holds for all A,B∈ Rn,n, although we have shown the result in Theorem7.15only for A,B ∈Kn,n.
The proof of Theorem7.15suggests that det(A)can be easily computed while transforming A∈ Kn,ninto its echelon form using elementary row operations.
Corollary 7.16 For A∈ Kn,nlet S1, . . . ,St ∈ Kn,nbe elementary matrices, such that A = St. . .S1A is in echelon form. Then either A has a zero row and hence det(A)=0, orA=Inand hencedet(A)=(det(S1))−1 · · ·(det(St))−1.
As shown in Theorem5.4, every matrixA∈ Kn,ncan be factorized asA=P LU, and hence det(A) = det(P)det(L)det(U). The determinants of the matrices on the right hand side are easily computed, since these are permutation and triangular matrices. AnLU-decomposition of a matrix Atherefore yields an efficient way to compute det(A).
MATLAB-Minute.
Look at the matriceswilkinson(n)forn=2,3,. . .,10in MATLAB. Can you find a general formula for their entries? Forn=2,3,. . .,10compute
A=wilkinson(n)
[L,U,P]=lu(A)(LU-decomposition; cp. the MATLAB-Minute above Defi- nition5.6)
det(L), det(U), det(P), det(P)∗det(L)∗det(U), det(A)
Which permutation is associated with the computed matrixP? Why isdet(A) an integer for oddn?
92 7 Determinants of Matrices
Theorem 7.18 For A∈Rn,n, n≥2, we have
Aadj(A) = adj(A)A = det(A)In.
In particular A is invertible if and only if det(A) ∈ R is invertible. In this case (det(A))−1 =det(A−1)and A−1=(det(A))−1adj(A).
Proof LetB= [bi j]have the entriesbi j =(−1)i+jdet(A(j,i)). ThenC= [ci j] = adj(A)Asatisfies
ci j = n k=1
bi kak j = n
k=1
(−1)i+kdet(A(k,i))ak j.
Letaℓbe theℓth column ofAand let
A(k,i):= [a1, . . . ,ai−1,ek,ai+1, . . . ,an] ∈ Rn,n,
whereek is thekth column of the identity matrix In. Then there exist permutation matricesPandQthat performk−1 row andi−1 column exchanges, respectively, such that
PA(k, i)Q=
1 ⋆ 0 A(k,i)
.
Using (1) in Lemma7.10we obtain
det(A(k,i))=det
1 ⋆
0 A(k,i) =det(PA(k,i)Q)
=det(P)det(A(k, i))det(Q)
=(−1)(k−1)+(i−1)det(A(k,i))
=(−1)k+idet(A(k,i)).
The linearity of the determinant with respect to the columns now gives
ci j = n k=1
(−1)i+k(−1)k+iak jdet(A(k,i))
=det([a1, . . . ,ai−1,aj,ai+1, . . . ,an])
=
0, i = j det(A), i = j
=δi j det(A),
and thus adj(A)A=det(A)In. Analogously we can show thatAadj(A)=det(A)In.
If det(A)∈ Ris invertible, then
In =(det(A))−1adj(A)A=A(det(A))−1adj(A),
i.e.,Ais invertible with A−1=(det(A))−1adj(A). If, on the other hand,Ais invert- ible, then
1=det(In)=det(A A−1)=det(A)det(A−1)=det(A−1)det(A), where we have used the multiplication theorem for determinants over R (cp. our comment following the proof of Theorem7.15). Thus, det(A) is invertible with (det(A))−1 =det(A−1), and again A−1=(det(A))−1adj(A). ⊓⊔ Example 7.19
(1) For
A= 4 1
2 1
∈Z2,2
we have det(A) = 2 and thus A is not invertible. But A is invertible when considered as an element ofQ2,2, since in this case det(A−1)=(det(A))−1= 12. (2) For
A=
t−1t−2 t t−1
∈(Z[t])2,2
we have det(A)=1. The matrixAis invertible, since 1∈Z[t]is invertible.
Note that if A ∈ Rn,n is invertible, then Theorem7.18shows that A−1 can be obtained by inverting only one ring element, det(A).
We now use Theorem7.18 and the multiplication theorem for matrices over a commutative ring with unit to prove a result already announced in Sect.4.2: In order to show that A ∈ Rn,n is the (unique) inverse of A ∈ Rn,n, only one of the two equations A A=In orAA=Inneeds to be checked.
Corollary 7.20 Let A∈Rn,n. If a matrixA∈Rn,nexists withA A=Inor AA=In, then A is invertible andA=A−1.
Proof IfA A =In, then the multiplication theorem for determinants yields 1=det(In)=det(A A)=det(A)det(A)=det(A)det(A),
i.e., det(A)∈ R is invertible with(det(A))−1 =det(A). Thus also Ais invertible and has a unique inverseA−1. Forn =1 this is obvious and forn≥2 it was shown in Theorem7.18. If we multiply the equationA A= Infrom the right with A−1we get A= A−1.
The proof starting fromAA=Inis analogous. ⊓⊔
94 7 Determinants of Matrices Let us summarize the invertibility criteria for a square matrix over a field that we have shown so far:
A∈G Ln(K) Theorem 5.2
⇐⇒ The echelon form of Ais the identity matrixIn Definition 5.10
⇐⇒ rank(A)=n
clear
⇐⇒ rank(A)=rank([A,b])=n for allb∈ Kn,1
Algorithm 6.6
⇐⇒ |L(A,b)| =1 for allb∈Kn,1
Theorem 7.18
⇐⇒ det(A)=0. (7.3)
Alternatively we obtain:
A∈/G Ln(K) Theorem 5.2
⇐⇒ The echelon form of Ahas at least one zero row
Definition 5.10
⇐⇒ rank(A) <n
clear
⇐⇒ rank([A,0]) <n
Algorithm 6.6
⇐⇒ L(A,0)= {0}
Theorem 7.18
⇐⇒ det(A)=0. (7.4)
In the fieldsQ,RandCwe have the (usual) absolute value| · |of numbers and can formulate the following useful invertibility criterion for matrices.
Theorem 7.21 If A∈Kn,nwith K ∈ {Q,R,C}isdiagonally dominant, i.e., if
|ai i|>
n
j=1 j=i
|ai j| for all i =1, . . . ,n,
thendet(A)=0.
Proof We prove the assertion by contraposition, i.e., by showing that det(A) =0 implies that Ais not diagonally dominant.
If det(A) = 0, then L(A,0) = {0}, i.e., the homogeneous linear system of equations Ax = 0 has at least one solution'x = ['x1, . . . ,'xn]T =0. Let'xm be an entry of'xwith maximal absolute value, i.e.,|'xm| ≥ |'xj|for all j = 1, . . . ,n. In particular, we then have|'xm|>0. Themth row of A'x=0 is given by
am1'x1+am2'x2+. . .+amn'xn=0 ⇔ amm'xm= − n
j=1 j=m
am j'xj.
We now take absolute values on both sides and use the triangle inequality, which yields
|amm| |'xm| ≤ n
j=1 j=m
|am j| |'xj| ≤ n
j=1 j=m
|am j||'xm|, hence |amm| ≤ n
j=1 j=m
|am j|,
so that Anot diagonally dominant. ⊓⊔
The converse of this theorem does not hold: For example, the matrix
A= 1 2
1 0
∈Q2,2, has det(A)= −2=0, butAis not diagonally dominant.
From Theorem7.18we obtain theLaplace expansion4of the determinant, which is particularly useful when Acontains many zero entries (cp. Example7.24below).
Corollary 7.22 For A∈ Rn,n, n≥2, the following assertions hold:
(1) For each i =1,2, . . . ,n we have det(A) =
n j=1
(−1)i+jai jdet(A(i,j)).
(Laplace expansion ofdet(A)with respect to the i th row A.) (2) For each j =1,2, . . . ,n we have
det(A) = n
i=1
(−1)i+jai jdet(A(i,j)).
(Laplace expansion ofdet(A)with respect to the j th column of A.)
Proof The two expansions for det(A)follow immediately by comparison of the diagonal entries in the matrix equations det(A)In = Aadj(A)and det(A)In =
adj(A)A. ⊓⊔
The Laplace expansions allows a recursive definition of the determinant: ForA∈ Rn,nwithn ≥2, let det(A)be defined as in (1) or (2) in Corollary7.22. We can choose an arbitrary row or column ofA. The formula for det(A)then contains only matrices of size(n−1)×(n−1). For each of these we can use the Laplace expansion again, now expressing each determinant in terms of determinants of(n−2)×(n−2)matrices.
We can do this recursively until only 1×1 matrices remain. For A= [a11] ∈ R1,1 we define det(A):=a11.
Finally we stateCramer’s rule,5which gives an explicit formula for the solution of a linear system in form of determinants. This rule is only of theoretical value, because in order to compute then components of the solution it requires the evaluation of n+1 determinants ofn×nmatrices.
4Pierre-Simon Laplace (1749–1827) published this expansion in 1772.
5Gabriel Cramer (1704–1752).
96 7 Determinants of Matrices Corollary 7.23 Let K be a field, A ∈ G Ln(K)and b ∈ Kn,1. Then the unique solution of the linear system of equations Ax =b is given by
'x= ['x1, . . . ,'xn]T = A−1b=(det(A))−1adj(A)b, with
'xi = det[a1, . . . ,ai−1,b,ai+1, . . . ,an]
det(A) , i =1, . . . ,n.
Example 7.24 Consider
A=
⎡
⎢⎢
⎣ 1 3 0 0 1 2 0 0 1 2 1 0 1 2 3 1
⎤
⎥⎥
⎦∈Q4,4, b=
⎡
⎢⎢
⎣ 1 2 1 0
⎤
⎥⎥
⎦∈Q4,1.
The Laplace expansion with respect to the last column yields
det(A)=1·det
⎛
⎝
⎡
⎣ 1 3 0 1 2 0 1 2 1
⎤
⎦
⎞
⎠=1·1·det 1 3
1 2 =1·1·(−1)= −1.
Thus,Ais invertible andAx =bhas a unique solution'x=A−1b∈Q4,1, which by Cramer’s rule has the following entries:
'x1=det
⎛
⎜⎜
⎝
⎡
⎢⎢
⎣ 1 3 0 0 2 2 0 0 1 2 1 0 0 2 3 1
⎤
⎥⎥
⎦
⎞
⎟⎟
⎠/det(A)= −4/(−1)=4,
'x2=det
⎛
⎜⎜
⎝
⎡
⎢⎢
⎣ 1 1 0 0 1 2 0 0 1 1 1 0 1 0 3 1
⎤
⎥⎥
⎦
⎞
⎟⎟
⎠/det(A)=1/(−1)= −1,
'x3=det
⎛
⎜⎜
⎝
⎡
⎢⎢
⎣ 1 3 1 0 1 2 2 0 1 2 1 0 1 2 0 1
⎤
⎥⎥
⎦
⎞
⎟⎟
⎠/det(A)=1/(−1)= −1,
'x4=det
⎛
⎜⎜
⎝
⎡
⎢⎢
⎣ 1 3 0 1 1 2 0 2 1 2 1 1 1 2 3 0
⎤
⎥⎥
⎦
⎞
⎟⎟
⎠/det(A)= −1/(−1)=1.
Exercises
7.1 A permutationσ∈Snis called anr -cycleif there exists a subset{i1, . . . ,ir} ⊆ {1,2, . . . ,n}withr≥1 elements and
σ(ik)=ik+1fork=1,2, . . . ,r−1, σ(ir)=i1, σ(i)=i for i∈ {/ i1, . . . ,ir}. We write anr-cycle asσ=(i1,i2, . . . ,ir). In particular, a transpositionτ ∈Sn
is a 2-cycle.
(a) Letn=4 and the 2-cyclesτ1,2=(1,2),τ2,3=(2,3)andτ3,4=(3,4)be given. Computeτ1,2◦τ2,3,τ1,2◦τ2,3◦τ1,2−1, andτ1,2◦τ2,3◦τ3,4.
(b) Letn≥4 andσ=(1,2,3,4). Determineσjfor j=2,3,4,5.
(c) Show that the inverse of the cycle(i1, . . . ,ir)is given by(ir, . . . ,i1).
(d) Show that two cycles with disjoint elements, i.e.(i1, . . . ,ir)and(j1, . . . ,js) with{i1, . . . ,ir} ∩ {j1, . . . ,js} =Ø, commute.
(e) Show that every permutationσ∈ Sn can be written as product of disjoint cycles that are, except for the order, uniquely determined byσ.
7.2 Prove Lemma7.10(1) using (7.1).
7.3 Show that the group homomorphism sgn : (Sn,◦)→({1,−1},·)satisfies the following assertions:
(a) The setAn = {σ∈ Sn|sgn(σ)=1}is a subgroup ofSn(cp. Exercise3.8).
(b) For allσ∈ Anandπ∈ Snwe haveπ◦σ◦π−1 ∈ An. 7.4 Compute the determinants of the following matrices:
(a) A= [en,en−1, . . . ,e1] ∈Zn,n, whereei is theith column of the identity matrix.
(b) B= bi j
∈Zn,nwith
bi j =
⎧⎪
⎨
⎪⎩
2 for |i−j| =0,
−1 for |i−j| =1, 0 for |i−j| ≥2.
(c)
C =
⎡
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
1 0 1 0 0 0 0
e 0 eπ 4 5 1 √
π e2 1 1731 √
6 √ 7 √
8√ 10 e3 0 −e π e 0 πe e4 0 10001 0 π−1 0 e2π e6 0 √
2 0 0 0 −1
0 0 1 0 0 0 0
⎤
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
∈R7,7.
98 7 Determinants of Matrices (d) The 4×4 Wilkinson matrix6 (cp. the MATLAB-Minute at the end of
Sect.7.2).
7.5 Construct matrices A,B ∈ Rn,n for some n ≥ 2 and with det(A+B) = det(A)+det(B).
7.6 LetR be a commutative ring with unit,n ≥ 2 and A ∈ Rn,n. Show that the following assertions hold:
(a) adj(In)=In.
(b) adj(A B)=adj(B)adj(A), ifAandB∈ Rn,nare invertible.
(c) adj(λA)=λn−1adj(A)for allλ∈ R.
(d) adj(AT)=adj(A)T.
(e) det(adj(A))=(det(A))n−1, ifAis invertible.
(f) adj(adj(A))=det(A)n−2A.
(g) adj(A−1)=adj(A)−1, ifAis invertible.
Can one drop the requirement of invertibility in (b) or (e)?
7.7 Let n ≥ 2 and A = [ai j] ∈ Rn,n withai j = xi+1yj for some x1, . . . ,xn, y1, . . . ,yn ∈R. Hence, in particular,xi+yj =0 for alli,j. (Such a matrixA is called aCauchy matrix.7)
(a) Show that
det(A)=
1≤i<j≤n(xj−xi)(yj−yi) n
i,j=1 (xi+yj) .
(b) Use (a) to derive a formula for the determinant of then×n Hilbert matrix (cp. the MATLAB-Minute above Definition5.6).
7.8 LetRbe a commutative ring with unit. Ifα1, . . . ,αn∈ R,n≥2, then
Vn :=! αij−1
"
=
⎡
⎢⎢
⎢⎣
1α1 · · ·αn1−1 1α2 · · ·αn2−1 ... ... ... 1αn · · ·αnn−1
⎤
⎥⎥
⎥⎦∈ Rn,n
is called aVandermonde matrix.8 (a) Show that
det(Vn)=
1≤i<j≤n
(αj−αi).
6James Hardy Wilkinson (1919–1986).
7Augustin Louis Cauchy (1789–1857).
8Alexandre-Théophile Vandermonde (1735–1796).
(b) LetK be a field and letK[t]≤n−1be the set of polynomials in the variable tof degree at mostn−1. Show that two polynomialsp,q ∈K[t]≤n−1are equal if there exist pairwise distinctβ1, . . . ,βn ∈K withp(βj)=q(βj).
7.9 Show the following assertions:
(a) LetK be a field with 1+1=0 and let A∈ Kn,n withAT = −A. Ifnis odd, then det(A)=0.
(b) IfA∈G Ln(R)withAT =A−1, then det(A)∈ {1,−1}. 7.10 LetK be a field and
A=
A11 A12 A21 A22
for some A11 ∈ Kn1,n1,A12 ∈ Kn1,n2,A21 ∈ Kn2,n1, A22 ∈ Kn2,n2. Show the following assertions:
(a) IfA11∈G Ln1(K), then det(A)=det(A11)det
A22−A21A−111A12 . (b) IfA22∈G Ln2(K), then det(A)=det(A22)det
A11−A12A−221A21 . (c) IfA21=0, then det(A)=det(A11)det(A22).
Can you show this also when the matrices are defined over a commutative ring with unit?
7.11 Construct matricesA11,A12,A21,A22∈Rn,n forn≥2 with det
A11 A12
A21 A22 =det(A11)det(A22)−det(A12)det(A21).
7.12 Let A = [ai j] ∈ G Ln(R)withai j ∈ Z fori,j = 1, . . . ,n. Show that the following assertions hold:
(a) A−1∈Qn,n.
(b) A−1∈Zn,nif and only if det(A)∈ {−1,1}.
(c) The linear system of equations Ax=bhas a unique solution'x∈Zn,1for everyb∈Zn,1if and only if det(A)∈ {−1,1}.
7.13 Show thatG= {A∈Zn,n| det(A)∈ {−1,1} }is a subgroup ofG Ln(Q).
Chapter 8
The Characteristic Polynomial and Eigenvalues of Matrices
We have already characterized matrices using their rank and their determinant. In this chapter we use the determinant map in order to assign to every square matrix a unique polynomial that is called the characteristic polynomial of the matrix. This polynomial contains important information about the matrix. For example, one can read off the determinant and thus see whether the matrix is invertible. Even more important are the roots of the characteristic polynomial, which are called the eigenvalues of the matrix.