G. Cigler), [email protected] (M. Jerman).
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.10.036
214
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
A matrix A is H-semiseparable if there is an X ∈ H such that X A has at least two different eigenvalues. By Uni denote the subgroup of all unitary matrices. P = [ p i j ]ni, j =1 is a permutation matrix if there is a permutation π of the set {1, . . . , n} such that the only non-zero elements of P = P (π ) are the elements p π (i ),i = 1. Let Sn denote the subgroup of all n × n permutation matrices. A monomial matrix is a matrix which can be written as a product D P of a diagonal matrix D ∈ Diag and a permutation matrix P ∈ Sn . By Mon denote the group of all invertible n × n monomial matrices. A subgroup H of GLn (C) is irreducible if the matrices from H do not posses a common invariant subspace. Since C is algebraically closed, H is irreducible if and only if it possesses a basis of Cn×n . In this article it is shown that each matrix A ∈ Cn×n is Mon-separable and Uni-separable. If H is an irreducible subgroup of GLn (C), it is proved that each non-zero matrix A ∈ Cn×n is H-semiseparable. 2. The monomial group Theorem 2.1. Let A ∈ Cn×n be a matrix with rank r n. Then for a suitable monomial matrix M the matrix M A has r distinct non-zero eigenvalues. Proof. By the definition of rank there are permutation matrices P and Q and an invertible r × r matrix A 1 such that
P AQ =
∗ . ∗
A1
∗
The main result in [2] provides an invertible diagonal matrix D 1 such that all eigenvalues of the product D 1 A 1 are distinct. Define the diagonal matrix D (x) = D 1 ⊕ xI ∈ Cn×n . Then, the matrix
D (0) P A Q =
D 1 A1 0
∗
0
has r distinct eigenvalues and by the continuity of the spectrum we can find an x = 0 such that D (x) P A Q has r distinct eigenvalues. The diagonal matrix D = D (x) is invertible, the matrix M = Q D P is monomial and the matrix M A which is similar to the matrix D P A Q , has r distinct eigenvalues. 2 Corollary 2.2. Each A ∈ Cn×n is Mon-separable. Remark 2.3. If B is an invertible matrix with non-zero leading principal minors, Ballantine’s result in [1] provides a diagonal matrix E such that all eigenvalues of E B are distinct and positive. If B were an invertible matrix with zero diagonal elements there would be no such diagonal matrix E since tr( E B ) = 0. In the proof of Theorem 2.1 we could choose suitable permutation matrices P and Q and achieve that the first r leading principal minors of the matrix P A Q are non-zero and its larger minors are zero. Then by [1, Theorem 2] there is a diagonal matrix D 1 such that all eigenvalues of D 1 A 1 are positive. Then, the continuity argument assures that the real parts of the r distinct non-zero eigenvalues of D (x) A are positive for small x. 3. Unitary group Theorem 3.1. Each matrix A ∈ Cn×n is Uni-separable. Proof. We shall prove the theorem by the induction on the size n of the matrix A. For an A ∈ Cn×n with rank A = r we must find a unitary matrix U ∈ Cn×n such that card(σ (U A ) \ {0}) = r. In the 1 × 1 case we can take U = [1]. Now suppose that the theorem holds for all matrices A ∈ Cm×m with m < n. Let A ∈ Cn×n be a matrix with rank A = r. The Schur decomposition provides a unitary matrix V such that the matrix
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
˜ = V AV ∗ = A
λ 0
u A
215
is upper-triangular. ˜ is not a linear combination of other rows and rank A = r − 1. By the If λ = 0, the first row in A induction hypothesis we can find a unitary matrix V ∈ C(n−1)×(n−1) such that card(σ ( V A ) \ {0}) = r − 1. There is a δ ∈ C, |δ| = 1, such that δλ ∈ / σ ( V A ). Form
U=
0 . V
δ 0
˜ ) \ {0}) = r. Then card(σ (U A In the remaining case when λ = 0,
˜ = [0 A
B ],
where B ∈ Cn×(n−1) and rank B = r. We can use a (unitary) permutation matrix P to rearrange the rows of B so that
˜= PA
0 0
u A
and rank A = r. By the induction hypothesis there is a unitary matrix V ∈ C(n−1)×(n−1) such that
card
σ V A \ { 0} = r .
Now we can take
U=
1 0
0 V
˜ ) \ {0}) = r. and again card(σ (U A
2
4. Irreducible groups Example 4.1. Let G ⊂ C4×4 be the group of permutation matrices corresponding to the permutations id, (12)(34), (13)(24), (14)(23) and let D ⊂ C4×4 be the group of all diagonal matrices with entries ±1 and determinant 1. Then, the semidirect product H = D · G is an irreducible matrix group. It is easy to check that each matrix from H has at most two different eigenvalues. Therefore matrix A = I is not H-separable. The next example shows that we can find an invertible matrix A and an infinite reducible group
G such that all products X A, X ∈ G, have the same single eigenvalue. Example 4.2. It is easy to check that the characteristic polynomial of a 3 × 3 complex matrix A with det A = 1 equals to
p A (x) = x3 − tr Ax2 + tr A −1 x − 1. For
A=
3 0 4 2 0 3 0 −1 0
we have p A (x) = (x − 1)3 . Note that A and A −1 have the same diagonal elements (3, 0, 0).
216
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
Define the following subgroup G of matrices
G=
1 0 0
0
0 0
α
; α = 0 .
α −1
0
For all X ∈ G the matrix X A has a single eigenvalue equal to 1. We can choose an identity matrix I of a suitable size and extend this example to arbitrary dimensions by replacing G with G ⊕ I and A with A ⊕ I . The previous examples lead to a natural question: Is there an irreducible group H ∈ Cn×n and a non-zero matrix A ∈ Cn×n such that for all X ∈ H the matrix X A has a single eigenvalue? Lemma 4.3. Suppose that H ⊂ Cn×n is an irreducible group and C ∈ Cn×n a matrix where n 2. If there exists a constant c ∈ C such that
tr( X C ) = c
for all X ∈ H,
then c = 0 and C = 0. Proof. Let us assume that tr( X C ) = c for all X ∈ H. Then, for all X , Y ∈ H
tr X (Y C − C ) = tr ( X Y )C − X C = 0. The irreducibility of H implies that the system of linear equations
tr( X Z ) = 0
for all X ∈ H
with the variable Z has only the trivial solution Z = 0. Therefore, Y C = C for all Y ∈ H. Suppose that C has a non-zero j-th row C j and pick
i = j. The elementary matrix E i j is a linear combination of some matrices Y 1 , . . . , Y k ∈ H, say E i j = k λk Y k . Then,
Eij C =
k
λk Y k C = λk (Y k C ) = λk C = λC . k
k
As the j-th row of the matrix E i j C is zero and the j-th row of C is non-zero, we get λ = 0. This leads to a contradiction since the i-th row of E i j C is non-zero. Therefore, C = 0 and c = 0. 2 Suppose that for all X ∈ H the matrices X A have a single eigenvalue. Since σ ( X (α A )) = ασ ( X A ) for all α ∈ C, we can replace the matrix A by a non-zero multiple α A. Also, I ∈ H, therefore we can assume that A is either a nilpotent matrix or det A = 1. Lemma 4.4. Let H ⊂ Cn×n be an irreducible group and A ∈ Cn×n a nonsingular matrix with det A = 1 such that for all X ∈ H the matrix X A has a single eigenvalue. Denote C∗ = C \ {0}. There exists a finite irreducible group G ⊂ Cn×n such that
C∗ G = C∗ H and for each X ∈ G the matrix X A has a single eigenvalue on the unit circle and det X = 1. Proof. We can assume that C∗ H = H. Since H is irreducible, we can find a basis X 1 , X 2 , . . . , X n2 ∈ H of Cn×n . The elements of a basis can be replaced by their non-zero multiples, therefore, we can assume that det X i = 1 for all i = 1, 2, . . . , n2 . It follows that G = { X ∈ H | det X = 1} is an irreducible subgroup of H such that for all X ∈ G the matrix X A has a single eigenvalue, say α X . As
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
217
1 = det( X A ) = αnX , we see that α X is an n-th root of unity. For a fixed matrix X ∈ G the matrix X A satisfies n2 linearly independent linear equations
tr X i ( X A ) = αi
for i = 1, . . . , n2 . 2
There are n different possibilities for each αi , hence we have at most nn different possibilities for the matrix X A. The nonsingularity of A implies that the group G is finite. 2 Theorem 4.5. Let n 2. Suppose that A ∈ Cn×n is a matrix and H ⊂ Cn×n is an irreducible matrix group such that for all X ∈ H the product X A has a single eigenvalue. Then A = 0. Proof. We shall consider two cases. If A is a nilpotent matrix, 0 is the only eigenvalue of X A for each X ∈ H. Therefore, tr( X A ) = 0 for all X ∈ H and by Lemma 4.3 A = 0. Now suppose that σ ( A ) = {α } and α = 0. Then, α −1 A is a unipotent matrix and we can assume that A = I + N for a nilpotent matrix N. By Lemma 4.4 we can also assume that H is a finite group and that det X = 1, σ ( X A ) = {α X } and |α X | = 1 for all X ∈ H. 1 −1 ) = tr( X A ). After applying a suitable Then, σ (( X A )−1 ) = {α − X } = {α X } and therefore tr(( X A ) similarity we can assume that all matrices in H are unitary. The unitarity of X implies
tr( X A ) = tr ( X A )−1 = tr A −1 X ∗ = tr X A −1
∗
,
hence
tr X A − A −1
∗
= 0.
We can again use Lemma 4.3 to see that A − ( A −1 )∗ = 0. Therefore, A is a unitary matrix, N = 0, A = I and each member of the group H has a single eigenvalue. Corollary 4.1.7 from [4] then implies the reducibility of H, which is a contradiction. 2 Corollary 4.6. Suppose that H is an irreducible subgroup of GLn (C). Then, each non-zero matrix A is H-semiseparable. References [1] C.S. Ballantine, Stabilization by a diagonal matrix, Proc. Amer. Math. Soc. 25 (1970) 728–734. [2] M. Choi, Z. Huang, C. Li, N. Sze, Every invertible matrix is diagonally equivalent to a matrix with distinct eigenvalues, Linear Algebra Appl. 436 (2012) 3773–3776. [3] X. Feng, Z. Li, T. Huang, Is every nonsingular matrix diagonally equivalent to a matrix with all distinct eigenvalues?, Linear Algebra Appl. 436 (2012) 120–125. [4] H. Radjavi, P. Rosenthal, Simultaneous Triangularization, Universitext, Springer-Verlag, New York, 2000.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On separation of eigenvalues by certain matrix subgroups ✩ Grega Cigler ∗ , Marjan Jerman Department of Mathematics, Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia
a r t i c l e
i n f o
Article history: Received 3 July 2013 Accepted 22 October 2013 Available online 14 November 2013 Submitted by C.-K. Li MSC: 15A18
a b s t r a c t Let A be an n × n complex matrix with rank r. It is shown that there are a monomial matrix M and a unitary matrix U such that each of the matrices M A and U A has r distinct non-zero eigenvalues. If H is an irreducible subgroup of GLn (C) and A = 0, it is shown that there is an X ∈ H such that X A has at least two distinct eigenvalues. © 2013 Elsevier Inc. All rights reserved.
Keywords: Distinct eigenvalues Irreducible group Monomial group Unitary group
1. Introduction In their article [3] the authors showed that every invertible complex 2 × 2 or 3 × 3 matrix is diagonally equivalent to a matrix with distinct eigenvalues. In [2] it was proved that the same statement holds for all square complex matrices regardless of their dimension. That is, for every invertible n × n complex matrix A there is an n × n diagonal invertible D such that D A has distinct eigenvalues. Note that the products P A Q and ( Q P ) A have the same spectrum since they are similar for any pair of invertible matrices P and Q . Here some similar questions are explored. Suppose that H is a subgroup of the general linear group GLn (C) and that A is an n × n complex matrix. We shall say that a matrix A ∈ Cn×n with rank r n is H-separable if there is a matrix X ∈ H such that X A has r distinct non-zero eigenvalues. ✩
*
The authors were supported by the Slovenian Research Agency. Corresponding author. E-mail addresses: [email protected] (G. Cigler), [email protected] (M. Jerman).
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.10.036
214
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
A matrix A is H-semiseparable if there is an X ∈ H such that X A has at least two different eigenvalues. By Uni denote the subgroup of all unitary matrices. P = [ p i j ]ni, j =1 is a permutation matrix if there is a permutation π of the set {1, . . . , n} such that the only non-zero elements of P = P (π ) are the elements p π (i ),i = 1. Let Sn denote the subgroup of all n × n permutation matrices. A monomial matrix is a matrix which can be written as a product D P of a diagonal matrix D ∈ Diag and a permutation matrix P ∈ Sn . By Mon denote the group of all invertible n × n monomial matrices. A subgroup H of GLn (C) is irreducible if the matrices from H do not posses a common invariant subspace. Since C is algebraically closed, H is irreducible if and only if it possesses a basis of Cn×n . In this article it is shown that each matrix A ∈ Cn×n is Mon-separable and Uni-separable. If H is an irreducible subgroup of GLn (C), it is proved that each non-zero matrix A ∈ Cn×n is H-semiseparable. 2. The monomial group Theorem 2.1. Let A ∈ Cn×n be a matrix with rank r n. Then for a suitable monomial matrix M the matrix M A has r distinct non-zero eigenvalues. Proof. By the definition of rank there are permutation matrices P and Q and an invertible r × r matrix A 1 such that
P AQ =
∗ . ∗
A1
∗
The main result in [2] provides an invertible diagonal matrix D 1 such that all eigenvalues of the product D 1 A 1 are distinct. Define the diagonal matrix D (x) = D 1 ⊕ xI ∈ Cn×n . Then, the matrix
D (0) P A Q =
D 1 A1 0
∗
0
has r distinct eigenvalues and by the continuity of the spectrum we can find an x = 0 such that D (x) P A Q has r distinct eigenvalues. The diagonal matrix D = D (x) is invertible, the matrix M = Q D P is monomial and the matrix M A which is similar to the matrix D P A Q , has r distinct eigenvalues. 2 Corollary 2.2. Each A ∈ Cn×n is Mon-separable. Remark 2.3. If B is an invertible matrix with non-zero leading principal minors, Ballantine’s result in [1] provides a diagonal matrix E such that all eigenvalues of E B are distinct and positive. If B were an invertible matrix with zero diagonal elements there would be no such diagonal matrix E since tr( E B ) = 0. In the proof of Theorem 2.1 we could choose suitable permutation matrices P and Q and achieve that the first r leading principal minors of the matrix P A Q are non-zero and its larger minors are zero. Then by [1, Theorem 2] there is a diagonal matrix D 1 such that all eigenvalues of D 1 A 1 are positive. Then, the continuity argument assures that the real parts of the r distinct non-zero eigenvalues of D (x) A are positive for small x. 3. Unitary group Theorem 3.1. Each matrix A ∈ Cn×n is Uni-separable. Proof. We shall prove the theorem by the induction on the size n of the matrix A. For an A ∈ Cn×n with rank A = r we must find a unitary matrix U ∈ Cn×n such that card(σ (U A ) \ {0}) = r. In the 1 × 1 case we can take U = [1]. Now suppose that the theorem holds for all matrices A ∈ Cm×m with m < n. Let A ∈ Cn×n be a matrix with rank A = r. The Schur decomposition provides a unitary matrix V such that the matrix
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
˜ = V AV ∗ = A
λ 0
u A
215
is upper-triangular. ˜ is not a linear combination of other rows and rank A = r − 1. By the If λ = 0, the first row in A induction hypothesis we can find a unitary matrix V ∈ C(n−1)×(n−1) such that card(σ ( V A ) \ {0}) = r − 1. There is a δ ∈ C, |δ| = 1, such that δλ ∈ / σ ( V A ). Form
U=
0 . V
δ 0
˜ ) \ {0}) = r. Then card(σ (U A In the remaining case when λ = 0,
˜ = [0 A
B ],
where B ∈ Cn×(n−1) and rank B = r. We can use a (unitary) permutation matrix P to rearrange the rows of B so that
˜= PA
0 0
u A
and rank A = r. By the induction hypothesis there is a unitary matrix V ∈ C(n−1)×(n−1) such that
card
σ V A \ { 0} = r .
Now we can take
U=
1 0
0 V
˜ ) \ {0}) = r. and again card(σ (U A
2
4. Irreducible groups Example 4.1. Let G ⊂ C4×4 be the group of permutation matrices corresponding to the permutations id, (12)(34), (13)(24), (14)(23) and let D ⊂ C4×4 be the group of all diagonal matrices with entries ±1 and determinant 1. Then, the semidirect product H = D · G is an irreducible matrix group. It is easy to check that each matrix from H has at most two different eigenvalues. Therefore matrix A = I is not H-separable. The next example shows that we can find an invertible matrix A and an infinite reducible group
G such that all products X A, X ∈ G, have the same single eigenvalue. Example 4.2. It is easy to check that the characteristic polynomial of a 3 × 3 complex matrix A with det A = 1 equals to
p A (x) = x3 − tr Ax2 + tr A −1 x − 1. For
A=
3 0 4 2 0 3 0 −1 0
we have p A (x) = (x − 1)3 . Note that A and A −1 have the same diagonal elements (3, 0, 0).
216
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
Define the following subgroup G of matrices
G=
1 0 0
0
0 0
α
; α = 0 .
α −1
0
For all X ∈ G the matrix X A has a single eigenvalue equal to 1. We can choose an identity matrix I of a suitable size and extend this example to arbitrary dimensions by replacing G with G ⊕ I and A with A ⊕ I . The previous examples lead to a natural question: Is there an irreducible group H ∈ Cn×n and a non-zero matrix A ∈ Cn×n such that for all X ∈ H the matrix X A has a single eigenvalue? Lemma 4.3. Suppose that H ⊂ Cn×n is an irreducible group and C ∈ Cn×n a matrix where n 2. If there exists a constant c ∈ C such that
tr( X C ) = c
for all X ∈ H,
then c = 0 and C = 0. Proof. Let us assume that tr( X C ) = c for all X ∈ H. Then, for all X , Y ∈ H
tr X (Y C − C ) = tr ( X Y )C − X C = 0. The irreducibility of H implies that the system of linear equations
tr( X Z ) = 0
for all X ∈ H
with the variable Z has only the trivial solution Z = 0. Therefore, Y C = C for all Y ∈ H. Suppose that C has a non-zero j-th row C j and pick
i = j. The elementary matrix E i j is a linear combination of some matrices Y 1 , . . . , Y k ∈ H, say E i j = k λk Y k . Then,
Eij C =
k
λk Y k C = λk (Y k C ) = λk C = λC . k
k
As the j-th row of the matrix E i j C is zero and the j-th row of C is non-zero, we get λ = 0. This leads to a contradiction since the i-th row of E i j C is non-zero. Therefore, C = 0 and c = 0. 2 Suppose that for all X ∈ H the matrices X A have a single eigenvalue. Since σ ( X (α A )) = ασ ( X A ) for all α ∈ C, we can replace the matrix A by a non-zero multiple α A. Also, I ∈ H, therefore we can assume that A is either a nilpotent matrix or det A = 1. Lemma 4.4. Let H ⊂ Cn×n be an irreducible group and A ∈ Cn×n a nonsingular matrix with det A = 1 such that for all X ∈ H the matrix X A has a single eigenvalue. Denote C∗ = C \ {0}. There exists a finite irreducible group G ⊂ Cn×n such that
C∗ G = C∗ H and for each X ∈ G the matrix X A has a single eigenvalue on the unit circle and det X = 1. Proof. We can assume that C∗ H = H. Since H is irreducible, we can find a basis X 1 , X 2 , . . . , X n2 ∈ H of Cn×n . The elements of a basis can be replaced by their non-zero multiples, therefore, we can assume that det X i = 1 for all i = 1, 2, . . . , n2 . It follows that G = { X ∈ H | det X = 1} is an irreducible subgroup of H such that for all X ∈ G the matrix X A has a single eigenvalue, say α X . As
G. Cigler, M. Jerman / Linear Algebra and its Applications 440 (2014) 213–217
217
1 = det( X A ) = αnX , we see that α X is an n-th root of unity. For a fixed matrix X ∈ G the matrix X A satisfies n2 linearly independent linear equations
tr X i ( X A ) = αi
for i = 1, . . . , n2 . 2
There are n different possibilities for each αi , hence we have at most nn different possibilities for the matrix X A. The nonsingularity of A implies that the group G is finite. 2 Theorem 4.5. Let n 2. Suppose that A ∈ Cn×n is a matrix and H ⊂ Cn×n is an irreducible matrix group such that for all X ∈ H the product X A has a single eigenvalue. Then A = 0. Proof. We shall consider two cases. If A is a nilpotent matrix, 0 is the only eigenvalue of X A for each X ∈ H. Therefore, tr( X A ) = 0 for all X ∈ H and by Lemma 4.3 A = 0. Now suppose that σ ( A ) = {α } and α = 0. Then, α −1 A is a unipotent matrix and we can assume that A = I + N for a nilpotent matrix N. By Lemma 4.4 we can also assume that H is a finite group and that det X = 1, σ ( X A ) = {α X } and |α X | = 1 for all X ∈ H. 1 −1 ) = tr( X A ). After applying a suitable Then, σ (( X A )−1 ) = {α − X } = {α X } and therefore tr(( X A ) similarity we can assume that all matrices in H are unitary. The unitarity of X implies
tr( X A ) = tr ( X A )−1 = tr A −1 X ∗ = tr X A −1
∗
,
hence
tr X A − A −1
∗
= 0.
We can again use Lemma 4.3 to see that A − ( A −1 )∗ = 0. Therefore, A is a unitary matrix, N = 0, A = I and each member of the group H has a single eigenvalue. Corollary 4.1.7 from [4] then implies the reducibility of H, which is a contradiction. 2 Corollary 4.6. Suppose that H is an irreducible subgroup of GLn (C). Then, each non-zero matrix A is H-semiseparable. References [1] C.S. Ballantine, Stabilization by a diagonal matrix, Proc. Amer. Math. Soc. 25 (1970) 728–734. [2] M. Choi, Z. Huang, C. Li, N. Sze, Every invertible matrix is diagonally equivalent to a matrix with distinct eigenvalues, Linear Algebra Appl. 436 (2012) 3773–3776. [3] X. Feng, Z. Li, T. Huang, Is every nonsingular matrix diagonally equivalent to a matrix with all distinct eigenvalues?, Linear Algebra Appl. 436 (2012) 120–125. [4] H. Radjavi, P. Rosenthal, Simultaneous Triangularization, Universitext, Springer-Verlag, New York, 2000.