3. Gaussian elimination 51
3.17. The swapping matrices T u,v
The λ-addition matrices Aλu,v from Definition 3.53 and the λ-scaling matrices Sλu from Definition 3.85 are two of the three kinds of matrices commonly called “ele-mentary matrices”. The third kind are the swapping matrices:
Definition 3.102. Letn∈ N. Letuandvbe two distinct elements of{1, 2, . . . ,n}. Then, Tu,v shall denote the n×n-matrix In−Eu,u −Ev,v+Eu,v+Ev,u. (Here, all the matrices of the formEp,q are meant to ben×n-matrices: that is,Ep,q =Ep,q,n,n
for all p ∈ {1, 2, . . . ,n} andq ∈ {1, 2, . . . ,n}.)
Again, the notation Tu,vis hiding the dependency onn, and we ought to write Tu,v,n instead; but we will not, because there will not be any real occasion for confusion.
Example 3.103. Letn =4. Then,
T1,3= In−E1,1−E3,3+E1,3+E3,1 =
0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1
and
T2,3= In−E2,2−E3,3+E2,3+E3,2 =
1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1
.
The pattern that you see on these examples is true in general:
Proposition 3.104. Let n ∈ N. Let u and v be two distinct elements of {1, 2, . . . ,n}. Then, the matrix Tu,v has the following entries:
• Its (u,u)-th and(v,v)-th entries are 0.
• All its other diagonal entries are 1.
• Its (u,v)-th and(v,u)-th entries are 1.
• All its remaining entries are 0.
Proof of Proposition 3.104. For each p ∈ {1, 2, . . . ,n}andq ∈ {1, 2, . . . ,n}, the matrix Ep,q is the n×n-matrix whose (p,q)-th entry is 1 and whose all other entries are 0 (indeed, this is how Ep,q was defined). Hence, we obtain the following two facts:
Fact 1: Adding Ep,q to an n×n-matrix C has the effect that the (p,q)-th entry ofC is increased by 1 (while all other entries remain unchanged).
Fact 2: Subtracting Ep,q from an n×n-matrix C has the effect that the (p,q)-th entry of C is decreased by 1 (while all other entries remain unchanged).
But recall that Tu,v = In −Eu,u −Ev,v+Eu,v+Ev,u. Thus, the matrix Tu,v is obtained from the matrix In by first subtracting Eu,u, then subtracting Ev,v, then adding Eu,v, and then adding Ev,u. Using Fact 1 and Fact 2, we can see how these subtractions and additions affect the entries of a matrix; thus, we can find all entries of the matrix Tu,v = In−Eu,u−Ev,v+Eu,v+Ev,u:
• The matrix In is the n×n-matrix whose diagonal entries79 are 1, and whose all other entries are 0.
• Subtracting Eu,u from In has the effect that the(u,u)-th entry is decreased by 1; thus, it becomes 1−1 = 0 (because it was 1 in In). Hence, the (u,u)-th entry of the matrix In−Eu,u is 0, all its other diagonal entries are 1, and all its remaining entries are 0.
• Subtracting Ev,v from In −Eu,u has the effect that the (v,v)-th entry is de-creased by 1; thus, it becomes 1−1 =0 (because it was 1 in In−Eu,u). Hence, the (u,u)-th and (v,v)-th entries of the matrix In −Eu,u −Ev,v are 0, all its other diagonal entries are 1, and all its remaining entries are 0.
• Adding Eu,v to In −Eu,u −Ev,v has the effect that the (u,v)-th entry is in-creased by 1; thus, it becomes 0+1=1 (because it was 0 in In−Eu,u−Ev,v).
Hence, the (u,u)-th and (v,v)-th entries of the matrix In −Eu,u−Ev,v+Eu,v
are 0, all its other diagonal entries are 1, its (u,v)-th entry is 1, and all its remaining entries are 0.
• Adding Ev,u to In−Eu,u−Ev,v+Eu,v has the effect that the(v,u)-th entry is increased by 1; thus, it becomes 0+1 = 1 (because it was 0 in In −Eu,u− Ev,v+Eu,v). Hence, the(u,u)-th and(v,v)-th entries of the matrix In−Eu,u− Ev,v+Eu,v+Ev,u are 0, all its other diagonal entries are 1, its (u,v)-th and (v,u)-th entries are 1, and all its remaining entries are 0. Since In−Eu,u− Ev,v+Eu,v+Ev,u = Tu,v, this rewrites as follows: The (u,u)-th and (v,v)-th entries of the matrixTu,v are 0, all its other diagonal entries are 1, its(u,v)-th and (v,u)-th entries are 1, and all its remaining entries are 0. This proves Proposition 3.104.
We can next see what happens to a matrix when it is multiplied byTu,v:
79This includes the(u,u)-th entry.
Proposition 3.105. Letn ∈Nandm∈ N. Letuandvbe two distinct elements of {1, 2, . . . ,n}. Let Cbe ann×m-matrix. Then, Tu,vCis then×m-matrix obtained fromC by swapping theu-th row with the v-th row.
Example 3.106. Let n = 3 and m = 2. Let C be the 3×2-matrix
a b a0 b0 a00 b00
. Then, Proposition 3.105 (applied to u = 1 and v = 3) claims that T1,3C is the 3×2-matrix obtained from C by swapping the 1-st row with the 3-rd row. A computation confirms this claim:
T1,3C=
0 0 1 0 1 0 1 0 0
a b a0 b0 a00 b00
=
a00 b00 a0 b0 a b
.
Proposition 3.107. Letn ∈Nandm∈ N. Letuandvbe two distinct elements of {1, 2, . . . ,n}. Let Cbe anm×n-matrix. Then,CTu,v is them×n-matrix obtained fromC by swapping theu-th column with the v-th column.
Corollary 3.108. Let n∈ N. Letuand vbe two distinct elements of{1, 2, . . . ,n}. Then:
(a)The matrixTu,v can be obtained fromIn by swapping theu-th row with the v-th row.
(b)The matrixTu,v can be obtained from In by swapping theu-th column with the v-th column.
I will prove Proposition 3.105 and Corollary 3.108 in Section 3.18 below.
We shall refer to the matrix Tu,v defined in Definition 3.102 as a “swapping ma-trix”.80 It is the third type of elementary matrices.
Here are a few more properties of swapping matrices:
Proposition 3.109. Let n ∈ N. Let u and v be two distinct elements of {1, 2, . . . ,n}. Then, (Tu,v)T =Tu,v.
Proposition 3.110. Let n ∈ N. Let u and v be two distinct elements of {1, 2, . . . ,n}.
(a)We have Tu,v = In.
(b)The matrix Tu,vis invertible, and its inverse is (Tu,v)−1 =Tu,v. (c)We have Tv,u =Tu,v.
These facts will be proven in Section 3.18.
80The letterTin “Tu,v” stands for “transposition”, but this is not really related to the transpose of a matrix. It is instead due to the fact that the word “transposition” means “changing the positions (of something)”, and in mathematics is often used for swapping two things.
3.18. (*) Some proofs about the swapping matrices
Proof of Proposition 3.105. Clearly, Tu,vC is an n×m-matrix.
Let x and y be two elements of {1, 2, . . . ,n}. Proposition 3.44 (applied to n, m, x and yinstead of m, p, u and v) shows that Ex,yC is the n×m-matrix whose x-th row is the y-th row ofC, and whose all other rows are filled with zeroes. Thus,
rowx Ex,yC
=rowyC (110)
(since thex-the row of Ex,yC is they-th row ofC) and rowi Ex,yC
=01×m for everyi ∈ {1, 2, . . . ,n} satisfyingi 6=x (111) (since all other rows of Ex,yC are filled with zeroes).
Now, forget that we fixedxandy. We thus have proven (110) and (111) for every two elementsx and yof{1, 2, . . . ,n}. In particular, everyy ∈ {1, 2, . . . ,n} satisfies
rowu Ev,yC
=01×m (112)
(by (111), applied to i=u and x=v (sinceu6=v)) and rowv Eu,yC
=01×m (113)
(by (111), applied to i=vand x=u (sincev6=u)).
But recall that matrices are added entry by entry. Thus, matrices are also added by row by row – i.e., ifU and V are two n×m-matrices, then any row ofU+V is the sum of the corresponding rows ofU and of V. In other words, ifU and V are twon×m-matrices, then
rowi(U+V) = rowiU+rowiV for everyi ∈ {1, 2, . . . ,n}. (114) Similarly,
rowi(U−V) = rowiU−rowiV for everyi ∈ {1, 2, . . . ,n}. (115) Using (114) and (115) repeatedly, we can show that
rowi(U−V−W+X+Y) = rowiU−rowiV−rowiW+rowiX+rowiY (116) for everyi∈ {1, 2, . . . ,n}.
We have
Tu,v
|{z}
=In−Eu,u−Ev,v+Eu,v+Ev,u
C= (In−Eu,u−Ev,v+Eu,v+Ev,u)C
= InC
|{z}
=C
−Eu,uC−Ev,vC+Eu,vC+Ev,uC
=C−Eu,uC−Ev,vC+Eu,vC+Ev,uC.
Hence, for eachi ∈ {1, 2, . . . ,n}, we have
Now, we must prove thatTu,vCis then×m-matrix obtained fromCby swapping theu-th row with the v-th row. In other words, we must prove the following three claims:
Now, we have proven Claim 1, Claim 2 and Claim 3; this completes the proof of Proposition 3.105.
The proof of Proposition 3.107 is analogous.
Proof of Corollary 3.108. (a) Proposition 3.105 (applied to m =n and C = In) shows that Tu,vIn is the n×n-matrix obtained from In by swapping theu-th row with the v-th row. In other words, Tu,vis then×n-matrix obtained from In by swapping the u-th row with the v-th row (since Tu,vIn =Tu,v). This proves Corollary 3.108(a).
(b) The proof of Corollary 3.108 (b) is similar, except that now we have to use Proposition 3.107.
Proof of Proposition 3.109. Proposition 3.18 (d) shows that any two n×m-matrices Aand B(wherem∈ N) satisfy
(A+B)T = AT+BT. (118)
Similarly, any two n×m-matrices Aand B (wherem∈ N) satisfy
(A−B)T = AT−BT. (119)
By repeated application of (118) and (119), we can show the following fact: If A, B, C, Dand E are fiven×m-matrices (for some m∈ N), then
(A−B−C+D+E)T = AT−BT −CT+DT+ET. (120) The definition ofTu,v yieldsTu,v = In−Eu,u−Ev,v+Eu,v+Ev,u. Hence,
(Tu,v)T = (In−Eu,u−Ev,v+Eu,v+Ev,u)T
= (In)T
| {z }
=In
(by Proposition 3.18(a))
− (Eu,u)T
| {z }
=Eu,u
(by Proposition 3.49, applied ton,uandu instead ofm,uandv)
− (Ev,v)T
| {z }
=Ev,v
(by Proposition 3.49, applied ton,vandv instead ofm,uandv)
+ (Eu,v)T
| {z }
=Ev,u
(by Proposition 3.49, applied tom=n)
+ (Ev,u)T
| {z }
=Eu,v
(by Proposition 3.49, applied ton,vandu instead ofm,uandv)
by (120), applied to m=n, A= In, B=Eu,u, C=Ev,v, D=Eu,vand E =Ev,u
= In −Eu,u−Ev,v+Ev,u+Eu,v
= In −Eu,u−Ev,v+Eu,v+Ev,u =Tu,v. This proves Proposition 3.109.
Proof of Proposition 3.110. (a) The definition of Tu,u yields Tu,u = In−Eu,u −Eu,u+ Eu,u +Eu,u = In. This proves Proposition 3.110(a).
(b) First proof: Corollary 3.108 (a) shows that Tu,v is the n×n-matrix obtained from In by swapping the u-th row with the v-th row. Proposition 3.105 (applied
to m = n and C = Tu,v) shows that Tu,vTu,v is the n×n-matrix obtained from Tu,v
by swapping the u-th row with the v-th row. Thus, in order to obtain the matrix Tu,vTu,v fromIn, we have to do the following procedure:
• swap theu-th row with the v-th row;
• then,againswap theu-th row with the v-th row.
But this procedure clearly returns the matrix In with which we started (since the second swap undoes the first swap). Thus,Tu,vTu,v= In.
FromTu,vTu,v = In and Tu,vTu,v = In (yes, this is one equality repeated twice), it follows that the matrixTu,vis an inverse ofTu,v. Hence, the matrixTu,v is invertible, and its inverse is (Tu,v)−1 =Tu,v. This proves Proposition 3.110(b).
Second proof: We can also prove Proposition 3.110 (b) using Proposition 3.52, provided that we can tolerate some rather lengthy computations.
Ifi, j, x and yare four elements of {1, 2, . . . ,n}, then
Ei,jEx,y =δj,xEi,y (121)
(where all three matricesEi,j,Ex,yandEi,yhave to be understood asn×n-matrices)81. In particular, everyi∈ {1, 2, . . . ,n}and y ∈ {1, 2, . . . ,n} satisfy
Ei,uEv,y =0n×n (122)
82 and
Ei,vEu,y =0n×n (123)
83. Furthermore, if i, x and yare three elements of{1, 2, . . . ,n}, then
Ei,xEx,y =Ei,y (124)
84.
81Proof of (121):Leti,j,xandybe four elements of{1, 2, . . . ,n}. Then, Proposition 3.52 (applied to n,n,iandjinstead ofm,p,uand v) shows thatEi,j,n,nEx,y,n,n = δj,xEi,y,n,n. Since we abbreviate the matricesEi,j,n,n,Ex,y,n,nandEi,y,n,nasEi,j,Ex,yandEi,y(respectively), this can be rewritten as Ei,jEx,y=δj,xEi,y. Thus, (121) is proven.
82Proof of (122): Leti∈ {1, 2, . . . ,n}andy ∈ {1, 2, . . . ,n}. Then, (121) (applied to j=uandx =v) yieldsEi,uEv,y= δu,v
|{z}
(since=0u6=v)
Ei,y=0Ei,y=0n×n, qed.
83Proof of (123): Leti∈ {1, 2, . . . ,n}andy ∈ {1, 2, . . . ,n}. Then, (121) (applied to j=vand x=u) yieldsEi,vEu,y= δv,u
|{z}
(since=0v6=u)
Ei,y=0Ei,y=0n×n, qed.
84Proof of (124):Leti,jandxbe three elements of{1, 2, . . . ,n}. Then, (121) (applied toj=x) yields Ei,xEx,y= δx,x
|{z}
(since=1x=x)
Ei,y=1Ei,y=Ei,y, qed.
Now, the definition ofTu,v yieldsTu,v= In−Eu,u−Ev,v+Eu,v+Ev,u. Hence, Tu,v
|{z}
=In−Eu,u−Ev,v+Eu,v+Ev,u
Tu,v
|{z}
=In−Eu,u−Ev,v+Eu,v+Ev,u
= (In−Eu,u−Ev,v+Eu,v+Ev,u) (In−Eu,u−Ev,v+Eu,v+Ev,u)
= In(In−Eu,u−Ev,v+Eu,v+Ev,u)
| {z }
=InIn−InEu,u−InEv,v+InEu,v+InEv,u
− Eu,u(In−Eu,u−Ev,v+Eu,v+Ev,u)
| {z }
=Eu,uIn−Eu,uEu,u−Eu,uEv,v+Eu,uEu,v+Eu,uEv,u
−Ev,v(In−Eu,u−Ev,v+Eu,v+Ev,u)
| {z }
=Ev,vIn−Ev,vEu,u−Ev,vEv,v+Ev,vEu,v+Ev,vEv,u
+ Eu,v(In−Eu,u−Ev,v+Eu,v+Ev,u)
| {z }
=Eu,vIn−Eu,vEu,u−Eu,vEv,v+Eu,vEu,v+Eu,vEv,u
+ Ev,u(In−Eu,u−Ev,v+Eu,v+Ev,u)
| {z }
=Ev,uIn−Ev,uEu,u−Ev,uEv,v+Ev,uEu,v+Ev,uEv,u
= follows that the matrixTu,vis an inverse ofTu,v. Hence, the matrixTu,v is invertible, and its inverse is (Tu,v)−1 =Tu,v. This proves Proposition 3.110(b)again.
(c)The definition ofTv,u yieldsTv,u = In−Ev,v−Eu,u +Ev,u+Eu,v= In−Eu,u− Ev,v+Eu,v+Ev,u. Comparing this with Tu,v = In −Eu,u −Ev,v+Eu,v+Ev,u, we obtain Tv,u =Tu,v. This proves Proposition 3.110(c).