linear algebra. session 2 - texas a&m universityroquesol/math_304_fall_2018... ·...

Matrices. Matrix Algebra

Linear Algebra. Session 2

Dr. Marco A Roque Sol

09/04/2018

Dr. Marco A Roque Sol Linear Algebra. Session 2


Matrices, matrix algebra

Applications of systems of linear equationsProblem 2.1.

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6




Applications of systems of linear equations

Problem 2.1.


x − y = −22x + 3y = 6






x − y = −22x + 3y = 6





Find the point

of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6





Find the point of intersection

of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6





Find the point of intersection of the lines x − y = −2 and

2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6





Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .

SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6





Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .Solution

The intersection point is the solution of the linear system{x − y = −2

2x + 3y = 6





Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point

is the solution of the linear system{x − y = −2

2x + 3y = 6





Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of

the linear system{x − y = −2

2x + 3y = 6





Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system

{x − y = −2

2x + 3y = 6






x − y = −22x + 3y = 6




Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.

Find a quadratic polynomial p(x) such that p(1) = 4, p(2) = 3,and p(3) = 4




Problem 2.2.

Find the point

of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection

of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection of the planes x − y = −2,

2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6

SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6Solution

The intersection point is the solution of the linear systemx − y = −2

2x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point

is the solution of the linear systemx − y = −2

2x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of

the linear systemx − y = −2

2x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.


x − y = −2

2x − y − z = 3x + y + z = 6

Problem 2.3.





Problem 2.2.


x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.





Solution

Suppose that p(x) = ax2 + bx + c , then

p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by the solution of the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Solution


p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c

are given by the solution of the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Solution


p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by

the solution of the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Solution


p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by the solution of

the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Solution


p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by the solution of the linear system

a + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Solution


p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by the solution of the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4




Problem 2.4.

Electrical network . Determine the amount of current in eachbranch of the network.




Problem 2.4.

Electrical network . Determine

the amount of current in eachbranch of the network.




Problem 2.4.

Electrical network . Determine the amount of current

in eachbranch of the network.




Problem 2.4.

Electrical network . Determine the amount of current in eachbranch of

the network.




Problem 2.4.

Electrical network . Determine the amount of current in eachbranch of the network.




Solution

To solve this problems, we will use three fundamental Laws comingfrom Physics, namely,

Kirchhofs law 1 ( Charge Conservation ):

At every node the sum of the incoming currents equals the sum ofthe outgoing currents.

Kirchhofs law 2 ( Energy Conservation ):

Around every loop the algebraic sum of all voltages is zero.

Ohm’s Law:

For every resistor the voltage drop E , the current i , and theresistance R satisfy E = iR




Solution

To solve this problems,

we will use three fundamental Laws comingfrom Physics, namely,





Ohm’s Law:





Solution

To solve this problems, we will use

three fundamental Laws comingfrom Physics, namely,





Ohm’s Law:





Solution

To solve this problems, we will use three fundamental Laws

comingfrom Physics, namely,





Ohm’s Law:





Solution






Ohm’s Law:





Solution



At every node

the sum of the incoming currents equals the sum ofthe outgoing currents.



Ohm’s Law:





Solution



At every node the sum of the incoming currents

equals the sum ofthe outgoing currents.



Ohm’s Law:





Solution



At every node the sum of the incoming currents equals the sum of

the outgoing currents.



Ohm’s Law:





Solution






Ohm’s Law:





Solution





Around every loop

the algebraic sum of all voltages is zero.

Ohm’s Law:





Solution





Around every loop the algebraic sum of all voltages

is zero.

Ohm’s Law:





Solution






Ohm’s Law:





Thus, applying these three laws to the above circuit we have

Node A: i1 + i2 = i3

Node B: i3 = i1 + i2

Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0




Thus,

applying these three laws to the above circuit we have



Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0




Thus, applying these three laws

to the above circuit we have



Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0




Thus, applying these three laws to the above circuit

we have



Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0







Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0






Node B:

i3 = i1 + i2

Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0







Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0







Left loop:

10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0







Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0







Left loop: 10− 10i1 − 40i3 = 0

Right loop:

20− 20i2 − 40i3 = 0







Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0




In this way we have the system:

i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine the amount of traffic between each ofthe four intersectionsof of the following diagram




In this way

we have the system:

i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.






i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.






i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine

the amount of traffic between each ofthe four intersectionsof of the following diagram





i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine the amount of traffic

between each ofthe four intersectionsof of the following diagram





i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine the amount of traffic between each of

the four intersectionsof of the following diagram





i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine the amount of traffic between each ofthe four intersectionsof

of the following diagram





i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.





650 400

610 −→ A x1 −→ B 640 −→

x4 x2

←− 520 D ←− x3 C ←− 600




Solution

At each intersection, the incoming traffic has to match theoutgoing traffic.

Intersection A : x4 + 610 = x1 + 450

Intersection B : x1 + 400 = x2 + 640

Intersection C : x2 + 600 = x3

Intersection D : x3 = x4 + 520

Which is equivalent to the system:x4 − x1 + 160 = 0x1 − x2 − 240 = 0x2 − x3 + 600 = 0x3 − x4 − 520 = 0




Solution






Which is equivalent to the system:

x4 − x1 + 160 = 0x1 − x2 − 240 = 0x2 − x3 + 600 = 0x3 − x4 − 520 = 0




Solution






Which is equivalent to the system:x4 − x1 + 160 = 0x1 − x2 − 240 = 0x2 − x3 + 600 = 0x3 − x4 − 520 = 0




Matrices

Let us start by solving an m × n system of linear equations

a11x1 + a12x2 + . . .+ a1nxn = b1a21x1 + a22x2 + . . .+ a2nxn = b2

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices

Let us start

by solving an m × n system of linear equations


......

am1x1 + am2x2 + . . .+ amnxn = bm


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices

Let us start by solving an m × n system

of linear equations


......

am1x1 + am2x2 + . . .+ amnxn = bm


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients,

b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, and

x ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns.

In this way, we can introduce new arrays ofnumbers to study the linear system

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers

to study the linear system

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




Matrices



......

am1x1 + am2x2 + . . .+ amnxn = bm


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

X =

x1x2...xm

B =

b1b2...bm




In this way, we have the following

Definition

An m× n matrix A , is an array of complex numbers ( m-rows andn-columns ),denoted by

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

In this context, an element in the i-row and j-column is of thematrix A denoted by aij .





Definition

An m× n matrix A ,

is an array of complex numbers ( m-rows andn-columns ),denoted by

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n






Definition

An m× n matrix A , is an array of complex numbers ( m-rows andn-columns ),

denoted by

A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n






Definition


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n






Definition


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

In this context,

an element in the i-row and j-column is of thematrix A denoted by aij .





Definition


A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n





An n-dimensional vector v, can be represented as a 1× n matrix(row vector) or as an n × 1matrix (column vector):

v =(x1, x2, x3, · · · , xn

)−→

(x1 x2 x3 · · · xn

)

v =(x1, x2, x3, · · · , xn

)−→

x1x2x3...xn




An m × n matrix A = (aij) can be regarded as a column ofn-dimensional row vectors or as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix

A = (aij) can be regarded as a column ofn-dimensional row vectors or as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix A = (aij)

can be regarded as a column ofn-dimensional row vectors or as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix A = (aij) can be regarded

as a column ofn-dimensional row vectors or as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix A = (aij) can be regarded as a column ofn-dimensional

row vectors or as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix A = (aij) can be regarded as a column ofn-dimensional row vectors or

as a row of m-dimensional columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




An m × n matrix A = (aij) can be regarded as a column ofn-dimensional row vectors or as a row of m-dimensional

columnvectors:

A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj





A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj





A =

v1

v2

v3...

vm

,

vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj





A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj





A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

),

wj =

a1ja2ja3j...

amj





A =

v1

v2

v3...

vm

, vi =(ai1 ai2 ai3 · · · ain

),

A =(

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

amj




Associated with any m × n matrix A, we have the following basicmatrices:

a) Transpose

Is the ( n ×m ) matrix, denoted by AT , and defined by

AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n

b) Complex Conjugate

Is the ( m × n ) matrix, denoted by A, and defined by




Associated with any

m × n matrix A, we have the following basicmatrices:

a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n






Associated with any m × n matrix A,

we have the following basicmatrices:

a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n







a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n







a) Transpose

Is the ( n ×m ) matrix, denoted by AT , and

defined by

AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n







a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n







a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n


Is the ( m × n ) matrix, denoted by A, and

defined by





a) Transpose


AT =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

)n×m

= (aji )m×n






A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n = (aij)m×n

c) Adjoint

Is the ( m × n ) matrix, denoted by A∗ = AT

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(a∗ij)n×m = (aji )m×n




A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn


c) Adjoint


, and

defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn





A =

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn


c) Adjoint


, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn





Basic Matrix Operations

Let A = (aij)m×n and B = (bij)m×n be two matrices, then wedefine

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition

A± B = (aij ± bij)m×n

e) Scalar Multiplication

rA = (raij)m×n






1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition

A± B =

(aij ± bij)m×n


rA = (raij)m×n






1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition



rA = (raij)m×n






1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition



rA =

(raij)m×n






1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition



rA = (raij)m×n




In particular we have the case of

Vector algebraLet

a =(a1, a2, a3, · · · , an

)and

b =(b1, b2, b3, · · · , bn

)be n-dimensional vectors, and r ∈ R

Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)





Vector algebra

Leta =

(a1, a2, a3, · · · , an

)and

b =(b1, b2, b3, · · · , bn


Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)





Vector algebraLet

a =(a1, a2, a3, · · · , an

)and

b =(b1, b2, b3, · · · , bn


Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)





Vector algebraLet

a =(a1, a2, a3, · · · , an

)and

b =(b1, b2, b3, · · · , bn


Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)

Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)





Vector algebraLet

a =(a1, a2, a3, · · · , an

)and

b =(b1, b2, b3, · · · , bn


Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)Dr. Marco A Roque Sol Linear Algebra. Session 2



Zero vector

r0 =(

0, 0, 0, · · · , 0)

Negative of a vector

−a =(−a1,−a2,−a3, · · · ,−an

)Vector difference

a− b =(a1 − b1, a2 − b2, a3 − b3, · · · , an − bn

)




Zero vector

r0 =(

0, 0, 0, · · · , 0)


−a =(−a1,−a2,−a3, · · · ,−an

)

Vector difference

a− b =(a1 − b1, a2 − b2, a3 − b3, · · · , an − bn

)




Zero vector

r0 =(

0, 0, 0, · · · , 0)


−a =(−a1,−a2,−a3, · · · ,−an

)Vector difference

a− b =(a1 − b1, a2 − b2, a3 − b3, · · · , an − bn

)




Given n-dimensional vectors, {v1, v2, v3, · · · , vk} and scalars{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk

is called a linear combination of vectors v1, v2, v3, · · · , vk.

Also, vector addition and scalar multiplication are called linearoperations

Definition. The dot product of n-dimensional vectors

x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn

The dot product is also called the scalar product .




Given

n-dimensional vectors, {v1, v2, v3, · · · , vk} and scalars{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk




x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn





Given n-dimensional vectors,

{v1, v2, v3, · · · , vk} and scalars{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk




x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn





Given n-dimensional vectors, {v1, v2, v3, · · · , vk} and

scalars{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk




x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn





Given n-dimensional vectors, {v1, v2, v3, · · · , vk} and scalars

{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk




x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn





Given n-dimensional vectors, {v1, v2, v3, · · · , vk} and scalars{r1, r2, r3, · · · , rk},

the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk




x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn

