Implementation

We look forward to implement by usage of orthogonal matrices because the condition number of them equals 1 [40]. In particular, we choose the Householder matrix [20], denoted as H ∈ IR^n×n, which is orthogonal and possesses following properties:

1. H = I − 2vv^T, where I is the (n × n) identity matrix, v ∈ IRⁿ is an unit vector.

2. -1 is one eigenvalue of H with corresponding eigenvector, v. Moreover, 1 of multi-plicity (n-1) are the other eigenvalues with corresponding eigen-space being perpen-dicular to v.

3. H^T = H, det(H) = −1, and tr(H) = n − 2.

4. Let H^′ = H2H1, where Hi = I − 2vⁱv_i^T, ||vⁱ|| = 1, i = 1, 2, then H’ is an orthog-onal matrix. Moreover, if v1 and v2 are LI, then the subspace perpendicular to span(v1, v2) is the eigen-space of H^′ corresponding to eigenvalue 1 of multiplicity (n − 2).

To implement, we need the following lemmas:

Lemma 5.5 Assume that A ∈ IR^n×n is orthogonal, diagonalizable, and has four distinct eigenvectors: p1 = pR+ ipI, p2 = pR − ip^I, p3 = qR+ iqI, and p4 = qR− iq^I, with corresponding distinct eigenvalues λ₁ = αp + iβp, λ₂ = αp − iβ^p, λ₃ = αq + iβq, and λ4 = αq+ iβq, where βp 6= 0 and β^q 6= 0. Then

(i) A can be represented as A = Ψ S 0 0 T

!

Ψ⁻¹, where S = αp βp

−β^p αp

!

∈ IR^2×2, T =diag[λ3, λ4, · · · , λⁿ] ∈ IR(n−2 )×(n−2 ), and Ψ =



pR...pI...p3...p4...···...p^n. (ii) pR⊥p^I and ||p^R|| = ||p^I|| = ^√1₂.

(iii) (pR, pI, qR, qI) is an orthogonal set.

Proof: (i) Since A is diagonalizable, let A = ^Pn_i=1λipiq^T_i = ^P2_i=1λipiq^T_i +^Pn_i=3λipiq^T_i , where pi and qi are right and left eigenvectors of A associated with A’s eigenvalues λi, respectively, for i = 1, 2, · · · , n. Consider the first summation only,

P2 w1 and w2 are complex LI vectors. In order to yield two real-valued LI vectors from w1

and w₂, let r₁ = ¹₂(w₁+ w₂) = pRαp− p^Iβp and r₂ = ⁻ⁱ₂ (w₁− w2) = pRβp+ pIαp. Hence for i = 1, 2, are both Householder matrices. Let α ± iβ denote two eigenvalues of H (else being 1) with corresponding eigenvectors, p = pR+ ipI and ¯p. Moreover, we choose LI unit vectors u3, u4 ∈ {u¹, u2}^⊥, then

(ii)From the definition of trace of a matrix, we have tr(H2H1) = n−2+2α. In addition, tr(H2H1) = tr^h(I − 2u²u^T₂)(I − 2u¹u^T₁)ⁱ = tr^hI − 2u¹u^T₁ − 2u²u^T₂ + 4(u^T₂u1)(u2u^T₁)ⁱ = n − 4 + 4 cos²θ, where cos θ = u^T₁u2 = tr(u^T₁u2) = tr(u^T₂u1). Combining these two results yields α = −1 + 2 cos²θ. Moreover, since det(H₂H₁) = 1, we have α² + β² = 1 ⇒ β =

√1 − α².

(iii)By (i), span(u1, u2) =span(pR, pI). Given that u3 ∈ {u¹, u2}^⊥, then u3 ∈ {p^R, pI}^⊥ ⇒ p^Hu3 = ¯p^Hu3 = 0. Hence H3Hp = (I − 2u³u^T₃)Hp = (I − 2u³u^T₃)(α + iβ)p = (α + iβ)p. Similarly, it is true for ¯p. By similar procedure, it is true for H4H3H.

From Lemma 5.6, we have the following two results:

Corollary 5.3 Recall that C is defined in Problem A in Section 5.1.

1. span(u1, u2) 6⊂ C^⊥ ⇔ span(p^R, pI) 6⊂ C^⊥ ⇔ C^T(p) = C^T(pR + ipI) 6= 0 ⇔ C^T(¯p) = C^T(pR− ip^I) 6= 0.

2. We categorize θ into the the following:

(a) θ ∈ (0,^π4) ∪ (^3π4, π) : α > 0.

(b) θ ∈ (^π₄,^3π₄ ) : α < 0.

(c) θ = ^π₄,^3π₄ : (α, β) = (0, 1).

(d) θ = ^π₂ : (α, β) = (−1, 0).

By using above lemmas, we obtain the next important result described by Theorem 5.2 as below:

Theorem 5.2 We consider that rank(C) =rank(B) = 1( other cases can be discussed similarly) and denote c^T and b as one row vector of C and one column vector of B, respectively. Under the following two cases of assumptions:

(i) If n is odd, i.e., n = 3, 5, 7, · · ·, we assume that {x, f} are LI, and {c^T, f} are LI.

(ii) If n is even, i.e., n = 4, 6, 8, · · ·, we assume that {x, f, c^T} are LI. Moreover, if f^Tx ≤ 0, we also assume that {x, f, b} are LI.

Then we can factorize A, which solves Problem A, into products of Householder matrice, therefore the condition number of A equals 1 [40].

Proof: (i)Let K = _||x||^{||f ||}. If f^Tx ≤ 0, then let u1 = _||Kx+f||^Kx+f ∈ IRⁿ; else, let u₁ = _||Kx−f||^Kx−f ∈ IRⁿ. Hence H1x = f, where H1 = I − 2u¹u1T. Then choose an unit vector u2 ∈ IRⁿ such that u2 ∈ f^⊥ , and u2 6∈ C^⊥. Since u2 ∈ f^⊥, we have H2H1x = f, where H2 = I − 2u²u2T. If f^Tx ≤ 0, then let u¹ = _||Kx+f||^Kx+f ∈ IRⁿ. By Lemma 5.6, we know that the two eigenvalues (6= 1) of H²H1 will have positive real parts; else, we require that |u^T2u1| = ξ^(2,1) < ^√1₂, hence the eigenvalues(6= 1) of H²H1 will have negative real parts, else if n = 3, we also require that b 6∈ span(u¹, u2).

When n = 3, we require that c^T 6∈ span(u¹, u2). Then we can construct A. If f^Tx ≤ 0, then let A = −H2H₁, hence all eigenvalues of A will have negative real parts, which implies that A is stabilizable. On the other hand, since u2 6∈ C^⊥, by Lemma 5.6 and Corollary 5.3, we have the two right eigenvectors(6= −1) of A not perpendicular to c^T. Moreover, since c^T 6∈ span(u¹, u2), the right eigenvector corresponding to eigenvalue -1 will not be perpendicular to c^T, which means A is observable; else, we let A = H2H1. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of real part (1) is not perpendicular to b. On the other hand, A is observable by same reason as the other case of f^Tx ≤ 0.

When n 6= 3, we need an iteration. Iterate this step over i = 3, 5, 7, · · · , n − 2.

Choose unit vectors ui and ui+1, which form a sub-basis of (u1, u2, · · · , uⁱ−1, f)^⊥ and c^Tui 6= 0. Therefore we have Hⁱ⁺¹· · · H²H1x = f, and by Lemma 5.6, that the eigenvec-tors of Hi−1· · · H²H1 corresponding to span(u1, u2, · · · , uⁱ−1) will still be eigenvectors of Hi+1· · · H²H1. If i = n − 2, we also require that c^T 6∈ span(u¹, u2, · · · , uⁿ−1). Moreover, if f^Tx ≤ 0, then we require that |u^Ti+1ui| = ξ^(i+1,i) > ^√1₂ and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ^(2,1); else, we require that |u^Ti+1ui| = ξ^(i+1,i) < ^√1₂ and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ^(2,1), else if i = n − 2, we also require that b 6∈ span(u1, u₂, · · · , uⁿ−1). As all iterations are done, we still need the final step to fully construct A. If f^Tx ≤ 0, then A = −Hⁿ−1· · · H²H1, where Hj = I − 2u^ju^T_j, ∀j = 1, 2, · · · , n − 1. By Lemma 5.6, we have all eigenvalues of A having negative real parts, which implies that A is stabilizable. On the other hand, since ui 6∈ c^⊥,

by Lemma 5.6 and Corollary 5.3, we have the n-1 eigenvectors of A corresponding to eigenvalues(6= −1) not perpendicular to c^T. Moreover, since c^T 6∈ span(u¹, u2, · · · , uⁿ−1), the right eigenvector of A corresponding to eigenvalue -1 will not be perpendicular to c^T, which means A is observable; else, we let A = Hn−1· · · H2H₁. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of positive real part, which equals 1 and this left eigenvector is perpendicular to span(u1, u2, · · · , uⁿ−1), is not per-pendicular to b. On the other hand, A is observable by same reason as the other case of f^Tx ≤ 0.

(ii)Let k and u1 as given in (i). Then choose an unit vector u2 ∈ IRⁿ such that u2 ∈ f^⊥ and u2 6∈ c^⊥. Since u2 ∈ f^⊥, we have H2H1x = f, where H2 = I − 2u²u2T. If f^Tx ≤ 0, we require that |u^T2u₁| = ξ(2,1) > ^√1₂, and by Lemma 5.6, the two eigenvalues (6= 1) of H2H1 will have positive real parts; else, we require that |u^T2u1| = ξ(2,1) < ^√1₂, hence the eigenvalues(6= 1) of H²H1 have negative real parts.

When n = 4, we choose an unit vector u3 ∈ {u¹, u2, f}^⊥ and require that c^T 6∈

span(u1, u2, f), which implies c^Tu3 6= 0. Finally, we can construct A. If f^Tx ≤ 0, then A = −H²H1, hence all eigenvalues of A have negative real parts, which implies that A is stabilizable. On the other hand, since u2 6∈ c^⊥and c^Tu3 6= 0, by Lemma 5.6 and Corollary 5.3, we know that the three right eigenvectors of A corresponding to eigenvalues (6= −1) is not perpendicular to c^T. Moreover, since the right eigenvector of A corresponding to eigenvalue -1 is perpendicular to span{u¹, u2, u3} and {x, f, c} are LI, we obtain that {u¹, c} are LI and therefore this right eigenvector is not perpendicular to c, which means A is observable; else, we let A = H2H1. By similar derivation of c, we conclude that the right eigenvector corresponding to eigenvalue 1 is not perpendicular to b. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of real parts (equals 1) is not perpendicular to b. On the other hand, A is observable by same reason as the other case of f^Tx ≤ 0.

When n 6= 4, we need an iteration. Iterate this step over i = 3, 5, · · · , n − 3.

Choose unit vectors ui and ui+1, which form a sub-basis of (u1, u2, · · · , uⁱ−1, f)^⊥ and c^Tui 6= 0. Therefore we have Hⁱ⁺¹· · · H²H1x = f, and by Lemma 5.6, that the eigen-vectors of Hi−1· · · H²H1 corresponding to span(u1, u2, · · · , uⁱ−1) are still eigenvectors of

Hi+1· · · H²H1. If i = n − 3, we also require that c^T 6∈ span(u¹, u2, · · · , uⁿ−2, f). Moreover, if f^Tx ≤ 0, then we require that |u^Ti+1ui| = ξ^(i+1,i) > ^√1₂ and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ^(2,1),

; else, we require that |u^Ti+1ui| = ξ^(i+1,i) < ^√1₂ and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ^(2,1). As all iterations are done, we choose unit vector un−1 ∈ {u1, u₂, · · · , uⁿ−2, f}^⊥. Finally, we can construct A. If f^Tx ≤ 0, then A = −Hⁿ−1· · · H2H₁, where Hj = I − 2u^ju^T_j, ∀j = 1, 2, · · · , n − 1. By Lemma 5.6, we have all eigenvalues of A having negative real parts, which implies that A is stabilizable. On the other hand, since ui 6∈ c^⊥ and c^T 6∈

span(u1, u2, · · · , uⁿ−2, f), by Lemma 5.6 and Corollary 5.3, we have the n-1 eigenvectors of A corresponding to eigenvalues (6= −1) not perpendicular to c^T. Moreover, since this right eigenvector of A corresponding to eigenvalue -1 is perpendicular to span(u1, · · · , uⁿ−1) and {x, f, c} are LI, we have {u1, c} are LI and thus this eigenvector is not perpendicular to c, which means A is observable; else, we let A = Hn−1· · · H²H1. By similar derivation of c, we conclude that the right eigenvector corresponding to eigenvalue 1 is not perpendic-ular to b. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of positive real part (equals 1) is not perpendicular to b. On the other hand, A is observable by same reason as the other case of f^Tx ≤ 0.