Download Revisions in Linear Algebra

Revisions on Linear Algebra. I
Revisions on Linear Algebra. II
Denition:
A linear function f is a mathematical function in which the variables appear
only in the rst degree, are multiplied by constants, and are combined only
by addition and subtraction. A linear equation is of the form f (x, y, · · · ) = 0
with f linear.
Solving Linear equations:
Find x such that:
Find (x, y) such that:
Tell me if those following functions or equations are linear. If not, do you
know how they are called ?
B.
A.
1
2
1
f (x, y) = 3xy
2
2
x4 + 3x2 − 4 = 0
−4
3
3
x
4
3x + y − 2 = 0
f (x) = 3x − 4
f (x) = 3x − 4
3
f (x) =
4
f (x, y) = 3x + y − 2
3
x
1
Write both equations in the form ax + by = k
2
Make the coefcients of one of the variables the same in both equations.
3
Add or substract (depending on signs) on equation from the other to
form a new equation in one variable.
Denition:
A matrix is a rectangular table of (real or complex) numbers. The order of a
matrix gives the number of rows and columns it has. A matrix with m rows
and n columns has order m × n.
Write down the order of the following matrices:
1
4
Solve the new equation obtained in step 3.
5
Put the value obtained in step 4 into one of the given equations to nd
the corresponding value of the other variables.
2
3
Revisions on Linear Algebra. V
is not square, but B =
1
−1
2
5
Denition:
The transpose of a matrix M , written M T is obtained by interchanging the
rows and columns. For example,
If M =
−1
5
3
−2
!
−3
!
1
−1
2
5
−5
8
2
A=
B=
C=
!
Revisions on Linear Algebra. VI
Denition:
A square matrix has the same number!of rows and columns (order of the
!
2
x − 2y − 3 = 0
3x + y = −1
Revisions on Linear Algebra. IV
Simultaneous linear equations are solved with the following steps
−3
−4=0
Revisions on Linear Algebra. III
form (m × m)). For instance, A =
3x = 4.
, then M T =
−1
3
5
−2
!
is.
Compute
1
2
(M T )T
MT
with M =
−1
5
3
−2
!
with
M=
8
2
4
3
1
2
!
Revisions on Linear Algebra. VII
Revisions on Linear Algebra. VIII
Denitions:
Example
Matrices can only be added, or substracted, if they are of the same
order. Simply add or substract corresponding elements. This gives a
matrix of the same order.
To multiply a matrix by a scalar (a number), multiply each element of
the matrix by the number.
Example
8
2
4
3
1
2
!
2
=
16
4
8
6
2
4
!
3
2
−1
6
2
4
2
With A =
1
A+B
2
A + CT
3
3B
−3
8
4
!
,B=
0
!
5
+
−1
2
5
1
−1 5
!
2 4
0
+
1
2
!
1
6
and C =
!
Matrix addition is commutative. Let A and B be two matrices of the
same order then:
A+B=B+A
Matrix addition is associative. Let A, B and c be three matrices of the
same order then:
−5
1
2
8
4
,
8
, compute (if you can):
Example
1

0

1


 4

1
 5
6
8
7
9
0


9

2
 = 5

3
4
13
17
7
9
6
8

5

3

2
Revisions on Linear Algebra. XII
The determinant of a 2 × 2 matrix A =
2
9
is not dened.
Denitions:
!
1
Memory aid: Row by Column

Revisions on Linear Algebra. XI
−5
6
!
matrix equals the number of rows in the second.
1
1
0
Multiplication of Matrices:
Multiply each row of the rst matrix by each column of the second matrix.
Two matrices can only be multiplied if the number of columns in the rst
A + (B + C) = (A + B) + C
2
4
Revisions on Linear Algebra. X
Denitions:
A=
8
.
Revisions on Linear Algebra. IX
Let
=

3

B=
 6
2

−8
4
0
5
1
9
0

2 

0
Find AB and BA, if possible.
Is the multiplication of matrices commutative?
ad − bc = det(A).
a
c
b
d
!
is the number
If det(A) = 0, A is called a singular matrix.
Compute the determinant of the matrices:
1
2
A=
B=
1
2
3
4
!
3
1
−2
4
.
!
Denition:
A square matrix with 1s on the diagonal and 0s elsewhere is called the
identity
matrix and denoted by I .
Revisions on Linear Algebra. XIV
Revisions on Linear Algebra. XIII
Denition:
If A is a square matrix, and there exists a matrix B so that
If A is invertible, we denote the inverse of A by A−1 , so
A−1 A = AA−1 = I.
AB = BA = I,
plays the rôle of the reciprocal or multiplicative inverse in ordinary
multiplication, since
A−1
we say
that A is invertible, and B is called an inverse of A. It can be shown
that an invertible matrix has exactly one inverse, so we refer to the inverse
of an invertible matrix.
aa−1 = a−1 a = 1,
for nonzero a ∈ R.
Show that if A is not square, it cannot have an inverse.
Revisions on Linear Algebra. XV
If A =
a
c
b
d
!
Revisions on Linear Algebra. XVI
then
A−1 =
1
det(A)
d
−c
−b
a
!
1
Exchange the elements of the diagonal.
2
Change the signs of the elements of the other diagonal.
3
Multiply by
1
det(A)
.
Prove the above result (i.e.
equal to I ).
compute AA−1 and A−1 A, and check
Revisions on Linear Algebra. XVII
If M =
1
2
3
3
5
2
3
!
Let M =
and N =
−1
3
−1
2
!
,
Compute (MN)−1 .
Compute N −1 M−1
Any comment on (MN)−1 and N −1 M −1 ?
1
−2
3
1
!
and A =
7
10
21
23
!
.
1
Compute M−1 A.
2
If MB = 2M + A, express B in matrix form.
they are
Revisions on Linear Algebra. XVIII
Lets come back to solving a system of linear equations:
x − 2y − 3 = 0
3x + y = −1
Solve this system by matrix methods.
Revisions on Linear Algebra. XIX
Revisions on Linear Algebra. XX
Denitions:
A vector (or column vector) is a matrix with a single column. A matrix with
a single row is a row vector. The entries of a vector are its components.
Solve
Solving simultaneous equations of the form:
2x + 3y = 4
10x + 4y = 9
Ax = b
with A a square matrix and x unknown vector and b vector of constants, is
given by
x = A−1 b
when A is invertible.
Revisions on Linear Algebra. XXI
Revisions on Linear Algebra. XXII
Solutions of linear equations in two variables correspond to intersection
points of lines.
Three ways in which lines can intersect:
Three possibilities for solutions of LS in two unknowns
(1) A unique solution
(0) No solution
(∞) Innitely many solutions
We call a linear system with no solutions
INCONSISTENT.
Solve:
1
2
Intersect at unique point
Parallel lines - no intersection point
Lines coincide
3
x+y=1
2x − y = 2
2x − y = −1
2x − y = 0
y−x=2
3y − 3x = 6
Revisions on Linear Algebra. XXIV
Revisions on Linear Algebra. XXIII
Denitions:
A scalar λ is called an eigenvalue of a square matrix A if it satises:
Ax = λx
with x a vector non null. x is then called an eigenvector. The eigenvalue λ is
found by solving the characteristic equation:
det(A − λI) = 0
Find the eigenvalues of
A=
2
0
1
3
!
Find the eigenvectors corresponding to λ = 2 and λ = 3.