Applications and discussions

The first application of our results in Section 2.2 is a generalization of Lemma 2.2.3.

Indeed, we are able to determine the value of k(A) for A = B ⊕ C with normal C.

Proposition 2.3.1. Let A = B ⊕ C, where C is an m-by-m normal matrix.

Then k(A) = k1(B) + k1(C). In this case, k(A) = k(B) + k(C) if and only if k1(B) = k(B) and k1(C) = k(C). In particular, if C = cIm for some scalar c, then k(A) = k1(B) + k1(cIm).

Proof. Let the normal C be unitarily similar to ⊕^mj=1[cj]. By [19, Lemma 2.9], we may assume that all the cj’s lie in ∂W (A). This shows that k1(C) = m immediately.

On the other hand, we also obtain k(A) = k1(B) + m by Lemma 2.2.3. Hence the asserted equality k(A) = k1(B) + k1(C) has been proven. The remaining assertions

hold trivially by this equality. 

An easy corollary of Proposition 2.3.1 is to determine when k (A) equals the size of A for a matrix A = B ⊕ C with normal C.

Corollary 2.3.2. Let A = B ⊕ C, where B is an n-by-n matrix and C is an m-by-m normal matrix. Then k(A) = n + m if and only if k1(B) = n and k1(C) = m.

Assume, moreover, that dim Hη = 1 for all η ∈ ∂W (B). Then k(A) = n + m if and only if k1(B) = n ≤ 2 and k¹(C) = m.

Proof. By Proposition 2.3.1, it is clear that k(A) equals the size of A if and only if k1(B) and k1(C) equal the sizes of B and C, respectively. In this case, the assumption on Hη implies that k1(B) = n ≤ 2 by [19, Proposition 2.10]. This completes the proof.



For a matrix A of the form B ⊕ C, we recall the decomposition Γ = Γ¹ ∪ Γ² at

the end of Section 2.2, where Γ1 = ∂W (A) \ (∂W (B) ∪ ∂W (C)) and Γ² = ∂W (A) ∩

∂W (B) ∩ ∂W (C). The next proposition gives a lower bound for k(A).

Proposition 2.3.3. Let A = B ⊕ C be an n-by-n (n ≥ 3) matrix. Then Γ is empty if and only if the numerical range of one summand is contained in the interior of the numerical range of the other. In particular, if Γ is nonempty, then k(A) ≥ 3.

Proof. If Γ = Γ1 ∪ Γ² is empty, then both Γ1 and Γ2 are empty. Since Γ1 is empty, ∂W (A) is contained in ∂W (B) ∪ ∂W (C). This implies that W (B) ∩ W (C) is nonempty, and thus W (B) = W (C), W (B) ⊆ int W (C) or W (C) ⊆ int W (B).

Moreover, Γ₂ = φ implies that W (B) 6= W (C). With this, we conclude that either W (B) ⊆ int W (C) or W (C) ⊆ int W (B). The converse is obvious. Hence we have proved the first assertion. Let Γ be nonempty, that is, either Γ₁ or Γ₂ is nonempty. If Γ1 is nonempty, then there is a line segment on the boundary of W (A). This shows that k(A) ≥ 3 by [19, Corollary 2.5]. On the other hand, if Γ2 is nonempty, then there is a (possibly degenerate) line segment on the common boundaries of the three numerical ranges W (A), W (B) and W (C). Using [19, Corollary 2.5] again, we may assume that the line segment is degenerate, say, to {ξ}. This implies immediately that dim_ξH(A) ≥ 2. Thus k(A) ≥ 3 by [19, Proposition 2.4]. 

As an application, when A is reducible, the next corollary gives a necessary and sufficient condition for k(A) = 2.

Corollary 2.3.4. Let A = B ⊕ C be an n-by-n (n ≥ 3) matrix. Then k(A) = 2 if and only if either k(B) = 2 and W (C) ⊆ int W (B), or k(C) = 2 and W (B) ⊆ int W (C).

Proof. If k(A) = 2, then Proposition 2.3.3 shows that Γ is empty, and thus the numerical range of one summand, say, B is contained in the interior of the numerical range of C. Hence k(C) = 2 by [19, Lemma 2.9]. The converse is obvious by [19,

22

Lemma 2.9] again. 

The following proposition determines exactly when k(A) equals the size of A for an irreducible matrix A. It is also stated in [2, Theorem 7] while the proof there is different from ours.

Proposition 2.3.5. Let A be an n-by-n (n ≥ 3) irreducible matrix. Then k(A) = n if and only if ∂W (A) contains a line segment l and there are n points (not necessarily distinct) in l ∪ (∂W (A) ∩ L), where L is the supporting line parallel to l such that their corresponding unit vectors form an orthonormal basis for Cⁿ.

Proof. We need only prove the necessity. Assume that A is an n-by-n (n ≥ 3) irreducible matrix with k(A) = n. If ∂W (A) contains no line segment, then dim Hξ= dim Eξ,l≤ n/2 for all ξ ∈ ∂W (A) by [19, Proposition 2.2]. If n is odd, say, n = 2m+1, then dim Hξ = dim Eξ,l ≤ m for all ξ ∈ ∂W (A). Since k(A) = n, it follows from [19, Theorem 2.7] that A is reducible, which is absurd. If n is even, say, n = 2m, then m ≥ 2 by our assumption that n ≥ 3. Since k(A) = n and ∂W (A) contains no line segment, A is unitarily similar to a matrix of the form

 and the latter is unitarily similar to

m

This contradicts the irreducibility of A. Hence ∂W (A) must contain a line segment.

We then apply [19, Theorem 2.7] again to complete the proof. 

An easy corollary of Proposition 2.3.5 is the following upper bound for k(A). This was given in [19, Proposition 2.10]. Here we give a simpler proof.

Corollary 2.3.6. If A is an n-by-n (n ≥ 3) matrix with dim Hξ = 1 for all ξ ∈ ∂W (A), then k(A) ≤ n − 1.

Proof. Assume that k(A) = n. It suffices to consider that A is reducible; this is because if otherwise, then Proposition 2.3.5 shows that ∂W (A) contains a line segment, which contradicts the assumption on Hξ. Let A = B ⊕ C. Then our assumption on Hξ implies that Γ is empty. By Proposition 2.3.3, we obtain that the numerical range of one summand is contained in the interior of the numerical range of the other summand. It follows from [19, Lemma 2.9] that the value of k(A) equals

k(B) or k(C). Thus k(A) ≤ n − 1 as asserted. 

We now combine Proposition 2.3.1, Corollary 2.3.2, Corollary 2.3.4, and Proposi-tion 2.3.5 to determine the value of k(A) for any 4-by-4 reducible matrix A. Corollary 2.3.4 shows exactly when the value of k(A) equals two. By Proposition 2.3.1, Corol-lary 2.3.2 and Proposition 2.3.5, we get a necessary and sufficient condition for the value of k(A) to be equal to four. In other words, the value of k(A) can be determined completely for any 4-by-4 reducible matrix A. To do this, we note that a 4-by-4 re-ducible matrix A can be written, after a unitary similarity, as (i) A = B ⊕ [c], where B is a 3-by-3 irreducible matrix and c is a complex number, (ii) A = B ⊕ [c], where B is a 3-by-3 reducible matrix and c is a complex number, or (iii) A = B ⊕ C, where B and C are 2-by-2 irreducible matrices. Proposition 2.3.7 below is to deal with case (i).

Recall that for a 3-by-3 irreducible matrix A, W (A) is of one of the following

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shapes (cf. [9]): an elliptic disc, the convex hull of a heart-shaped region, in which case ∂W (A) contains a line segment, and an oval region.

Proposition 2.3.7. Let A = B ⊕ [c], where B is a 3-by-3 irreducible matrix and c is a complex number. Then k(A) = 4 if and only if c /∈ int W (B) and {a¹, a2, b} ⊆

∂W (A), where W (B) is the convex hull of a heart-shaped region, in which case ∂W (B) contains a line segment [a1, a2] contained in the supporting line L1 of W (B) and L2

is the supporting line of W (B) passing through b and parallel to L1.

Proof. By Corollary 2.3.2, we see that k(A) = 4 is equivalent to k1(B) = 3 and k₁([c]) = 1. Since a necessary and sufficient condition for k₁([c]) = 1 is that c /∈ int W (B), it remains to show that k¹(B) = 3 if and only if {a¹, a2, b} ⊆ ∂W (A) and W (B) satisfies the asserted properties. If k₁(B) = 3, then k(B) = 3. Hence it follows from Proposition 2.3.5 that ∂W (A) contains {a¹, a2, b}, and W (B) is as

asserted. The converse is trivial. 

For case (ii), let A = B ⊕ [c], where B is a 3-by-3 reducible matrix. After a unitary similarity, B can be written as C ⊕ [b], where C is a 2-by-2 matrix, so that k(A) = k1(C) + k1([b] ⊕ [c]) by Proposition 2.3.1. The following proposition gives a necessary and sufficient condition for k(A) to be equal to four.

Proposition 2.3.8. Let A = C ⊕ [b] ⊕ [c], where C is a 2-by-2 matrix, and b and c are complex numbers. Then k(A) = 4 if and only if both b and c are in ∂W (A) and k1(C) = 2.

Proof. By Corollary 2.3.2, it is obvious that k (A) = 4 if and only if k1(C) = 2 and k1([b] ⊕ [c]) = 2. Moreover, it is also clear that k¹([b] ⊕ [c]) = 2 is equivalent to both of b and c being in ∂W (A). Hence the proof is complete. 

To prove for case (iii), let A = B ⊕ C, where B and C are 2-by-2 irreducible

matrices. Since W (A) is the convex hull of the union of the two elliptic discs W (B) and W (C), either W (B) equals W (C), or Γ consists of at most four (possibly degen-erate) line segments. With this, we are now ready to give a necessary and sufficient condition for k(A) = 4.

Proposition 2.3.9. Let A = B ⊕ C, where B and C are 2-by-2 irreducible matri-ces. Then k(A) = 4 if and only if Γ consists of at least three line segments (including the possibly degenerate ones), or Γ consists of exactly two (possibly degenerate) line segments such that k1(B) = k1(C) = 2.

Proof. If Γ consists of more than four (possibly degenerate) line segments, then the two elliptic discs W (B) and W (C) are identical. Hence k(A) = 4 by direct computations. If Γ consists of four or three (possibly degenerate) line segments, then the endpoints of the major axes of the two elliptic discs W (B) and W (C) are in

∂W (A). Hence k(A) = 4. If Γ consists of exactly two (possibly degenerate) line segments such that k1(B) = k1(C) = 2, then k(A) = 4 by Theorem 2.2.6. Therefore we have proved the sufficient condition for k(A) = 4. Next assume that k(A) = 4 and either Γ consists of exactly two (possibly degenerate) line segments such that the equalities k₁(B) = k₁(C) = 2 fail, or Γ consists of at most one (possibly degenerate) line segment. Since property Λ holds in each case, we must have k1(B) = k1(C) = 2 by Theorem 2.2.6. This shows that we need only consider the latter. If Γ consists of exactly one (possibly degenerate) line segment, then Γ1 is empty and Γ2is a singleton.

Hence we may assume that W (B) is contained in W (C) and the intersection of W (B) and W (C) is Γ. This shows that k1(B) = 1 and k1(C) = 2, which is a contradiction.

If Γ is empty, then it follows from Proposition 2.3.3 that the numerical range of one summand, say, B is contained in the interior of the numerical range of the other summand C. By Corollary 2.3.4 and [5, Lemma 4.1], we see that k(A) = k(C) = 2,

which is absurd. This completes the proof. 

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As a final application of Theorem 2.2.6, it is obvious that the convex hull of the union of W (A) and W (A + aI_n) has property Λ for any a 6= 0. Hence we obtain the following proposition.

Proposition 2.3.10. Let A be an n-by-n matrix and a be a nonzero complex num-ber. Then k(A ⊕ (A + aIⁿ)) = k1(A) + k1(A + aIn). In this case, k(A ⊕ (A + aIⁿ)) = 2k(A) if and only if k1(A + aIn) = k1(A) = k(A).

We conclude this paper by stating the following open questions concerning this topic. Is it true that the equality k(A) = k1(B) + k1(C) holds for a matrix A of the form B ⊕C even if property Λ fails? We note that although property Λ fails, the men-tioned formula may still be correct (cf. Proposition 2.3.1). Another natural example of the failure of property Λ is that both W (B) and W (C) have the same numerical range. Is it true that k (B ⊕ C) = k(B) + k(C) in this case? In particular, can we determine the value of k (A ⊕ A) (cf. Proposition 2.3.10)? The following proposition gives a partial answer for k (A ⊕ A) if we assume, in addition, that dim H^ξ = 1 for all ξ ∈ ∂W (A).

Proposition 2.3.11. If A is an n-by-n matrix with dim Hξ = 1 for all ξ ∈

∂W (A), then

k

m

M

j=1

A

!

= m · k (A) .

Proof. Obviously, the inequality k ⊕^mj=1A
≥ m · k (A) holds. To prove the re-versed inequality, we consider, for convenience, the case m = 2. Let ξ1 ∈ ∂W (A ⊕ A).

Then dim Hξ1(A ⊕ A) = 2 by our assumption on H^ξ(A). Hence the subspace Hξ1(A ⊕ A) is spanned by the two unit vectors x₁⊕ 0 and 0 ⊕ x1, where ξ₁ = hAx1, x₁i. Let z1

be a unit vector in Hξ1(A ⊕ A). Then z¹ = (α1x1⊕ α²x1) / q

|α¹|²+ |α²|², where α1

and α₂ are in C. Similarly for ξ2 ∈ ∂W (A ⊕ A). That is, the subspace Hξ2(A ⊕ A)

is spanned by the two unit vectors x2 ⊕ 0 and 0 ⊕ x², where ξ2 = hAx², x2i. More-over, if z₂ is a unit vector in H_ξ2(A ⊕ A), then z2 = (β₁x₂⊕ β2x₂) /

q

|β1|²+ |β2|², where β1 and β2 are in C. Obviously, the orthogonality of z1 and z2 is equivalent to

α₁β¯₁ + α₂β¯₂
hx1, x₂i = 0, that is,

This shows that k(A ⊕ A) ≤ 2k(A) immediately by the definition of k(A).

For general m, a similar argument as above yields that

* orthogonal to each other is at most m. We infer from this and the above equality that the reversed inequality k ⊕^mj=1A
≤ m · k (A) holds. Therefore we have the asserted

equality. 

At the end of this section, we apply Proposition 2.3.11 to the quadratic matrices.

Recall that an n-by-n quadratic matrix A is unitarily similar to a matrix of the form

aIn1 ⊕ bIⁿ2 ⊕

Corollary 2.3.12. If A is an n-by-n quadratic matrix of the above form and D is not missing, then k(A) = 2 · # ({λ ∈ σ (D) : λ = kDk}).

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Proof. If D > 0, then D is unitarily similar to diag (d1, ..., dn3) , where d1 = · · · = d_p = kDk ≡ d > dp+1 ≥ · · · ≥ dn3 ≥ 0 (1 ≤ p ≤ n3). Hence A is unitarily similar to a matrix of the form aIn1 ⊕ bIⁿ2 ⊕^pj=1B ⊕ⁿj=p+1³ Bj, where n1+ n2+ 2n3 = n,

B ≡



 a d 0 b



, and Bj ≡



 a dj

0 b



, j = p + 1, . . . , n3.

Since the set {a, b} and all of the numerical ranges W (B^j), j = p + 1, . . . , m, are contained in the interior of W (B), it follows from [19, Lemma 2.9] that k(A) = k(⊕^pj=1B). Since dim Hξ(B) = 1 for all ξ ∈ ∂W (B), we have k(A) = p · k(B) by Proposition 2.3.11. Obviously, k(B) = 2 by [5, Lemma 4.1]. Thus k(A) = 2p as

asserted. 

We remark that in the preceding proof the equality k(⊕^pj=1B) = 2p can also be established directly. Indeed, the inequality k(⊕^pj=1B) ≥ 2p holds trivially and we can infer from [5, Lemma 4.1] that k(⊕^pj=1B) = 2p.