Basic Linear Algebra

32 Pages·2016·0.323 MB·English

by  Jürgen Müller

Basic linear algebra
Friedrich-Schiller-Universit¨at Jena, SS 2016
Ju¨rgen Mu¨ller
Contents
1 Linear equations 1
(1.1) Example: Linear optimization. . . . . . . . . . . . . . . . . . . . . 1
(1.2) Lines in R2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
(1.3) Linear equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
(1.4) Theorem: Gauß algorithm. . . . . . . . . . . . . . . . . . . . . . . 5
(1.5) Solving linear equations. . . . . . . . . . . . . . . . . . . . . . . . . 5
(1.6) Example: Geometric interpretation. . . . . . . . . . . . . . . . . . 7
(1.7) Inversion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
(1.8) Example: A simple cipher. . . . . . . . . . . . . . . . . . . . . . . . 9
2 Vector spaces 10
(2.1) Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
(2.2) Vector spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
(2.3) Subspaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
(2.4) Linear combinations. . . . . . . . . . . . . . . . . . . . . . . . . . . 12
(2.5) Linear independence. . . . . . . . . . . . . . . . . . . . . . . . . . . 12
(2.6) Bases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
(2.7) Theorem: Characterization of bases. . . . . . . . . . . . . . . . . . 14
(2.8) Theorem: Steinitz’s base change theorem. . . . . . . . . . . . . . . 14
(2.9) Dimension. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
(2.10) Theorem: Dimension formula for subspaces. . . . . . . . . . . . . . 16
3 Linear maps 16
(3.1) Linear maps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
(3.2) Theorem: Linear maps and bases. . . . . . . . . . . . . . . . . . . 17
(3.3) Theorem: Dimension formula for linear maps. . . . . . . . . . . . . 17
(3.4) Linear maps and matrices. . . . . . . . . . . . . . . . . . . . . . . . 18
(3.5) Matrix products. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
(3.6) Base change. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
(3.7) Matrix rank. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
(3.8) Gaussian normal forms, revisited. . . . . . . . . . . . . . . . . . . . 22
4 Determinants 23
(4.1) Oriented volumes. . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
(4.2) Permutations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
(4.3) Determinants of matrices. . . . . . . . . . . . . . . . . . . . . . . . 24
(4.4) Theorem: Laplace expansion. . . . . . . . . . . . . . . . . . . . . . 26
(4.5) Corollary: Cramer’s rule. . . . . . . . . . . . . . . . . . . . . . . . 26
(4.6) Example: Computation of determinants. . . . . . . . . . . . . . . . 27
(4.7) Example: Vandermonde matrix.. . . . . . . . . . . . . . . . . . . . 28
(4.8) Determinants of linear maps. . . . . . . . . . . . . . . . . . . . . . 28
5 References 30
1
1 Linear equations
(1.1) Example: Linear optimization. A peasant has got an area of 40ha=
400.000m2 of farm land, and a cow shed providing space to keep at most 10
cows. He is able to work 2400h per year, where to nurse a cow he needs 1ha of
farmlandtogrowfeedinggrassand200hworkingtime,whiletogrowwheaton
1ha of farm land he needs 50h working time. He earns 260e per cow, and 130e
per hectare of wheat. Now he wants to maximize his yearly profit, by choosing
the appropriate number x ∈ R of cows to keep and area y ∈ R of land to grow
wheat on; note that we might just weigh the cows instead of counting them.
Thus we have [x,y]∈D, where D ⊆R2 is given by the following constraints:
y ≥ 0 (1)
x ≥ 0 (2)
x ≤ 10 (3)
x+y ≤ 40 (4)
200x+50y ≤ 2400 (5)
Hence D is a convex polygon, see Table 1, where P is the intersection of the
lines {[x,y] ∈ R2;x = 10} and {[x,y] ∈ R2;200x+50y = 2400} defined by (3)
and (5), respectively, thus P = [10,8]. Moreover, Q is the intersection of the
lines {[x,y] ∈ R2;x+y = 40} and {[x,y] ∈ R2;200x+50y = 2400} defined by
(4)and(5),respectively. Thuswehavetosolveasystemoflinearequations:
(cid:26)
x + y = 40
200x + 50y = 2400
Dividingthesecondequationby50,andsubtractingthefirstequationyieldsthe
equation 3x = 8, hence x = 8, and subtracting the latter from the first yields
3
y =40− 8 = 112, thus Q= 1 ·[8,112]. Note that the system has coefficients
3 3 3
in Z, and we allow for solutions in R2, but the solution is in Q2\Z2.
We have to maximize the cost function ϕ: R2 → R: [x,y] (cid:55)→ 260x+130y on
D. Since D is compact and ϕ is continuous, the maximum c :=max{ϕ(x,y)∈
0
R;[x,y] ∈ D} is attained. To find it, we for c ∈ R consider the line L :=
c
ϕ−1(c):={[x,y]∈R2;ϕ(x,y)=c}. It is geometrically seen that c =max{c∈
0
R;L ∩D =(cid:54) ∅}, hence c is determined by the condition Q ∈ L and we have
c 0 c0
ϕ−1(c )∩D = {Q}. Thus we have c = 260· 8 +130· 112 = 16640 = 5546,66.
0 0 3 3 3
In conclusion, the peasant at best earns 5546,66e per year, which happens if
he keeps 8 = 2,66 cows and grows wheat on 112ha = 37,33ha of farm land.
3 3
Note that hence the problem of finding the maximum c has been reduced to
0
essentially solving a system of linear equations.
(1.2)LinesinR2. Havingchosenacoordinatesystem,geometricalobjects
like lines L in the Euclidean plane R2 can be described algebraically. There
are various ways to do so, where these descriptions are by no means unique,
2
Table 1: Geometric picture of constraints and cost function.
(1)
x
(4) P = [10,8] 10(3)(5)
Q = [8/3,112/3] 5
y L5546,66... L5200 40 30L3640 L260020 10 0(2)
but are all equivalent, inasmuch it is possible to switch from either of these
representations to any other one:
i) L can be given as {[x,y] ∈ R2;ax+by = c}, where a,b,c ∈ R such that
[a,b](cid:54)=[0,0] are fixed, that is the points satisfying a certain linear equation.
ii) L can be given in parameterized form as {[x ,y ]+t·[u,v]∈R2;t∈R},
0 0
where [x ,y ],[u,v] ∈ R2 such that [u,v] (cid:54)= [0,0] are fixed; that is [x ,y ] is a
0 0 0 0
fixed point belonging to L, and [u,v] describes the direction into which L runs.
iii)Lcanbegivenasbyspecifyingtwopoints[x ,y ](cid:54)=[x ,y ]∈R2belonging
0 0 1 1
toit;notethathereitbecomesclearthatwemakeuseoftheaxiomofEuclidean
geometry saying that two distinct points determine a unique line in the plane.
Here, the expression ‘[x,y]+t·[u,v]’ is comprised of a scalar multiplication
and an addition, both performed entrywise on tuples. This is the algebraic
translation of the geometric processes of dilating and negating ‘point vectors’,
and of adding two ‘point vectors’, respectively.
For example, the lines in (1.1) are described as follows:
(i) (ii) (iii)
(1) y =0 t·[1,0] [0,0],[1,0]
(2) x=0 t·[0,1] [0,0],[0,1]
(3) x=10 [10,0]+t·[0,1] [10,0],[10,1]
(4) x+y =40 [20,20]+t·[1,−1] [40,0],[0,40]
(5) 4x+y =48 [12,0]+t·[1,−4] [12,0],[0,48]
ϕ 260x+130y =c [0, c ]+t·[1,−2] [ c ,0],[0, c ]
130 260 130
Forexample,forline(5)wehavethelinearequation4x+y =48,henceforx =0
0
we get y = 48, and for y = 0 we get x = 12, thus from [x ,y ]−[x ,y ] =
0 1 1 1 1 0 0
[12,−48] we get the parameterizations {[0,48] + t · [1,−4] ∈ R2;t ∈ R}, or
3
equivalently {[12,0]+t·[1,−4] ∈ R2;t ∈ R}. Conversely, given the latter, for
any point [x,y] on the line we have x = 12+t and y = −4t, for some t ∈ R,
which yields y =−4(x−12)=−4x+48, or equivalently 4x+y =48.
(1.3) Linear equations. Given coefficients a ∈ R, for i ∈ {1,...,m} and
ij
j ∈ {1,...,n}, known ‘output’ quantities y ,...,y ∈ R, and unknown inde-
1 m
terminate ‘input’ quantities x ,...,x ∈ R, where m,n ∈ N, the associated
1 n
system of linear equations is given as follows:
 (cid:80)n a x = a x + a x + ··· + a x = y
 (cid:80)njj==11a12jjxjj = a1211x11 + a1222x22 + ··· + a12nnxnn = y12
. .
. .
 (cid:80)n a x =. a x + a x + ··· + a x =. y
j=1 mj j m1 1 m2 2 mn n m
We combine the above quantities to columns v := [x ,...,x ]tr ∈ Rn×1 and
1 n
w :=[y ,...,y ]tr ∈Rm×1, respectively, where by ‘tr’ we just indicate that the
1 m
rows in question are considered as columns. Moreover, we write the coefficients
as a (m×n)-matrix with entries a ∈R, that is as a rectangular scheme
ij
 
a ... a
11 1n
A=[a ] = .. .. ∈Rm×n
ij ij  . . 
a ... a
m1 mn
with m rows and n columns; if m=n then A is called quadratic. Then the
system can be written as A·v = w, where the matrix product ‘A·v’ on the
left hand side is just defined as the column [(cid:80)n a x ,...,(cid:80)n a x ]tr ∈
j=1 1j j j=1 mj j
Rm×1. The set of solutions is defined as L(A,w) := {v ∈ Rn×1;Av = w}; if
L(A,w) (cid:54)= ∅ the system is called solvable. The aim is to understand when a
system is solvable, and, in this case, how the set of solutions can be described.
To solve the system we consider the extended matrix [A|w] ∈ Rm×(n+1) ob-
tainedbyconcatenatingthecolumnsofAwiththecolumnw. Thenmultiplying
equationiwith0(cid:54)=a∈R,addingthea-foldofequationj toequationi(cid:54)=j,for
a∈R, andinterchangingequationsiandj, translatesintorow operationson
[A|w],namelymultiplyingrowientrywisewith0(cid:54)=a∈R,addingentrywisethe
a-fold of row j to row i, and interchanging rows i and j, respectively. Since we
are replacing equations by consequences of given equations, the set of solutions
might become larger, but since all steps are invertible, the set of solutions ac-
tually remains the same, as desired. The aim then is to arrive at an equivalent
system, whose solutions can be read off readily. For example:
i) We reconsider the system turning up in (1.1); note that this is an algebraic
translation of the geometrical problem of finding the intersection of two lines in
the plane: We have m=n=2 and A·[x,y]tr =w, where
(cid:20) (cid:21)
1 1 40
[A|w]= ∈R2×(2+1),
200 50 2400
4
and suitable row operations, yielding L(A,w)={[8,112]tr}, are:
3 3
(cid:20) 1 1 40 (cid:21) (cid:20) 1 1 40 (cid:21) (cid:20) 1 1 40 (cid:21) (cid:20) . 1 112 (cid:21)
[A|w](cid:55)→ (cid:55)→ (cid:55)→ (cid:55)→ 3
4 1 48 3 . 8 1 . 8 1 . 8
3 3
ii) For the system

3x + 6x + 2x + 10x = 2
 1 2 3 4
10x + 16x + 6x + 30x = 6
1 2 3 4
 5x + 14x + 4x + 14x = 10
1 2 3 4
we have m=3 and n=4, and A·[x ,x ,x ,x ]tr =w where
1 2 3 4
 
3 6 2 10 2
[A|w]= 10 16 6 30 6 ∈R3×(4+1).
5 14 4 14 10
Adding the (−3)-fold of row 1 to row 2, and the (−2)-fold of row 1 to row 3
yields
   
3 6 2 10 2 3 6 2 10 2
[A|w](cid:55)→ 1 −2 . . . (cid:55)→ 1 −2 . . . .
5 14 4 14 10 −1 2 . −6 6
Next,addingthe(−3)-foldofrow2torow1,addingrow2torow3,anddividing
row 1 by 2, and dividing row 3 by −6 yields
     
3 6 2 10 2 . 12 2 10 2 . 6 1 5 1
 1 −2 . . . (cid:55)→ 1 −2 . . . (cid:55)→ 1 −2 . . . .
−1 2 . −6 6 . . . −6 6 . . . 1 −1
Finally, adding the (−5)-fold of row 3 to row 1, and interchanging rows 1 and 2
yields
   
. 6 1 5 1 1 −2 . . .
 1 −2 . . . (cid:55)→ . 6 1 . 6 .
. . . 1 −1 . . . 1 −1
Henceweinferx =−1,andwemaychoosex =t∈Rfreely,thenwegetx =
4 2 3
6−6t and x = 2t, implying that L(A,w) = {[0,0,6,−1]tr+t·[2,1,−6,0]tr ∈
1
R4×1;t∈R}, a line in R4×1.
5
(1.4) Theorem: Gauß algorithm. Let A ∈ Rm×n, where m,n ∈ N. Then
using row operations A can be transformed into Gaussian normal form
 
... 1 ∗∗∗ . ∗∗∗ . ∗∗∗ . ∗∗∗ . ∗∗∗
 ... . ... 1 ∗∗∗ . ∗∗∗ . ∗∗∗ . ∗∗∗ 
 
 ... . ... . ... 1 ∗∗∗ . ∗∗∗ . ∗∗∗ 
 
 ... . ... . ... . ... 1 ∗∗∗ . ∗∗∗ 
A(cid:48) = .. .. ∈Rm×n,
 . . 
 
 ... . ... . ... . ... . ... 1 ∗∗∗ 
 
 ... . ... . ... . ... . ... . ... 
 . . 
. .
. .
having r = r(A) ∈ {0,...,min{m,n}} non-zero rows, being called the Gauß
rankofA. The‘1’soccurintheuniquelydeterminedpivot columns1≤j <
1
j <···<j ≤nofA;hencetheGaussiannormalformisuniquelydetermined.
2 r
Proof. Here,weonlyshowexistence,beingsufficientfortheproblemofsolving
systems of linear equations, and postpone the question of uniqueness to (3.8):
We proceed row by row: Looking at row k ∈ N, by induction we may assume
that k−1∈N rows have already been processed; let j :=0. We consider the
0 0
submatrix A(cid:101)consisting of rows [k,...,m] and columns [jk−1+1,...,n] of A.
We may assume that A(cid:101)(cid:54)=0, and let jk ∈{jk−1+1,...,n} be the first column
containing an entry a := a (cid:54)= 0, where i ∈ {k,...,m}. Then interchanging
i,jk
rowsk andi,andmultiplyingrowk bya−1,yieldsamatrixsuchthata =1.
k,jk
Adding the (−a )-fold of row k to row i, for all i∈{1,...,m}, cleans up all
i,jk
of column j , yielding a matrix such that a =0 for all k (cid:54)=i∈{1,...,m}. (cid:93)
k i,jk
(1.5) Solving linear equations. a)LetA=[a ] ∈Rm×n,wherem,n∈N,
ij ij
and w = [y ,...,y ]tr ∈ Rm×1. We consider the system of linear equations in
1 m
the indeterminates x ,...,x ∈ R given by A·[x ,...,x ]tr = w; the latter is
1 n 1 n
called homogeneous if w = 0, otherwise it is called inhomogeneous. Hence
the set of solutions is L(A,w):={v ∈Rn×1;Av =w}, where the homogeneous
system associated with A we also write L(A):=L(A,0).
Since for 0∈Rn×1 we have A·0=0∈Rm×1, we infer 0∈L(A), in particular
anyhomogeneoussystemissolvable. Moreover,itisimmediatefromtheexplicit
description A·[x ,...,x ]tr = [(cid:80)n a x ,...,(cid:80)n a x ]tr ∈ Rm×1 of the
1 n j=1 1j j j=1 mj j
matrixproduct,forall[x ,...,x ]tr ∈Rn×1,thatwehaveA(v+v(cid:48))=Av+Av(cid:48)
1 n
and A(av) = a·Av, for all v,v(cid:48) ∈ Rn×1 and a ∈ R, where addition of tuples
and taking scalar multiples is performed entrywise. Hence we infer that L(A)
is closed with respect to these operations.
Moreover, if v,v(cid:48) ∈ L(A,w) then we have A(v − v(cid:48)) = w − w = 0, that is
v−v(cid:48) ∈L(A); conversely, for all u∈L(A) we have A(v+u)=w+0=w, that
is v+u∈L(A,w). Hence if L(A,w)(cid:54)=∅, then we have L(A,w)=v +L(A):=
0
{v +u∈Rn×1;u∈L(A)}, wherev ∈L(A,w)isafixedparticular solution.
0 0
6
Hence solving the system amounts to prove solvability, and in this case to find
a particular solution and the solutions L(A) of the associated homogeneous
system. To decide solvability, we apply the Gauß algorithm to the extended
matrix[A|w]∈Rm×(n+1),whichshowsthatL(A,w)(cid:54)=∅ifandonlyifn+1isnot
apivotcolumn,whichisequivalenttosayingthatfortheassociatedGaußranks
we have r(A)=r([A|w]). In this case the solutions v =[x ,...,x ]tr ∈L(A,w)
1 n
are described as follows:
b)Let[A|w]haveGaussiannormalform[A(cid:48)|w(cid:48)]∈Rm×(n+1) withpivotcolumns
1 ≤ j < ··· < j ≤ n, where r := r(A) ∈ N . By the above analysis we have
1 r 0
L(A,w)={v ∈Rn×1;Av =w}={v ∈Rn×1;A(cid:48)v =w(cid:48)}=L(A(cid:48),w(cid:48)).
The n−r entries x ∈R, where j ∈J :={1,...,n}\{j ,...,j } are the non-
j 1 r
pivot columns, can be chosen arbitrarily, thus are considered as parameters.
Then the r entries x ∈ R, for the pivot columns {j ,...,j }, are uniquely
determined by x :=jyk(cid:48) −(cid:80) a(cid:48) x , for k ∈{1,...,r1}. r
jk k j∈J kj j
To facilitate the explicit description of solutions, for j ∈ {1,...,n} let e :=
j
[0,...,0,1,0,...,0]tr ∈Rn×1 be the unit tuple with entry j non-zero; then we
have [x ,...,x ]tr = (cid:80)n x e . Now, the particular solution given by letting
1 n j=1 j j
x :=0, for all j ∈J, equals v :=(cid:80)r y(cid:48)e ∈Rn×1.
j 0 k=1 k jk
Considering the homogeneous system, specifying the parameters to be unit tu-
ples on J, we get the basic solutions v := e −(cid:80)r a(cid:48) e ∈ Rn×1, for
j j k=1 kj jk
j ∈ J, and thus L(A) = {(cid:80) x v ∈ Rn×1;x ∈ R for j ∈ J}, hence
j∈J j j j
L(A,w) = v +L(A). Note that any element of L(A) can be written uniquely
0
as an R-linear combination of the basic solutions {v ∈Rn×1;j ∈J}.
j
c) From a more general point of view we make the following observations: We
have L(A,w) (cid:54)= ∅ for all w ∈ Rm×1 if and only if r = m; hence in this case we
have m = r ≤ n. Moreover, if L(A,w) (cid:54)= ∅ for some w ∈ Rm×1, then we have
|L(A,w)| = 1 if and only if |L(A)| = 1, which holds if and only if n−r = 0;
hence in this case we have m≥r =n.
Thus in the case m=n of quadratic systems the following are equivalent:
i) We have r =n, that is the Gaussian normal form is the identity matrix.
ii) We have |L(A)|=1, that is L(A)={0}.
iii) There is w ∈Rn×1 such that |L(A,w)|=1.
iv) For all w ∈Rn×1 we have |L(A,w)|=1.
v) For all w ∈Rn×1 we have L(A,w)(cid:54)=∅.
d) Note that in practice it is usually not necessary to compute the Gaussian
normal form in order to solve a system of linear equations, but an intermediate
steptypicallyissufficient, aslongasitalreadyexhibitsanechelon form; such
a form is also indicated in example (i) below. Here are a couple of examples;
recallalsothosein(1.1)and(1.3),theformerbeingofthetypedescribedin(c):
7
i) The Gauß algorithm yields:
 
1 0 4 0 1 0 1
 2 1 0 2 1 1 1 
 
[A|w|w(cid:101)]:= 3 1 1 1 1 2 2 
 
 6 2 5 3 3 3 3 
7 2 12 4 5 2 2
 1 . 4 . 1 . 1   1 . . −4 −1 4 1 
3 3 3
 . 1 −8 2 −1 1 −1   . 1 . 14 5 −5 −1 
(cid:55)→ . . 1 1 1 −1 . (cid:55)→ . . 1 31 31 −31 . 
 3 3 3   3 3 3 
 . . . . . . 1   . . . . . . 1 
. . . . . . . . . . . . . .
HencewehaveL(A,w)=∅,whileL(A,w)(cid:54)=∅. ThelatterisgivenasL(A,w)=
(cid:101)
v +L(A),whereaparticularsolutioncanbechosenasv :=[4,−5,−1,0,0]tr ∈
0 0 3 3 3
R5×1,whiletheassociatedhomogeneoussystemhasthesetofsolutionsL(A)=
{sv +tv ∈ R5×1;s,t ∈ R}, where in turn the basic solutions are given as
4 5
v :=[4,−14,−1,1,0]tr ∈R5×1 and v :=[1,−5,−1,0,1]tr ∈R5×1.
4 3 3 3 5 3 3 3
ii)TheGaußalgorithmyieldsforgenericrighthandsidew :=[a,b,c]tr ∈R3×1;
we will come back to this point of view in (1.7):
   
1 1 5 a 1 1 5 a
[A|w]:=  6 −1 16 b  (cid:55)→  . −7 −14 −6a+b 
3 1 11 c . −2 −4 −3a+c
 1 1 5 a   1 . 3 a+b 
7
(cid:55)→  . 1 2 6a−b  (cid:55)→  . 1 2 6a−b 
7 7
. −2 −4 −3a+c . . . −9a−2b+7c
7
Hence the system A·[x ,x ,x ]tr =[a,b,c]tr is solvable if and only if 9a+2b−
1 2 3
7c = 0. For the homogeneous system we get L(A) = {t·[−3,−2,1]tr;t ∈ R},
hence if c= 9a+2b we get L(A,[a,b,c]tr)={[a+b,6a−b,0]tr+t·[−3,−2,1]tr;t∈
7 7 7
R}, a line in R3×1; for example L(A,[3,11,7]tr)={[2−3t,1−2t,t]tr;t∈R}.
(1.6) Example: Geometric interpretation. For n∈{2,3} we discuss L:=
L(A,w), where A ∈ Rm×n and w ∈ Rm×1, in dependence of r := r(A) ∈
{0,...,n}. We may assume that the system [A|w] ∈ Rm×(n+1) is in Gaussian
normal form, hence we may also assume that m=r+1.
(cid:2) (cid:3)
If r = 0, then we get . (cid:15) , where (cid:15) ∈ {0,1}, hence we have L (cid:54)= ∅ if and
only if (cid:15)=0, and in this case we have L=Rn×1. If r =n, then L has at most
one element. The intermediate cases are more interesting:
a) If n=2 and r =1, then for some s,a∈R and (cid:15)∈{0,1} we get
(cid:20) (cid:21) (cid:20) (cid:21)
1 s a . 1 a
or .
. . (cid:15) . . (cid:15)
8
Hence we have L (cid:54)= ∅ if and only if (cid:15) = 0. In this case, in the first case we
have L = {[a−sy,y]tr = [a,0]tr +y·[−s,1]tr ∈ R2×1;y ∈ R}, a line in R2×1;
it intersects the first coordinate axis in [a,0]tr, and the second coordinate axis
in [0,a]tr if s (cid:54)= 0, while it is parallel to it if s = 0, coinciding with it if and
s
only if a = 0. In the second case we have L = {[x,a]tr = [0,a]tr+x·[1,0]tr ∈
R2×1;x∈R}, a line in R2×1; it intersects the second coordinate axis in [0,a]tr,
andisparalleltothefirstcoordinateaxis,coincidingwithitifandonlyifa=0.
b) If n=3 and r =2, then for some s,t,a,b∈R and (cid:15)∈{0,1} we get
     
1 . s a 1 s . a . 1 . a
 . 1 t b  or  . . 1 b  or  . . 1 b .
. . . (cid:15) . . . (cid:15) . . . (cid:15)
HencewehaveL=(cid:54) ∅ifandonlyif(cid:15)=0. Inthiscase,wehaveL={[a−sz,b−
tz,z]tr = [a,b,0]tr+z·[−s,−t,1]tr ∈ R3×1;z ∈ R} and L = {[a−sy,y,b]tr =
[a,0,b]tr +y ·[−s,1,0]tr ∈ R3×1;y ∈ R} and L = {[x,a,b]tr = [0,a,b]tr +x·
[1,0,0]tr ∈R3×1;x∈R}, respectively, lines in R3×1.
c) If n=3 and r =1, then for some s,t,a∈R and (cid:15)∈{0,1} we get
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
1 s t a . 1 s a . . 1 a
or or .
. . . (cid:15) . . . (cid:15) . . . (cid:15)
Hence we have L =(cid:54) ∅ if and only if (cid:15) = 0. In this case, we have L = {[a−
sy−tz,y,z]tr = [a,0,0]tr+y·[−s,1,0]tr+z·[−t,0,1]tr ∈ R3×1;y,z ∈ R} and
L = {[x,a−zs,z]tr =[0,a,0]tr+x·[1,0,0]tr+z·[0,−s,1]tr ∈ R3×1;x,z ∈ R}
and L = {[x,y,a]tr = [0,0,a]tr +x·[1,0,0]tr +y·[0,1,0]tr ∈ R3×1;x,y ∈ R},
respectively, hyperplanes in R3×1.
(1.7)Inversion. Weagaincomebacktothesystemoflinearequationsin(1.1),
(cid:20) (cid:21)
1 1
givenbyA:= ∈R2×2, whichisofthetypedescribedin(1.5)(c). We
200 50
have seen how to solve the specific system A·[x,y]tr = [40,2400]tr. Now any
system with the same matrix A and some right hand side w = [c,d]tr ∈ R2×1
has a unique solution, and it is immediate that we can similarly solve it, but
at the cost of doing the explicit computation for any w separately. Hence we
are tempted to do this only once, for a generic right hand side, that is for the
system A·[x,y]tr = [c,d]tr, where now c,d ∈ R are indeterminates. Note that
example (ii) in (1.5) already points in the same direction.
To see how L(A,[c,d]tr) depends on [c,d], we redo the above row operations:
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
1 1 c 1 1 c 1 1 c
(cid:55)→ (cid:55)→
200 50 d 4 1 d 3 . −c+ d
50 50
(cid:20) 1 1 c (cid:21) (cid:20) . 1 4c − d (cid:21) (cid:20) 1 . −c + d (cid:21)
(cid:55)→ (cid:55)→ 3 150 (cid:55)→ 3 150
1 . −c + d 1 . −c + d . 1 4c − d
3 150 3 150 3 150