Summary Linear Algebra and Its Applications, Global Edition, ISBN: 9781292092232 Linear Algebra

Lineaire algebra – blok 1
Week 1, les 1
1.2: systems of linear equations
X1 + 2x2 = 3 standard form
 Linear system:
R1 : 1 equation = line
2 equations = intersection point // parallel  n.a.
R2: x=0 R2
x=1 line
x=2 point
x=3 n.a.
R3: x=0 R3
x=1 plane
x=2 line
x=3 point
x= 4 n.a.
many solutions
consistent
1 solution
inconsistent
no solutions
- Example
X1 – 2x2 = -1
-x1 + 3x2 = 3
van elkaar afhalen
x2 = 2  invullen in eerste formule voor x1
Een andere manier van noteren is in een augmented matrix (variables):
[ 1 -2 -1
-1 3 3 ]
Coefficient matrix = hetzelfde als een augmented matrix, alleen dan zonder de dingen achter het
‘’=’’-teken  naam van de matrix moet/staat er altijd bij.
[ 1 -2 -1 [1 -2 -1 +2row2 [1 0 3 x1 = 3
-1 3 3 ] +row1 012 ] 012] x2 = 2
Op het moment dat er een 0 bovenaan de matrix staat, wissel deze dan om zodat er (het liefst een
1) een ander getal links bovenin staat. DUS:
1. Eliminate 1e variabel in 2 van de 3 rijen – alleen 1 e rij overhouden
2. Eliminate 2e variabel in 2 van de 3 rijen – alleen 2 e rij overhouden
3. Eliminate 3e variabel in 2 van de 3 rijen – alleen 3 e rij overhouden

,1.2: row reduction and echelon forms
Two linear equations are called equivalent if they have the same solution set.
 Elementary row operations
1. Replacement: replace one row by the sum of itself and a multiple of another row
2. Interchange: interchange 2 rows
3. Scaling: multiply all entries in a row by a non-zero constant
- Row equivalent matrixes = als je de ene kan verbouwen/omschrijven tot de andere  row
operations are reversible  if the augmented matrixes of two linear systems are row
equivalent, then the two systems have the same solution set. SCHEMATISCH:
X X X X
O X X X
O O X X
O O O X
O O O O

Pivots ^^^^^^^^^^^^ also non-zero number
1 0 0 0 0
O 1 0 0 0
O O 1 0 0
O O O 1 0
O O O O 1


- A matrix is an echelon form if:
1. Non-zero rows above rows of all zeros
2. Each leading entry of a row is in a column tot the right of the leading entry of the row above it.
3. All entries in the column below a leading entry are zeros.

- A matrix is a reduced echelon form if:
1. Non-zero rows above rows of all zeros
2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
3. All entries in the column below a leading entry are zeros.
4. All leading entries are 1
5. Each leading entry is the only non-zero entry in its column.
Linear system  augmented matrix  (by row reduction) matrix in echelon form  (by row
reduction)
0 solutions: pivot in last column
1 solution: no pivot in last column
: pivots in all other columns
∞ solutions: no pivot in last column
: at least 1 other column without a pivot
Matrix in reduced echelon form (to determine solution set precisely)
- Theorem: every matrix is row equivalent to one and only one reduced echelon form. PROOF:
A matrix is row equivalent to many matrixes in echelon form but they have the same pivot position.

different pivot positions so it can’t be row equivalent.

,Week 1, les 2
1.3: vector equations
R2 has a set of points : (2,1)
R2 has a set of vectors : [21]  eind van de vector = coordinaat
- Vector space: (R2, +, x) kan je doen met vector, niet met coordinaat
Definition:
U = [ab] & v = [cd]
u + v = [a+c b+d]
(when R2 = vector space ^^^)
u = v if a=c en b=d
o = [00] in R2  R3 [000] etc. zero vector
e1 = [10] e2 = [01] in R2  unit vector
1 – [11]
- Algebraic properties of (Rn, +, x)
1. U+v=v+u
2. (u + v) + w = u +(v + w)
3. U+0=u
4. U + (-u) = 0
5. C (u + v) = cu + cv
6. (c+d)u = cu + du
7. C(du) = (cd)u
8. 1u = u
Given vectors v1 and v2, Vp e Rn we can make other vectors using addition and scalar
multiplication. We get vectors y of the form:
Y = c1v1 + … + cpvp for certain C’s
Y is called a linear combination of v1 … vp with weights c1…cp
- Definition
 Given vectors v1…vp
 Span {v1, v2, v3} = {c1v1 + cpvp}
All linear combinations
 Span {[2 1 ]} – all vectors on the line through 0 and [2 1 ]
 Span {[ 2 1 ] [ 1 2 ]} = R2
 R3 = span {e1, e2, e3}
 Span {o} = {o}
 Span o doorgehaald = {o}

, 1.4: the matrix equation
- Definition
Let A ben a mxn matrix (m = boven naar beneden). N columns from Rm
Let x =[x1 … xn]
Ax = [a1, a2, an] [x1, x2, xn]  same size!
Ax = b has a solution x b e span {a1, a2, an}  system with an augmented matrix [a1, a2, anb]
has a solution.
- Algebraic properties
1. A(u + v) = Au + Av
2. A(cu) = c (Au)
3. Ao = o
4. Ae1 = A [1 0] : a1
5. Ae2 = a2 etc
6. A1 = a1 + an = [row sum of row 1 of A , row sum of row n of A]
Under which condition does Ax = b have a solution for every b?  A has a pivot in every row 
no pivot in last column = solution
Ax = b has a solution for every b  b e span {a1, an}  span {a1, an} = Rm  [a1, a2, anb] 
echolon form
- Equivalent are (theorem)
1. Ax = b has a solution for every b
2. Span {a1, …, an} = Rm
3. A is row equivalent with an echelon form with a pivot in every row.