Calculating Determinants of Block Matrices

Home , Block matrix, Square matrix, Zero matrix

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the right side of Eq. (1.2) are the (N k) (N k) matrix D, and the k k Schür complement with respect to D [14, 15].− × − × While Eq. (1.2) has proven immensely useful, there remain many applications in which this result is not ideally suited. For example, there are situations in which physics dictates that we partition a matrix in a form different than in Eq. (1.1). The description of intermediate-energy quark matter in quantum chromodynamics, for instance, involves propagators which possess four Dirac indices, three color indices and three flavor indices. While one could, in principle, partition the resulting 36 36 matrix as shown in Eq. (1.2), such a partitioning would treat the Dirac indices× asymmetrically, eliminating two of the indices while retaining two others. While such a division is of no consequence in computational applications, it is quite undesirable in analytical descriptions. Thus, we are led to desire a generalization of Eq. (1.2) which holds for a block matrix of more general partitioning. The remainder of this paper is organized as follows. In Sec. 2, a direct, “brute force” calculation of the determinant of an N N complex block matrix is presented, which manifests clearly the emergence of a Schür× complement structure. In Sec. 3, an alternate proof of our result is given, which is somewhat more elegant, but perhaps less intuitive, by means of induction on N. Finally, in Sec. 4, some applications of our result are discussed, including the cases N = 2 and N = 3, as well as a 48 48 matrix with N = 6, which arises in the study of quark matter. × 2. Direct Calculation . Theorem 2.1. Let S be an (nN) (nN) complex matrix, which is partitioned into N 2 blocks, each of size n n: × × S S S 11 12 ··· 1N S21 S22 S2N S =  . . ···. .  . (2.1) . . .. .   S S S   N1 N2 ··· NN    Then the determinant of S is given by

N (N−k) det(S)= det(αkk ), (2.2) kY=1 where the α(k) are deﬁned by

(0) αij = Sij (k) T −1 α = S σ − S˜ s − , k 1, (2.3) ij ij − i,N k+1 k N k+1,j ≥ T and the vectors σij and sij are

T s = S S S , σT = S S S . (2.4) ij ij i+1,j ··· Nj ij ij i,j+1 ··· iN Proof. In order to facilitate computing the determinant of S, we deﬁne a lower triangular auxiliary matrix I 0 0 ··· U21 I 0 U =  . . ···. .  . (2.5) . . .. .   U U I  N1 N2 ···   2  Calculating Determinants of Block Matrices

Forming the product SU gives

S + S U + S U S + S U + S U S 11 12 21 ··· 1N N1 12 13 32 ··· 1N N2 ··· 1N S21 + S22U21 + S2N UN1 S22 + S23U32 + S2N UN2 S2N SU =  . ··· . ··· ···. .  ......   S + S U + S U S + S U + S U S   N1 N2 21 ··· NN N1 N2 N3 32 ··· NN N2 ··· NN   (2.6)

Now, let us assume that the blocks of U can be chosen such that the product SU is upper triangular. In this case, we obtain (N 2 N)/2 constraint equations (the lower triangular blocks of SU = 0) with (N 2 N−)/2 unknowns (the blocks of U). The N k equations arising from the kth column− of SU are − S + S U + + S U = 0, k+1,k k+1,k+1 k+1,k ··· k+1,N Nk S + S U + + S U = 0, (2.7) k+2,k k+2,k+1 k+1,k ··· k+2,N Nk . . S + S U + + S U = 0. N,k N,k+1 k+1,k ··· NN Nk

In order to simplify our notation, we now deﬁne the block vectors sij (Eq. 2.4)) and

T u = U U U , (2.8) ij ij i+1,j ··· Nj which represent the portions of the kth columns of S and U, respectively, which lie below, and inclusive of, the block with indices (i, j). We also let S˜k represent the k k block matrix formed from the lower-right corner of S, × S − − S − − S − N k+1,N k+1 N k+1,N k+2 ··· N k+1,N SN−k+2,N−k+1 SN−k+2,N−k+2 SN−k+2,N ˜  ···  Sk = . . . . . (2.9) . . .. .    S − S − S   N,N k+1 N,N k+2 ··· N,N    With these deﬁnitions, we can rewrite Eqs. (2.7) in the matrix form:

S˜ − u = s . (2.10) N k k+1,k − k+1,k Solving for the auxiliary block vector yields

−1 u = S˜ − s . (2.11) k+1,k − N k k+1,k T We now deﬁne σij as the block row vector of S lying to the right, and inclusive of, the position (i, j) (Eq. (2.4)). Inspecting Eq. (2.6), we can express the kth diagonal element of SU in the form

T (SU)kk = Skk + σk,k+1uk+1,k, T −1 = S σ S˜ − s , kk − k,k+1 N k k+1,k = SÑ−k+1/Skk, (2.12) where SÑ−k+1/Skk denotes the Schür complement of SÑ−k+1 with respect to Skk. (k) (N−k) Next, defining the matrices αij as shown in Eq. (2.3), we write (SU)kk = αkk . 3 Calculating Determinants of Block Matrices

Since SU is upper triangular by design, the determinant is simply the product of the determinants of its diagonal blocks. This, together with the fact that det(U)=1 gives

N (N−k) det(S)= det(αkk ). (2.13) kY=1 ˜−1 In order to express det(S) in terms of the blocks of S, we must now evaluate Sk , which appears in Eq. (2.3). However, rather than approaching this problem directly, (k) we will focus on the matrices αij , and seek to ﬁnd a recursive relationship between matrices of consecutive values of k. Thus, we begin by writing

(k+1) T −1 α = S α − S˜ s − ij ij − i,N k k+1 N k,j T −1 S − − σ S S σT N k,N k N−k,N−k+1 N−k,j S i,N−k i,N−k s = ij [ +1 ] s S˜ N−k+1,j , − N−k+1,N−k k h i h i (2.14) where we have partitioned the block vectors and matrix into sections of block length 1 and N. Next, we evaluate the inverse matrix above by making use of the Banachiewic identity [16, 17]

− − − − − A B −1 (A−BD 1C) 1 −(A−BD 1C) 1BD 1 [ ] = −1 −1 −1 −1 −1 −1 −1 . (2.15) C D −D C(A−BD C) D [I+C(A−BD C) BD ] h i ˜−1 Identifying this expression with the partitioned form of Sk in Eq. (2.14), the Sch¨ur complement with respect to SN−k,N−k becomes −1 T −1 A BD C = S − − σ − − S˜ s − − − N k,N k − N k,N k+1 k N k+1,N k (k) = αN−k,N−k. (2.16) Thus, evaluating the inverse matrix in Eq. (2.14), we have

(k+1) S σT α = S [ i,N−k i,N−k+1 ] ij ij − α(k) −1 − α(k) −1σT S˜−1 ( N−k,N−k) ( N−k,N−k) N−k,N−k+1 k −S˜−1s α(k) −1 S˜−1 I s α(k) −1σT S˜−1 × k N−k+1,N−k( N−k,N−k) k + N−k+1,N−k( N−k,N−k) N−k,N−k+1 k h i SN−k,j s , × N−k+1,j h i S −σT S˜−1s α(k) −1 [ i,N−k i,N−k+1 N−k+1,N−k]( − − ) S k N k,N k = ij σT S˜−1− S −σT S˜−1s α(k) −1σT S˜−1 − i,N−k+1 k [ i,N−k i,N−k+1 k N−k+1,N−k]( N−k,N−k) N−k,N−k+1 k SN−k,j s , · N−k+1,j − h i α(k) (α(k) ) 1 i,N−k N−k,N−k SN−k,j = S − − s , ij σT S˜ 1−α(k) α(k) −1σT S˜ 1 N−k+1,j − i,N−k+1 k i,N−k( N−k,N−k) N−k,N−k+1 k · (k) (k) −1 T −1 h i = S α − (α − − ) S − σ − S˜ s − ij − i,N k N k,N k N k,j − i,N k+1 k N k+1,j (k) (k) −1 T ˜−1 +αi,N−k(αN−k,N−k) σN−k,N−k+1Sk sN−k+1,j , T −1 = S σ − S˜ s − ij − i,N k+1 k N k+1,j h (k) (k) −i1 T −1 α − (α − − ) S − σ − − S˜ s − , − i,N k N k,N k N k,j − N k,N k+1 k N k+1,j (k) (k) (k) −1 h(k) i = α α − (α − − ) α − . (2.17) ij − i,N k N k,N k N k,j 4 Calculating Determinants of Block Matrices

(k) We can therefore express the αij in the recursive form:

(0) αij = Sij , (k+1) (k) (k) (k) −1 (k) α = α α − (α − − ) α − , k 1. (2.18) ij ij − i,N k N k,N k N k,j ≥

(0) Given this recursive relationship, the matrix αij can be read oﬀ directly from S, and (k) all higher order αij can be calculated iteratively. Finally, the determinant may be computed via Eq. (2.13).

3. Proof by Induction . Rather than calculating the determinant of S directly, as in the prior section, here we begin by establishing a recursive relationship between a sort of generalized Sch¨ur complement of S. To this end, we ﬁrst prove the following lemma. Lemma 3.1. Let S be a complex block matrix of the form

S S S 11 12 ··· 1N S21 S22 S2N S =  . . ···. .  , (3.1) . . .. .   S S S   N1 N2 ··· NN   

and let us deﬁne the set of block matrices α(0), α(1), , α(N−1) , where α(k) is an (N k) (N k) block matrix with blocks ··· − × −

(0) αij = Sij −1 (k+1) (k) (k) (k) (k) α = α α − α − − α − , k 1. (3.2) ij ij − i,N k N k,N k N k,j ≥

Then the determinants of consecutive α(k) are related via

(k) (k+1) (k) det(α ) = det(α ) det(αN−k,N−k). (3.3)

Proof. Let us partition α(k) in the form

α(k) α(k) α(k) 11 ··· 1,N−k−1 1,N−k ...... α(k) =  . . .  , (3.4) (k) (k) (k)  αN−k−1,1 αN−k−1,N−k−1 αN−k−1,N−k   (k) ··· (k) (k)   α − α − − − α − −   N k,1 ··· N k,N k 1 N k,N k   5  Calculating Determinants of Block Matrices

and let us denote the upper-left block of α(k) by α˜(k). The Sch¨ur complement of α(k) with respect to α˜ (k) is then

(k) (k) α α − − 11 ··· 1,N k 1 (k) (k) . . . α /α˜ =  . .. .  (k) (k) α − − α − − − −   N k 1,1 ··· N k 1,N k 1   (k) α1,N−k −1 (k) (k) (k) . α − α − − − α − −  .  , − N k,1 ··· N k,N k 1 N k,N k . (k) h i h i α − − −   N k 1,N k (k) (k) −1 (k)   = α α − (α − − ) α − , ij − i,N k N k,N k N k,j = α(k+1). (3.5)

Applying the 2 2 block identity (Eq. (1.2)), we can therefore write the determinant of α(k) as ×

(k) (k+1) (k) det(α ) = det(α ) det(αN−k,N−k). (3.6)

Theorem 3.2. Given a complex block matrix of the form (2.1), and the matrices (k) αij deﬁned in Eq. (3.2), the determinant of S is given by

N (N−k) det(S)= det(αkk ). (3.7) kY=1

Proof. Recall from the deﬁnition in Eq. (3.2) that α(0) = S. Starting with det(S) and applying Eq. (3.6) repeatedly yields

det(S) = det(α(0)), (1) (0) = det(α ) det(αNN ), (2) (1) (0) = det(α ) det(αN−1,N−1) det(αNN ), . . (3.8) − − = det(α(N 1)) det(α(N 2)) det(α(0) ), (3.9) 22 ··· NN where the sequence of equalities terminates because α(N−1) is a 1 1 block matrix, (N−1) (N−×1) which cannot be partitioned further. Thus, writing α = α11 we obtain the result

N (N−k) det(S)= det(αkk ). (3.10) kY=1

6 Calculating Determinants of Block Matrices

4. Applications . Having derived an expression for the determinant of an N N complex block matrix, it will be useful to examine the result for a few speciﬁc values× of N. We choose to consider N = 2 and N = 3, as the ﬁrst is the well-known result of Eq. (1.2) and the second gives a clear picture of what sort of objects our result actually involves. Larger values of N quickly become cumbersome to write down, but the procedure for their calculation will be made clear. Lastly, we present a “real world” application of our result by calculating the determinant of a 48 48 matrix, which arises in the study of quark matter. × 4.1. 2 2 Block Matrices. In the case N = 2, Eq. (2.13) reduces to × (1) (0) det(S) = det(α11 ) det(α22 ), (4.1) while Eqs. (2.18) reduce to − α(0) = S α(1) = S S S 1S . (4.2) ij ij ij ij − i2 22 21 Thus, we obtain the result − det(S) = det(S S S 1S ) det(S ). (4.3) 11 − 12 22 21 22 Note that this expression is identical to Eq. (1.1). It is instructive to combine the two determinants in this expression and rewrite it in the alternate forms −1 det(S) = det(S11S22 S12S22 S21S22), − − = det(S S S S S 1S ). (4.4) 22 11 − 22 12 22 21 In these forms, it is clear that when either S12 or S21 commute with S22 our expression reduces to: S S = S S : det(S) = det(S S S S ), 12 22 22 12 22 11 − 12 21 S S = S S : det(S) = det(S S S S ). (4.5) 21 22 21 21 11 22 − 12 21 Alternatively, for anti-commuting matrices, which often arise in the study of fermionic systems, the sign in (4.5) becomes positive. 4.2. 3 3 Block Matrices. In the case N = 3, Eq. (2.13) reduces to × (2) (1) (0) det(S) = det(α11 ) det(α22 ) det(α33 ), (4.6) while Eqs. (2.18) reduce to − α(0) = S , α(1) = S S S 1S , ij ij ij ij − i3 33 31 − α(2) = S S S 1S ij ij − i3 33 3j − − −1 − S S S 1S S S S 1S S S S 1S . − i2 − i3 33 32 22 − 23 33 32 2j − 23 33 3j (4.7) Thus, we obtain the result

− det(S) = det S S S 1S 11 − 13 33 31 −1 S S S−1S S S S−1S S S S−1S − 12 − 13 33 32 22 − 23 33 32 21 − 23 33 31 − det S S S 1S det(S ). (4.8) × 22 − 23 33 32 33 Analogously to the N = 2 case, the commutation of certain blocks (e.g., S33 and S3j , (1) (1) α12 with α22 ) allows this expression to be simpliﬁed. 7 Calculating Determinants of Block Matrices

4.3. Eigenenergies of high-density quark matter. Having considered the general form of the determinant of 2 2 and 3 3 block matrices, we now consider a true application of our result by calculating× the× eigenenergies of quark matter in the two ﬂavor Nambu–Jona-Lasinio model [2, 18, 19]. In this model, the energies are the roots of the equation det S = 0, where

0 k/ + µγ M ∆γ5τ2λ2 S = ∗ , (4.9) ∆ γ τ−λ k/ µγ0 M − 5 2 2 − − and where M is the effective quark mass, µ is the chemical potential, ∆ is the pairing 0 T ν gap, k/ Eγ γ k, where k = (kx, ky, kz) is the quark momentum and the γ (ν =0...≡3) are− the 4· 4 Dirac matrices: × I 0 0 σ γ0 = , γi = i , (4.10) 0 I σ 0 − − i T with σ = (σx, σy, σz) representing the vector of Pauli matrices: 0 1 0 i 1 0 σ = , σ = , σ = . (4.11) x 1 0 y i −0 z 0 1 − 0 1 2 3 In addition, γ5 iγ γ γ γ , τ2 is the second Pauli matrix in flavor space, and λ2 is the second Gell-Mann≡ matrix in color space: 0 i 0 − λ2 = i 0 0 . (4.12) 0 0 0   Thus, while we have written Eq. (4.9) in 2 2 block form, each block is itself a 24 24 matrix (2 (flavor) 3 (color) 4 (Dirac)).× × Before constructing× the determinant× from Eq. (2.13), we must choose how we wish to partition S. We could, for instance, choose to begin with the 2 2 block form shown in Eq. (4.9), in which case the α(k) would be 24 24 matrices, which× we would partition further, repeating the process until we have× eliminated all indices. While this choice has the advantage of requiring the construction of only a single α(k) (α(1)) for the first step (Eq. (4.1)), the price is that correspondingly more steps are required to finally obtain det S. As a middle ground, balancing the number of α(k)’s which must be constructed in each step with the number of steps, we will partition S into a 6 6 block matrix, with each block of size 8 8. We achieve this partitioning by writing× out the color indices explicitly, while leaving× the Dirac and flavor indices intact. Thus, the non-zero blocks of S become S = S = S = k/ + µγ0 M, 11 22 33 − S = S = S = k/ µγ0 M, 44 55 66 − − S24 = S15 = i∆γ5τ2, (4.13) − ∗ S = S = i∆ γ τ . 42 − 51 5 2 (1) (5) (k+1) We now must construct α α . From Eq. (2.18), we see that αij will (k) (k···) (k) be equal to αij unless both αi,N−k and αN−k,j are non-zero. As a result, since S = S = 0 for i, j = 6, we find i6 6j 6 α(1) = S , 1 i, j 5. (4.14) ij ij ≤ ≤ 8 Calculating Determinants of Block Matrices

Next, we note that S15 and S51 are the only non-zero blocks with a row or column (2) (1) index of 5. Thus, the only αij which diﬀers from αij is

− α(2) = S S S 1S . (4.15) 11 11 − 15 55 51 −1 A straightforward application of the Banachiewicz identity (Eq. (2.15)) yields S55 :

0 − k/ µγ + M S 1 = − , (4.16) 55 (E µ)2 E2 − − k where E √k2 + M 2. Substituting this expression into Eq. (4.15) yields k ≡ k/ µγ0 + M α(2) = k/ + µγ0 M ( i∆γ τ ) − ( i∆γ τ ), 11 − − − 5 2 (E µ)2 E2 − 5 2 − − k k/ µγ0 M = k/ + µγ0 M ∆ 2 − − , (4.17) − − | | (E µ)2 E2 − − k 2 µ µ where we have used the fact that τ2 = I and γ5γ = γ γ5. Meanwhile, the rest of (2) (1) − the αij = αij = Sij , so that

0 2 k/−µγ0−M (2) k/ + µγ M ∆ − 2− 2 if i, j =1, (E µ) Ek αij = − − | | (4.18) (Sij else (1 i, j 4). ≤ ≤

(3) (2) Constructing α , we note that α24 and S42 are the only non-zero blocks in the fourth row or column of α(2). As a result, we ﬁnd

−1 α(3) = α(2) α(2) α(2) α(2), 22 22 − 24 44 42 −1 = S11 ( S15)(S55) ( S51), − − − − = S S S 1S , 11 − 15 55 51 (2) = α11 . (4.19)

(3) (2) The rest of the αij are equal to αij so we ﬁnd

0 2 k/−µγ0−M (3) k/ + µγ M ∆ − 2− 2 if i, j =1 or i, j =2, (E µ) Ek αij = − − | | (4.20) (Sij else (1 i, j 3). ≤ ≤ (3) Since α33 = S33 is the only non-zero block in the third row/column, we have

α(4) = α(3), 1 i, j 2. (4.21) ij ij ≤ ≤

(4) (4) (5) Finally, since α12 = α21 = S12 = S21 = 0, we ﬁnd that the sole block of α is

(5) (4) (3) α11 = α11 = α11 , k/ µγ0 M = k/ + µγ0 M ∆ 2 − − . (4.22) − − | | (E µ)2 E2 − − k 9 Calculating Determinants of Block Matrices

Having constructed the necessary determinants, Eq. (2.13) gives

(5) (4) (3) (2) (1) (0) det(S) = det(α11 ) det(α22 ) det(α33 ) det(α44 ) det(α55 ) det(α66 ), k/ µγ0 M 2 = det k/ + µγ0 M ∆ 2 − − det(k/ + µγ0 M) f,D − − | | (E µ)2 E2 f,D − − − k det(k/ µγ0 M)3, (4.23) × f,D − − where the subscript of f,D on the determinant indicates that the remaining determinant is to be taken over flavor and Dirac indices. Thus, we have eliminated 6 of the original 48 indices, and have only the Dirac and flavor indices remaining. In fact, the flavor indices are now trivial, having vanished upon squaring the matrix τ2 (see Eq. (4.17)), so that

k/ µγ0 M 4 det(S) = det k/ + µγ0 M ∆ 2 − − det(k/ + µγ0 M)2 D − − | | (E µ)2 E2 D − − − k det(k/ µγ0 M)6. (4.24) × D − −

Computing det(k/ µγ0 M) yields ± − (E µ) M σ k det(k/ µγ0 M) = det ± − − · , ± − σ k (E µ) M · − ± − 2 = (E µ)2 + M 2 + (σ k)2 , − ± · 2 = (E µ)2 E2 , ± − k 2 2 = [E + (Ek µ)] [E (Ek µ)] , (4.25) ± − ∓ where we have used the fact that (σ k)2 = k2. · Finally, computing the remaining determinant from Eq. (4.24), we ﬁnd

k/ µγ0 M det k/ + µγ0 M ∆ 2 − − − − | | (E µ)2 E2 − − k 2 E µ 2 ∆ 2 2 = (E + µ) ∆ 2 − + 1 | | E2 , − − | | (E µ)2 E2 − (E µ)2 E2 k ( − − k − − k ) [E2 (E + µ)2 ∆ 2]2[E2 (E µ)2 ∆ 2]2 = − k − | | − k − − | | . (4.26) (E E µ)2(E + E µ)2 − k − k − Inserting Eq. (4.25) and (4.26) into Eq. (4.24), we obtain the result

Finally, then, we can read oﬀ the eigenenergies, which are the absolute values of the 10 Calculating Determinants of Block Matrices

roots of det(S) = 0:

E = E + µ (multiplicity 8), 1 | k | E = E µ (multiplicity 8), 2 | k − | E = (E + µ)2 + ∆ 2 (multiplicity 16), (4.28) 3 k | | 2 2 E4 = p(Ek µ) + ∆ (multiplicity 16). − | | Indeed, these are the correctp eigenenergies, as reported previously by Rossner [2]. 5. Acknowledgements. This research was supported in part by NSF grant PHY09-69790. The author would like to thank Gordon Baym for his guidance through- out the research which inspired this work.

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