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Block Kronecker linearizations of matrix polynomials and their backward errors

Abstract

We introduce a new family of strong linearizations of matrix polynomials—which we call “block Kronecker pencils”—and perform a backward stability analysis of complete polynomial eigenproblems. These problems are solved by applying any backward stable algorithm to a block Kronecker pencil, such as the staircase algorithm for singular pencils or the QZ algorithm for regular pencils. This stability analysis allows us to identify those block Kronecker pencils that yield a computed complete eigenstructure which is exactly that of a slightly perturbed matrix polynomial. The global backward error analysis in this work presents for the first time the following key properties: it is a rigorous analysis valid for finite perturbations (i.e., it is not a first order analysis), it provides precise bounds, it is valid simultaneously for a large class of linearizations, and it establishes a framework that may be generalized to other classes of linearizations. These features are related to the fact that block Kronecker pencils are a particular case of the new family of “strong block minimal bases pencils”, which are robust under certain perturbations and, so, include certain perturbations of block Kronecker pencils.

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Notes

  1. 1.

    We emphasize that this bijection is not a linear map since the fields of the linear spaces corresponding to the domain and the codomain are different. Nevertheless, it has some obvious linear properties that can be used.

  2. 2.

    The reader should take into account that in [48] the characteristic polynomial is defined as \(\det (M_\eta M_\eta ^T-\lambda I)\) and the change of variable is slightly different.

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Author information

Correspondence to Froilán M. Dopico.

Additional information

This work was partially supported by the “Ministerio de Economía, Industria y Competitividad of Spain” and “Fondo Europeo de Desarrollo Regional (FEDER) of EU” through Grants MTM-2012-32542, MTM-2015-68805-REDT, MTM-2015-65798-P, MTM2017-90682-REDT, by the Belgian network DYSCO (Dynamical Systems, Control, and Optimization), funded by the Interuniversity Attraction Poles Programme initiated by the Belgian Science Policy Office, and by the Engineering and Physical Sciences Research Council of UK through Grant EP/I005293.

Appendices

Appendix A. The minimal bases of strong block minimal bases pencils

In this appendix, we state and prove Lemma A.1, which establishes, first, the relationship between the vectors in the rational right null spaces of any of the strong block minimal bases pencils \(\mathcal {L}_{}(\lambda )\) introduced in Definition 3.1 and of the corresponding matrix polynomial \(Q(\lambda )\) in (3.2), and, second, the relationship between the right minimal bases of \(\mathcal {L}_{}(\lambda )\) and \(Q(\lambda )\). In this paper Lemma A.1 is only used in the proof of Theorem 3.6, but we emphasize that is very useful for proving the recovery procedures of eigenvectors and minimal bases of block Kronecker pencils in [25, Section 7] and that is a fundamental result in the theory of strong block minimal bases linearizations.

Lemma A.1

Let \(\mathcal {L}_{}(\lambda )\) be a strong block minimal bases pencil as in (3.1), let \(N_1(\lambda )\) be a minimal basis dual to \(K_1 (\lambda )\), let \(N_2(\lambda )\) be a minimal basis dual to \(K_2 (\lambda )\), let \(Q(\lambda )\) be the matrix polynomial defined in (3.2), and let \(\widehat{N}_2 (\lambda )\) be the matrix appearing in (3.3). Then the following hold:

  1. (a)

    If \(h(\lambda ) \in \mathcal {N}_r (Q)\), then

    $$\begin{aligned} z(\lambda ) := \begin{bmatrix} N_1 (\lambda )^T \\ -\widehat{N}_2 (\lambda ) M(\lambda ) N_1 (\lambda )^T \end{bmatrix} h(\lambda ) \, \in \mathcal {N}_r (\mathcal {L}). \end{aligned}$$
    (A.1)

    Moreover, if \(0 \ne h(\lambda ) \in \mathcal {N}_r (Q)\) is a vector polynomial, then \(z(\lambda )\) is also a vector polynomial and

    $$\begin{aligned} \deg (z (\lambda )) = \deg ( N_1 (\lambda )^T \, h(\lambda )) = \deg ( N_1 (\lambda )) + \deg (h(\lambda )). \end{aligned}$$
    (A.2)
  2. (b)

    If \(\{h_1(\lambda ), \ldots , h_p(\lambda )\}\) is a right minimal basis of \(Q(\lambda )\), then

    $$\begin{aligned} \left\{ \begin{bmatrix} N_1 (\lambda )^T \\ -\widehat{N}_2 (\lambda ) M(\lambda ) N_1 (\lambda )^T \end{bmatrix} h_1(\lambda ), \ldots , \begin{bmatrix} N_1 (\lambda )^T \\ -\widehat{N}_2 (\lambda ) M(\lambda ) N_1 (\lambda )^T \end{bmatrix} h_p(\lambda ) \right\} \end{aligned}$$

    is a right minimal basis of \(\mathcal {L}_{}(\lambda )\).

Proof

(a) It can be checked, via a direct multiplication, that the matrix \(X(\lambda )\) in (3.4) satisfies \(X(\lambda ) = \widehat{N}_2 (\lambda ) M(\lambda ) N_1 (\lambda )^T\). Then, from (3.4), we get that

$$\begin{aligned} (U_2(\lambda )^{-T} \oplus I_{m_1}) \, \mathcal {L}_{}(\lambda ) \, (U_1(\lambda )^{-1} \oplus I_{m_2}) \begin{bmatrix} 0 \\ I \\ -X(\lambda ) \end{bmatrix} = \begin{bmatrix} 0 \\ Q(\lambda ) \\ 0 \end{bmatrix}, \end{aligned}$$

where the sizes of the identity and zero blocks are conformable with the partition of the last matrix in (3.4). By using the structure of \(U_1(\lambda )^{-1} \oplus I_{m_2}\) (recall (3.3)), the multiplication of the last two factors in the left-hand side of the previous equation leads to

$$\begin{aligned} (U_2(\lambda )^{-T} \oplus I_{m_1}) \, \mathcal {L}_{}(\lambda ) \, \begin{bmatrix} N_1(\lambda )^T \\ -X(\lambda ) \end{bmatrix} = \begin{bmatrix} 0 \\ Q(\lambda ) \\ 0 \end{bmatrix}. \end{aligned}$$
(A.3)

This equation implies that \(z(\lambda ) \in \mathcal {N}_r (\mathcal {L})\) if \(h(\lambda ) \in \mathcal {N}_r (Q)\), and also that \(z(\lambda )\) is a vector polynomial if \(h(\lambda )\) is, because \(N_1(\lambda )\) and \(X(\lambda )\) are matrix polynomials.

It only remains to prove the degree shift property (A.2) to conclude the proof of part (a). First, take into account that all the row degrees of the minimal basis \(N_1 (\lambda )\) are equal and that its highest degree coefficient has full row rank. Therefore,

$$\begin{aligned} \deg ( N_1 (\lambda )^T \, g(\lambda )) = \deg ( N_1 (\lambda )) + \deg (g(\lambda )), \end{aligned}$$
(A.4)

for any vector polynomial \(g(\lambda ) \ne 0\). The same argument applied to the minimal basis \(K_2 (\lambda )\) proves that

$$\begin{aligned} \deg ( K_2 (\lambda )^T \, y(\lambda )) = \deg ( K_2 (\lambda )) + \deg (y(\lambda )) = 1 + \deg (y(\lambda )), \end{aligned}$$
(A.5)

for any vector polynomial \(y(\lambda ) \ne 0\). Next, observe that

$$\begin{aligned} \deg (z(\lambda )) = \max \{\deg ( N_1 (\lambda )^T h(\lambda )), \deg (X(\lambda ) h(\lambda ))\}. \end{aligned}$$
(A.6)

Therefore (A.2) follows trivially if \(X(\lambda ) h(\lambda ) = 0\). Finally, assume that \(X(\lambda ) h(\lambda ) \ne 0\) and \(h(\lambda ) \in \mathcal {N}_r (Q)\). Then use \(\mathcal {L}_{}(\lambda ) z(\lambda ) = 0\), and perform the multiplication corresponding to the first block of \(\mathcal {L}_{}(\lambda ) z(\lambda )\), using the expressions of \(z(\lambda )\) in (A.1) and \(\mathcal {L}_{}(\lambda )\) in (3.1), to get

$$\begin{aligned} M(\lambda ) N_1 (\lambda )^T h(\lambda ) = K_2(\lambda )^T X(\lambda ) h(\lambda ). \end{aligned}$$

This equality implies, together with (A.5), that

$$\begin{aligned} 1 + \deg (X(\lambda ) h(\lambda ))&= \deg (K_2(\lambda )^T X(\lambda ) h(\lambda )) \le \deg (M(\lambda )) + \deg ( N_1 (\lambda )^T h(\lambda )) \\&\le 1 + \deg ( N_1 (\lambda )^T h(\lambda )), \end{aligned}$$

and, so, \(\deg (X(\lambda ) h(\lambda )) \le \deg ( N_1 (\lambda )^T h(\lambda ))\). This inequality, together with (A.4) and (A.6) thus prove (A.2).

(b) Let us consider the matrix product

$$\begin{aligned} B(\lambda ):= \begin{bmatrix} N_1 (\lambda )^T \\ -\widehat{N}_2 (\lambda ) M(\lambda ) N_1 (\lambda )^T \end{bmatrix} [h_1(\lambda ) \cdots h_p (\lambda )], \end{aligned}$$

and let us prove that their columns are a minimal basis of the rational subspace they span by applying a version of Theorem 2.2 for columns. Note that for all \(\lambda _0 \in \overline{{\mathbb F}}\), \(B(\lambda _0)\) has full column rank since \(N_1 (\lambda _0)^T\) and \([h_1(\lambda _0) \cdots h_p (\lambda _0)]\) have both full column rank, since the columns of \(N_1 (\lambda )^T \) and \([h_1(\lambda ) \cdots h_p (\lambda )]\) are minimal bases. Next, observe that (A.2) implies that the highest column degree coefficient matrix \(B_{hc}\) of \(B(\lambda )\) has as a submatrix the highest column degree coefficient matrix \(C_{hc}\) of \(C(\lambda ) := N_1 (\lambda )^T [h_1(\lambda ) \cdots h_p (\lambda )]\). Since the column degrees of \(N_1 (\lambda )^T\) are all equal, we have that \(C_{hc}\) is the product of the highest column degree coefficient matrices of \(N_1 (\lambda )^T\) and \([h_1(\lambda ) \cdots h_p (\lambda )]\), which have both full column rank because the columns of both matrices are minimal bases. So \(C_{hc}\) has full column rank, as well as \(B_{hc}\). This implies that the columns of \(B(\lambda )\) are a minimal basis of a rational subspace \(\mathcal {S}\). In addition, \(\mathcal {S} \subseteq \mathcal {N}_r (\mathcal {L}_{}(\lambda ))\) by part (a). Finally, note that \(\mathcal {S} = \mathcal {N}_r (\mathcal {L})\) because \(\dim (\mathcal {N}_r (Q)) = \dim (\mathcal {N}_r (\mathcal {L}))\), since \(\mathcal {L}_{}(\lambda )\) is a strong linearization of \(Q(\lambda )\) by Theorem 3.3(b) and, then, Theorem 4.1 in [17] holds. \(\square \)

Appendix B. Proof of Lemma 5.4

In this appendix, we assume that \(\varepsilon \ne 0\) and \(\eta \ne 0\) according to Remark 5.1. We first reduce in Lemma B.1 the problem of computing \(\sigma _{\min }(T)\) to the problem of computing the minimum singular value of a matrix of size \(2 \varepsilon \eta \times (2 \varepsilon \eta + \varepsilon + \eta )\), which is much smaller than the size of T.

Lemma B.1

Let T be the matrix defined in (5.13) and

$$\begin{aligned} \widehat{T} := \left[ \begin{array}{c|c} I_{\varepsilon }\otimes E_{\eta } &{}E_{\varepsilon }\otimes I_{\eta }\\ \hline I_{\varepsilon }\otimes F_{\eta }&{}F_{\varepsilon }\otimes I_{\eta } \end{array} \right] \>, \end{aligned}$$
(B.1)

where \(\lambda F_k - E_k := L_k(\lambda )\) is the pencil in (2.3). Then \(\sigma _{\min } (T) = \sigma _{\min } (\widehat{T})\).

Proof

Since the Kronecker product is associative [45, Chapter 4], we may write the matrix T as

$$\begin{aligned} \begin{aligned} T=&\left[ \begin{array}{c|c} E_{\eta }\otimes I_{m}\otimes I_{\varepsilon }\otimes I_{n} &{}I_{\eta }\otimes I_{m}\otimes E_{\varepsilon }\otimes I_{n}\\ \hline F_{\eta }\otimes I_{m}\otimes I_{\varepsilon }\otimes I_{n}&{}I_{\eta }\otimes I_{m}\otimes F_{\varepsilon }\otimes I_{n} \end{array} \right] \\ =&\left[ \begin{array}{c|c} (E_{\eta }\otimes I_{m})\otimes I_{\varepsilon }&{}I_{\eta m}\otimes E_{\varepsilon }\\ \hline (F_{\eta }\otimes I_{m})\otimes I_{\varepsilon }&{}I_{\eta m}\otimes F_{\varepsilon } \end{array} \right] \otimes I_{n}=: \widetilde{T}\otimes I_{n}. \end{aligned} \end{aligned}$$
(B.2)

Thus, \(\sigma _{\min }(T) = \sigma _{\min } (\widetilde{T})\) by [45, Theorem 4.2.15]. Following Van Loan [76], let us perform a perfect shuffle on the matrix \(\widetilde{T}\) on the right of (B.2) to swap the order of the Kronecker products of its blocks. That is, there exist permutation matrices S, \(R_1^T\) and \(R_2^T\) of sizes \(\varepsilon \eta m\times \varepsilon \eta m\), \(\varepsilon (\eta +1)m\times \varepsilon (\eta +1) m\) and \((\varepsilon +1)\eta m\times (\varepsilon +1)\eta m\), respectively, such that

$$\begin{aligned}&\left[ \begin{array}{c|c} S&{}\\ \hline &{}S \end{array} \right] \left[ \begin{array}{c|c} (E_{\eta }\otimes I_{m})\otimes I_{\varepsilon }&{}I_{\eta m}\otimes E_{\varepsilon }\\ \hline (F_{\eta }\otimes I_{m})\otimes I_{\varepsilon }&{}I_{\eta m}\otimes F_{\varepsilon } \end{array} \right] \left[ \begin{array}{c|c} R_1^{T}&{}\\ \hline &{}R_2^{T} \end{array} \right] \\&\quad = \left[ \begin{array}{c|c} I_{\varepsilon }\otimes (E_{\eta }\otimes I_{m})&{}E_{\varepsilon }\otimes I_{\eta m}\\ \hline I_{\varepsilon }\otimes (F_{\eta }\otimes I_{m})&{}F_{\varepsilon }\otimes I_{\eta m} \end{array} \right] = \left[ \begin{array}{c|c} I_{\varepsilon }\otimes E_{\eta } &{}E_{\varepsilon }\otimes I_{\eta }\\ \hline I_{\varepsilon }\otimes F_{\eta }&{}F_{\varepsilon }\otimes I_{\eta } \end{array} \right] \otimes I_{m}=\widehat{T}\otimes I_{m}. \end{aligned}$$

Using again [45, Theorem 4.2.15], we get \(\sigma _{\min } (T)= \sigma _{\min } (\widetilde{T}) = \sigma _{\min } (\widehat{T})\). \(\square \)

Lemma B.2 reduces the problem of computing the minimum singular value of \(\widehat{T}\) in (B.1) to compute the largest singular value of a matrix smaller than \(\widehat{T}\), essentially with half its size, and with a simpler structure.

Lemma B.2

Let \(\widehat{T}\) be the matrix in (B.1). Then

$$\begin{aligned} \sigma _\mathrm{min}(\widehat{T}) = \sqrt{2-\sigma _\mathrm{max}(W_{\varepsilon ,\eta }}), \end{aligned}$$
(B.3)

where \(W_{\varepsilon ,\eta } = I_\varepsilon \otimes E_\eta F_\eta ^T+E_\varepsilon F_\varepsilon ^T\otimes I_\eta \in \mathbb {R}^{\varepsilon \eta \times \varepsilon \eta }\) and \(\sigma _\mathrm{max}(W_{\varepsilon ,\eta })\) denotes its maximum singular value.

Proof

The singular values of \(\widehat{T}\) are the square roots of the eigenvalues of

$$\begin{aligned} \widehat{T} \widehat{T}^T = \begin{bmatrix} 2I_{\varepsilon \eta }&\quad W_{\varepsilon ,\eta } \\ W_{\varepsilon , \eta }^T&\quad 2I_{\varepsilon \eta } \end{bmatrix} = 2 \, I_{2 \varepsilon \eta } + \begin{bmatrix} 0&\quad W_{\varepsilon ,\eta } \\ W_{\varepsilon , \eta }^T&\quad 0 \end{bmatrix}, \end{aligned}$$

where \(W_{\varepsilon ,\eta } = I_\varepsilon \otimes E_\eta F_\eta ^T+E_\varepsilon F_\varepsilon ^T\otimes I_\eta \). It is well known (see, for instance, [68, Theorem I.4.2]) that the eigenvalues of \([0, \, W_{\varepsilon ,\eta } \, ; \, W_{\varepsilon , \eta }^T, \, 0]\) are \(\pm \sigma _1 (W_{\varepsilon ,\eta }), \ldots , \pm \sigma _{\varepsilon \eta } (W_{\varepsilon ,\eta })\), where \(\sigma _1 (W_{\varepsilon ,\eta }) \ge \cdots \ge \sigma _{\varepsilon \eta } (W_{\varepsilon ,\eta })\) are the singular values of \(W_{\varepsilon ,\eta }\). Therefore, the eigenvalues of \(\widehat{T} \widehat{T}^T\) are \(2 \pm \sigma _1 (W_{\varepsilon ,\eta }), \ldots , 2 \pm \sigma _{\varepsilon \eta } (W_{\varepsilon ,\eta })\), which implies the result. Observe that \(\widehat{T} \widehat{T}^T\) is positive semidefinite and, thus, its eigenvalues are nonnegative. \(\square \)

The advantage of the matrix \(W_{\varepsilon ,\eta }\) is that has a bidiagonal block Toeplitz structure with very simple blocks. This comes from the fact that

$$\begin{aligned} E_k F_k^T = \begin{bmatrix} 0 \\ 1&\quad 0 \\&\quad 1&\quad 0 \\&\quad&\quad \ddots&\quad \ddots \\&\quad&\quad&\quad 1&\quad 0 \end{bmatrix} =: J_k \in \mathbb {R}^{k \times k} \quad (\text{ with }\quad J_1 := 0_{1\times 1}), \end{aligned}$$

which implies

$$\begin{aligned} W_{\varepsilon ,\eta } = I_\varepsilon \otimes E_\eta F_\eta ^T+E_\varepsilon F_\varepsilon ^T\otimes I_\eta = \underbrace{ \begin{bmatrix} J_\eta \\ I_\eta&\quad J_\eta \\&\quad I_\eta&\quad J_\eta \\&\quad&\quad \ddots&\quad \ddots \\&\quad&\quad&\quad I_\eta&\quad J_\eta \end{bmatrix}}_{\displaystyle \varepsilon \; \text{ block } \text{ columns }} \left. \phantom {\begin{array}{l} l\\ l\\ l\\ l\\ l \end{array}}\!\! \!\!\!\right\} \varepsilon \text{ block } \text{ rows } \,. \end{aligned}$$
(B.4)

This structure will allow us to compute explicitly the largest singular value of \(W_{\varepsilon ,\eta }\). Without loss of generality, we assume that \(\varepsilon \ge \eta \), since, otherwise, \(W_{\varepsilon ,\eta }\) is transformed into \(W_{\eta ,\varepsilon }\) with a perfect shuffle permutation, i.e., by interchanging the order of the Kronecker products in the summands of \(W_{\varepsilon ,\eta }\). In this situation, note that if \(\eta =1\), then \(W_{1,1} = 0_{1\times 1}\) and \(W_{\varepsilon ,1} = J_\varepsilon \) for \(\varepsilon > \eta =1\). Therefore,

$$\begin{aligned} \sigma _{\max }( W_{1,1}) = 0 \quad \text{ and } \quad \sigma _{\max }( W_{\varepsilon ,1}) = 1, \quad \text{ if }\quad \varepsilon > \eta = 1. \end{aligned}$$
(B.5)

If \(\eta > 1\), then \(\sigma _{\max }( W_{\varepsilon ,\eta })\) can be computed with the help of Lemma B.3, where we show that \(W_{\varepsilon ,\eta }\) is permutationally equivalent to a direct sum involving the following two types of matrices

$$\begin{aligned} M_k := \begin{bmatrix} 1&\quad 1 \\&\quad 1&\quad 1 \\&\quad&\quad \ddots&\quad \ddots \\&\quad&\quad&\quad 1&\quad 1 \\&\quad&\quad&\quad&\quad 1 \end{bmatrix}\in \mathbb {R}^{k\times k}\quad \text{ and }\quad G_k := \begin{bmatrix} 1 \\ 1&1 \\&\ddots&\ddots \\&1&1\\&&1 \end{bmatrix}\in \mathbb {R}^{(k+1)\times k}. \end{aligned}$$
(B.6)

Lemma B.3

Let \(W_{\varepsilon , \eta }\) be the matrix in (B.4), let \(M_k\) and \(G_k\) be the matrices in (B.6), and assume that \(\varepsilon \ge \eta \). Then, there exist two permutation matrices \(P_1\) and \(P_2\) such that

$$\begin{aligned} P_1W_{\varepsilon ,\eta }P_2 = \underbrace{(M_\eta \oplus M_\eta \oplus \cdots \oplus M_\eta )}_{\varepsilon -\eta \text{ times }}\oplus (G_{\eta -1}\oplus G_{\eta -1}^T)\oplus \cdots \oplus (G_1\oplus G_1^T) \oplus 0_{1\times 1}. \end{aligned}$$
(B.7)

Proof

If \(\eta =1\), then the result follows trivially from the discussion in the two lines above (B.5) with the convention \(G_0 \oplus G_0^T := 0_{1 \times 1}\). Therefore, we assume in the rest of the proof that \(\eta > 1\). Observe that the \(0_{1\times 1}\) block in (B.7) is a consequence of the fact that the first row and the last column of \(W_{\varepsilon ,\eta }\) are both zero. Thus, permuting the first row to the last row position produces the \(0_{1\times 1}\) block.

We first point out that every nonzero row or column of \(W_{\varepsilon , \eta }\) contains only one or two 1’s and if there are two 1’s, their indices differ exactly by \(\eta -1\). We now construct the permutations \(P_1\) and \(P_2\) that yield (B.7). Let us use for this the MATLAB index notation to indicate which permutations are “extracting” the different blocks of the direct sum decomposition (B.7). It is easy to see that the \(2\times 1\) submatrix of \(W_{\varepsilon , \eta }\) with row indices \((2,\eta +1)\) and column index 1 yields \(G_1\):

$$\begin{aligned} G_1 = W_{\varepsilon , \eta } (2:\eta -1:\eta +1, 1) \end{aligned}$$

and this block is “decoupled” from the rest of the matrix \(W_{\varepsilon , \eta }\) since the remaining elements in the corresponding rows and column are zero. In a similar manner, one extracts for \(i=1,\ldots , \eta -1\) the following \((i+1)\times i\) decoupled blocks

$$\begin{aligned} G_i = W_{\varepsilon , \eta } (i+1:\eta -1:i\eta +1, i:\eta -1:(i-1)\eta +1), \quad 1 \le i\le \eta -1, \end{aligned}$$

each starting from the element \((i+1,i)\) in the leading block \(J_\eta \) and ending at the leading 1 of the block \(I_\eta \) at the \((i+1,i)\) block-entry. In a similar fashion one also extracts for \(i=1,\ldots , \eta -1\) the “transposed” matrices \(G_i^T\) backwards from the trailing block \(J_\eta \), i.e., each \(G_i^T\) starting from the element \((\eta -i +1, \eta -i)\) in the trailing block \(J_\eta \) and ending (backwards) at the trailing 1 of the block \(I_\eta \) at the \((\varepsilon -i +1, \varepsilon -i)\) block-entry. In MATLAB index notation this amounts to

$$\begin{aligned} G_i^T = W_{\varepsilon ,\eta } (\varepsilon \eta -i+1:1-\eta : (\varepsilon -i+1)\eta , \, \varepsilon \eta -i:1-\eta :(\varepsilon -i)\eta ). \end{aligned}$$

So far, we have “extracted” \(\eta -1\) trailing 1s of the \(\eta -1\) trailing blocks \(I_\eta \). This allows us to find the remaining \((\varepsilon -\eta )\) blocks \(M_\eta \) in (B.7) as follows: each of them starts from the trailing 1 in the block \(I_\eta \) at the \((i+1,i)\) block-entry and ends at the leading 1 of the block \(I_\eta \) at the \((i+\eta ,i+\eta -1)\) block-entry. In MATLAB index notation this amounts to

$$\begin{aligned}&M_\eta ^{(i)} = W_{\varepsilon ,\eta } ((i+1)\eta :\eta -1: (\eta +i-1)\eta +1, i\eta :\eta -1:(\eta +i-2)\eta +1),\\&1\le i\le \varepsilon -\eta . \end{aligned}$$

Finally, it is also easy to verify that the dimensions and the number of 1s of the direct sum decomposition in the right-hand side of (B.7) match those of \(W_{\varepsilon ,\eta }\). This completes the proof. \(\square \)

Now, we are in the position of computing \(\sigma _{\max }(W_{\varepsilon ,\eta })\).

Proposition B.4

Let \(W_{\varepsilon ,\eta }\) be the matrix in (B.4). Then

$$\begin{aligned} \sigma _{\max }(W_{\varepsilon ,\eta }) = \left\{ \begin{array}{ll} 2\cos \frac{\pi }{2\min \{\varepsilon ,\eta \}+1}, &{} \text{ if } \varepsilon \ne \eta , \\ 2\cos \frac{\pi }{2\eta }, &{} \text{ if } \varepsilon = \eta . \end{array} \right. \end{aligned}$$
(B.8)

Proof

As explained after the equation (B.4), we may assume without loss of generality that \(\varepsilon \ge \eta \). In addition, if \(\eta =1\), then the result follows immediately from (B.5). Thus, the rest of the proof assumes \(\varepsilon \ge \eta > 1\).

Let us consider first the case \(\varepsilon =\eta > 1\). Lemma B.3 implies that \(\sigma _{\max }(W_{\eta ,\eta }) = \max \{ \sigma _{\max }(G_{\eta -1}),\ldots , \sigma _{\max }(G_{2}), \sigma _{\max }(G_{1}) \}\). In addition, since \(G_k\) is a submatrix of \(G_{k+1}\), we have that \(\sigma _{\max }(G_{\eta -1}) \ge \cdots \ge \sigma _{\max }(G_{2}) \ge \sigma _{\max }(G_{1})\) [45, Corollary 3.1.3]. Therefore, \(\sigma _{\max }(W_{\eta ,\eta }) = \sigma _{\max }(G_{\eta -1})\). The singular values of \(G_{\eta -1}\) are the square roots of the eigenvalues of

$$\begin{aligned} G_{\eta -1}^T G_{\eta -1} = \begin{bmatrix} 2&\quad 1 \\ 1&\quad 2&\quad 1 \\&\quad 1&\quad \ddots&\quad \ddots \\&\quad&\quad \ddots&\quad 2&\quad 1\\&\quad&\quad&\quad 1&\quad 2 \end{bmatrix} \in \mathbb {R}^{(\eta -1) \times (\eta -1)}, \end{aligned}$$

which are known at least from the 1940s [34, p. 111]. They are

$$\begin{aligned} \lambda _j = 2\left( 1 - \cos \frac{\pi j}{\eta } \right) , \quad \text{ for } j=1,2,\ldots ,\eta -1. \end{aligned}$$

Therefore the maximum of these eigenvalues is

$$\begin{aligned} \lambda _{\eta -1} = 2\left( 1 - \cos \frac{\pi (\eta -1)}{\eta } \right) = 2\left( 1 + \cos \frac{\pi }{\eta } \right) = 4 \cos ^2 \frac{\pi }{2\eta }. \end{aligned}$$

The result follows from \(\sigma _{\max }(W_{\eta ,\eta }) = \sigma _{\max }(G_{\eta -1}) = \sqrt{\lambda _{\eta -1}}\).

Next, we consider the case \(\varepsilon>\eta > 1\). In this situation, Lemma B.3 implies that \(\sigma _{\max }(W_{\varepsilon ,\eta }) = \max \{ \sigma _{\max }(M_{\eta }), \sigma _{\max }(G_{\eta -1}),\ldots , \sigma _{\max }(G_{1}) \} = \sigma _{\max }(M_{\eta }),\) where we have used again that \(G_k\) is a submatrix of \(G_{k+1}\) and that \(G_{\eta -1}\) is a submatrix of \(M_\eta \). The singular values of \(M_\eta \) are the square roots of the eigenvalues of \(M_\eta M_\eta ^T\), i.e., the square roots of the roots of the characteristic equation

$$\begin{aligned} \det (\lambda I- M_\eta M_\eta ^T) = \det \begin{bmatrix} (\lambda -2)&\quad -1 \\ -1&\quad (\lambda -2)&\quad -1 \\&\quad -1&\quad \ddots&\quad \ddots \\&\quad&\quad \ddots&\quad (\lambda -2)&\quad -1\\&\quad&\quad&\quad -1&\quad (\lambda -1) \end{bmatrix} =0. \end{aligned}$$

With the change of variable \(\lambda =2\mu +2\), the equation above becomes

$$\begin{aligned} \det \begin{bmatrix} 2\mu&\quad -1 \\ -1&\quad 2\mu&\quad -1 \\&\quad -1&\quad \ddots&\quad \ddots \\&\quad&\quad \ddots&\quad 2\mu&\quad -1\\&\quad&\quad&\quad -1&\quad 2\mu +1 \end{bmatrix}=U_\eta (\mu )+U_{\eta -1}(\mu ) =0, \end{aligned}$$
(B.9)

where \(U_\ell (\mu )\) is the degree-\(\ell \) Chebyshev polynomial of the second kind [58]. The first equality in (B.9) can be obtained directly from [48, eq. (11)] by applying the recurrence relation of the Chebyshev polynomials of the second kind.Footnote 2 Alternatively this fact can also be easily established from results found in [39]. Observe that Gershgorin Circle Theorem [38, Theorem 7.2.1] implies that the eigenvalues of \(M_\eta M_\eta ^T\) satisfy \(0\le \lambda \le 4\). Therefore, the roots of (B.9) satisfy \(-1\le \mu \le 1\). Moreover, we also have that 1 and \(-1\) are not roots of (B.9) since \(U_\eta (1)+U_{\eta -1}(1) = 2\eta +1\ne 0\) and \(U_\eta (-1)+U_{\eta -1}(-1)=(-1)^\eta \ne 0\). Thus, the roots of (B.9) satisfy \(-1< \mu < 1\). With the change of variable \(\mu = \cos \theta \), we get the equation

$$\begin{aligned} U_\eta (\mu )+U_{\eta -1}(\mu ) = \frac{1}{\sin \theta }(\sin (\eta +1)\theta +\sin \eta \theta ) = 2 \, \frac{\cos \frac{\theta }{2}}{\sin \theta } \, \sin \frac{(2\eta +1)\theta }{2} =0, \end{aligned}$$

whose roots are \(\theta _j = 2\pi j/(2\eta + 1)\), \(j=1,\ldots , \eta \) in the interval \(0< \theta < \pi \). We finally obtain that the eigenvalues of \(M_\eta M_\eta ^T\) are

$$\begin{aligned} \lambda _j = 2 + 2 \cos \frac{2j\pi }{2\eta +1},\quad \text{ for } j=1,2,\ldots ,\eta . \end{aligned}$$
(B.10)

The largest one is \(\lambda _1\), which implies

$$\begin{aligned} \sigma _{\max }(W_{\varepsilon ,\eta }) = \sigma _{\max } (M_\eta ) = \sqrt{ 2 + 2 \cos \frac{2\pi }{2\eta +1} } = 2\cos \frac{\pi }{2\eta +1}. \end{aligned}$$

\(\square \)

Finally, Lemma 5.4 follows from combining Lemmas B.1 and B.2, Proposition B.4 and an elementary trigonometric identity. Observe that \(\sigma _{\min } (T) \ne 0\), which implies that T has full row rank.

Appendix C. Proof of Theorem 5.14

Taking into account that \(L_\varepsilon (\lambda ) \otimes I_n\) and \(\Lambda _\varepsilon (\lambda )^T \otimes I_n\) are dual minimal bases with all their row degrees equal, respectively, to 1 and \(\varepsilon \), part (a) is an immediate consequence of Theorem 5.12. Part (b) can also be seen as a consequence of Theorem 5.13 (except the obvious equality \(C_{0}(\Lambda _\varepsilon (\lambda )^T \otimes I_n) = I_{(\varepsilon + 1) n}\)), although it can be deduced directly because the matrices \(C_{0}(\Lambda _\varepsilon (\lambda )^T \otimes I_n)\) and \(C_{1}(\Lambda _\varepsilon (\lambda )^T \otimes I_n)\) are very simple.

In order to prove part (c), we first note that \(C_{\varepsilon -1}(L_\varepsilon (\lambda ) \otimes I_n)=C_{\varepsilon -1}(L_\varepsilon (\lambda ) )\otimes I_n\) and \(C_{\varepsilon }(L_\varepsilon (\lambda ) \otimes I_n)=C_{\varepsilon }(L_\varepsilon (\lambda ))\otimes I_n\). So, it suffices to look at \(C_{\varepsilon -1}(L_\varepsilon (\lambda ))\) and \(C_{\varepsilon }(L_\varepsilon (\lambda ))\). We then point out that there exist diagonal sign scalings, \(S_1, S_2, S_3, S_4,\) (and hence orthogonal matrices) which get rid of all negative signs in \(C_{\varepsilon -1}(L_\varepsilon (\lambda ))\) and \(C_{\varepsilon }(L_\varepsilon (\lambda ))\), and that with the notation at the beginning of Sect. 5.1 lead to:

$$\begin{aligned} S_1C_{\varepsilon - 1} (L_\varepsilon (\lambda ))S_2 =: \hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda )) = \underbrace{ \begin{bmatrix} F_{\varepsilon } \\ E_{\varepsilon }&\quad \ddots \\&\quad \ddots&\quad F_{\varepsilon } \\&\quad&\quad E_{\varepsilon } \end{bmatrix}}_{\displaystyle \varepsilon \; \text{ block } \text{ columns }} \left. \phantom {\begin{array}{l} l\\ l\\ l\\ l\\ l \end{array}} \! \!\!\!\right\} \varepsilon +1 \; \text{ block } \text{ rows }, \end{aligned}$$

and

$$\begin{aligned} S_3C_{\varepsilon } (L_\varepsilon (\lambda ))S_4 =: \hat{C}_{\varepsilon } (L_\varepsilon (\lambda )) = \underbrace{ \begin{bmatrix} F_{\varepsilon } \\ E_{\varepsilon }&\quad \ddots \\&\quad \ddots&\quad F_{\varepsilon } \\&\quad&\quad E_{\varepsilon } \end{bmatrix}}_{\displaystyle \varepsilon +1 \; \text{ block } \text{ columns }} \left. \phantom {\begin{array}{l} l\\ l\\ l\\ l\\ l \end{array}} \! \! \!\!\! \!\!\!\right\} \varepsilon +2 \text{ block } \text{ rows } \,. \end{aligned}$$

Clearly we can as well look at the singular values of the matrices \(\hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) and \(\hat{C}_{\varepsilon } (L_\varepsilon (\lambda ))\) since they are orthogonally equivalent to \(C_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) and \(C_{\varepsilon } (L_\varepsilon (\lambda ))\), respectively. We then show that there exist row and column permutations (and hence orthogonal transformations) that put \(\hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) and \(\hat{C}_{\varepsilon } (L_\varepsilon (\lambda ))\) in the following block diagonal forms

$$\begin{aligned} {P_1} \hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda )) {P_2}&= M_{\varepsilon }\oplus M_{\varepsilon }^T\oplus \cdots \oplus M_1\oplus M_1^T, \end{aligned}$$
(C.1)
$$\begin{aligned} {P_3} \hat{C}_{\varepsilon } (L_\varepsilon (\lambda )) {P_4}&= M_{\varepsilon }\oplus M_{\varepsilon }^T\oplus \cdots \oplus M_1\oplus M_1^T \oplus G^T_\varepsilon , \end{aligned}$$
(C.2)

where \(M_k\) and \(G_k\) were defined in (B.6). Since a formal proof of (C.1) and (C.2) is long, we simply sketch the main ideas. Notice that each of the matrices \(\hat{C}_{\varepsilon - 1}(L_\varepsilon (\lambda ))\) and \(\hat{C}_{\varepsilon }(L_\varepsilon (\lambda ))\) have one or two 1’s in each column or row. Moreover, note that \(\hat{C}_{\varepsilon - 1}(L_\varepsilon (\lambda ))\) has exactly \(2 \varepsilon \) columns with only one “1” and exactly \(2 \varepsilon \) rows with only one “1”, while \(\hat{C}_{\varepsilon }(L_\varepsilon (\lambda ))\) has exactly \(2 (\varepsilon +1)\) columns with only one “1” and exactly \(2 \varepsilon \) rows with only one “1”. Then, starting from the leading column in \(\hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) with a single “1”, one can then reconstruct a staircase \(M_\varepsilon \) and starting from its trailing column with a single “1”, one can reconstruct a staircase \(M_\varepsilon ^T\). The corresponding index selection in Matlab notation for these two submatrices is:

$$\begin{aligned} M_\varepsilon= & {} \hat{C}_{\varepsilon -1}(L_\varepsilon (\lambda ))(\varepsilon +1:\varepsilon +1:\varepsilon ^2+\varepsilon , 1:\varepsilon +2:\varepsilon ^2+\varepsilon -1),\\ M_\varepsilon ^T= & {} \hat{C}_{\varepsilon -1}(L_\varepsilon (\lambda ))(1:\varepsilon +1:\varepsilon ^2, 2:\varepsilon +2:\varepsilon ^2+\varepsilon ). \end{aligned}$$

After permuting these two blocks out of \(\hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) one continues in a similar way to recover all other blocks \(M_k\) and \(M_k^T\), for \(k=\varepsilon -1, \ldots , 1\). For the matrix \(\hat{C}_{\varepsilon } (L_\varepsilon (\lambda ))\), the procedure is similar, except that in the first step, one extracts

$$\begin{aligned} G_\varepsilon ^T = \hat{C}_{\varepsilon } (L_\varepsilon (\lambda )) (\varepsilon +1:\varepsilon +1:\varepsilon ^2+\varepsilon , 1:\varepsilon +2:\varepsilon ^2+2\varepsilon +1) \end{aligned}$$

starting from the “1” in the leading column. The rest of the extraction is similar to the one for the matrix \(\hat{C}_{\varepsilon -1} (L_\varepsilon (\lambda ))\).

So the smallest singular values of \(\hat{C}_{\varepsilon - 1} (L_\varepsilon (\lambda ))\) and \(\hat{C}_{\varepsilon } (L_\varepsilon (\lambda ))\) are those of the diagonal blocks with the smallest singular values. This turns out to be \(M_\varepsilon \) for both matrices, since the smallest singular value of the full-row rank matrix \(G_\varepsilon ^T=\left[ M_\varepsilon | e_\varepsilon \right] \) is larger than that of \(M_\varepsilon \) [45, Corollary 3.1.3] and, according to (B.10), \(\sigma _\mathrm{min} (M_\varepsilon )< \sigma _\mathrm{min} (M_{\varepsilon -1})< \cdots < \sigma _\mathrm{min} (M_{1})\). The smallest singular value of \(M_\varepsilon \) is the square root of the smallest eigenvalue given in (B.10):

$$\begin{aligned} \sigma _{\min }(M_\varepsilon ) = \sqrt{2+2\cos \left( \frac{2\varepsilon \pi }{2\varepsilon +1} \right) }=2 \sin \left( \frac{\pi }{4\varepsilon +2} \right) . \end{aligned}$$

The inequality \(2 \sin (\frac{\pi }{4\varepsilon +2})\ge \frac{3}{2\varepsilon +1} \ge \frac{3}{2(\varepsilon +1)}\) follows then from the inequality \(\sin (x) \ge 3x/\pi \) for \(0 \le x \le \pi /6\) since we assumed \(\varepsilon \ge 1\).

The proof of part (d) follows from the equality \(C_{0}(\Lambda _\varepsilon (\lambda )^T \otimes I_n) = I_{(\varepsilon + 1) n}\) and the fact that an obvious column permutation P allows us to prove that \(C_{1}(\Lambda _\varepsilon (\lambda )^T \otimes I_n) \, {P} = I_n \oplus (I_{\varepsilon n} \otimes [1,1]) \oplus I_n\). Therefore, the singular values of \(C_{1}(\Lambda _\varepsilon (\lambda )^T \otimes I_n)\) are 1 (with multiplicity 2n) and \(\sqrt{2}\) (with multiplicity \(\varepsilon n\)).

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Dopico, F.M., Lawrence, P.W., Pérez, J. et al. Block Kronecker linearizations of matrix polynomials and their backward errors. Numer. Math. 140, 373–426 (2018). https://doi.org/10.1007/s00211-018-0969-z

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Mathematics Subject Classification

  • 65F15
  • 65F35
  • 15A18
  • 15A22
  • 15A54
  • 93B18
  • 93B40
  • 93B60