Linear Algebra. Session 4 - Texas A&M Universityroquesol/Math_304_Fall_2019_Session… · Dr. Marco A Roque Sol Linear Algebra. Session 4. Determinants II Applications Vector Spaces

Determinants IIApplications

Vector Spaces

Linear Algebra. Session 4

Dr. Marco A Roque Sol

09/12/2019

Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Determinants II

Example 4.1

Find the determinant of the following matrix 1 2 34 5 67 8 9

Solution

Expansion by the 1st row

1 ∗ ∗∗ 5 6∗ 8 9

∗ 2 ∗4 ∗ 67 ∗ 9

∗ ∗ 34 5 ∗7 8 ∗



Vector Spaces

Determinants II

1 ∗ ∗∗ 5 6∗ 8 9

∗ 2 ∗4 ∗ 67 ∗ 9

∗ ∗ 34 5 ∗7 8 ∗

det(A) = (1)

∣∣∣∣ 5 68 9

∣∣∣∣+ (−2)

∣∣∣∣ 4 67 9

∣∣∣∣+ (3)

∣∣∣∣ 4 57 8

∣∣∣∣ =

(5 · 9− 6 · 8)− 2(4 · 9− 6 · 7) + 3(4 · 8− 5 · 7) = 0

Example 4.2

Find the determinant of the following matrix

A =

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1



Vector Spaces

Determinants II

Solution

Using elementary operations, we have

|A| =

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102



Vector Spaces

Determinants II

Example 4.3


A =

2 3 3 10 4 3 −32 −1 −1 −30 −4 −3 2

Solution

|A| =

∣∣∣∣∣∣∣∣2 3 3 10 4 3 −32 −1 −1 −30 −4 −3 2

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣2 3 3 12 −1 −1 −30 4 3 −30 −4 −3 2

∣∣∣∣∣∣∣∣ =



Vector Spaces

Determinants II

−

∣∣∣∣∣∣∣∣2 3 3 12 −1 −1 −30 4 3 −30 −4 −3 2

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣2 3 3 10 −4 −4 −40 4 3 −30 −4 −3 2

∣∣∣∣∣∣∣∣ =

−

∣∣∣∣∣∣∣∣2 3 3 10 −4 −4 −40 4 3 −30 0 0 −1

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣2 3 3 10 −4 −4 −40 0 −1 70 0 0 −1

∣∣∣∣∣∣∣∣ =

= (−2)(−4)(−1)(−1) = 8



Vector Spaces

Determinants II

Example 4.4


A =

1 4 4 10 1 −2 23 3 1 40 1 −3 2

Solution

|A| =

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 23 3 1 40 1 −3 2

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 20 −9 −11 10 1 −3 2

∣∣∣∣∣∣∣∣ =



Vector Spaces

Determinants II

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 20 −9 −11 10 1 −3 2

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 20 0 −29 190 0 −1 0

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 20 0 −29 190 0 29 0

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 4 4 10 1 −2 20 0 −29 190 0 0 19

∣∣∣∣∣∣∣∣ =

= (1)(1)(−29)(19) = −(29)(19)



Vector Spaces

Determinants II

Example 4.5


A =

1 0 0 30 1 −2 0−2 3 −2 30 −3 3 3

Solution

|A| =

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 0−2 3 −2 30 −3 3 3

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 3 −2 90 −3 3 3

∣∣∣∣∣∣∣∣ =



Vector Spaces

Determinants II

|A| =

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 3 −2 90 −3 3 3

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 0 4 90 0 −3 3

∣∣∣∣∣∣∣∣ =

|A| =

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 0 4 90 0 −3 3

∣∣∣∣∣∣∣∣ = 4

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 0 1 9/40 0 −3 3

∣∣∣∣∣∣∣∣ =

|A| = 4

∣∣∣∣∣∣∣∣1 0 0 30 1 −2 00 0 1 9/40 0 0 39/4

∣∣∣∣∣∣∣∣ =

= (4)(1)(1)(1)(39/4) = 39Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Determinants II

Example 4.6


C =

∣∣∣∣∣∣∣∣2 2 0 3−5 3 2 11 −1 0 −32 0 0 1

∣∣∣∣∣∣∣∣Solution

Expand the determinant by the 3rd column:

|C| =

∣∣∣∣∣∣∣∣2 2 0 3−5 3 2 11 −1 0 −32 0 0 1

∣∣∣∣∣∣∣∣ = −2

∣∣∣∣∣∣2 2 31 −1 −32 0 1

∣∣∣∣∣∣ =



Vector Spaces

Determinants II

Add 2 times the 2nd row to the 1st row

|C| = −2

∣∣∣∣∣∣2 2 31 −1 −32 0 1

∣∣∣∣∣∣ = −2

∣∣∣∣∣∣4 0 −31 −1 −32 0 1

∣∣∣∣∣∣ =

Expand the determinant by the 2nd column

|C| = −2

∣∣∣∣∣∣4 0 −31 −1 −32 0 1

∣∣∣∣∣∣ = (−2)(−1)

∣∣∣∣4 −32 1

∣∣∣∣ = (2)(10) = 20



Vector Spaces

Determinants II

Example 4.7

For what values of a will the following system have a uniquesolution?

x +2y +z = 1−x +4y +2z = 22x −2y +az = 3

Solution

The system has a unique solution if and only if the coefficientmatrix is invertible.

A =

1 2 1−1 4 22 −2 a

To this end, let’s find det(A).Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Determinants II

det(A) =

∣∣∣∣∣∣1 2 1−1 4 22 −2 a

∣∣∣∣∣∣Add −2 times the 3rd column to the 2nd column:

det(A) =

∣∣∣∣∣∣1 2 1−1 4 22 −2 a

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 1−1 0 22 −2− 2a a

∣∣∣∣∣∣Expand the determinant by the 2nd column:

det(A) =

∣∣∣∣∣∣1 0 1−1 0 22 −2− 2a a

∣∣∣∣∣∣ =



Vector Spaces

Determinants II

∣∣∣∣∣∣1 0 1−1 0 22 −2− 2a a

∣∣∣∣∣∣ = (−1)(−2− 2a)

∣∣∣∣ 1 1−1 2

∣∣∣∣ =

−(−2− 2a)3 = 6(1 + a)

Thus A is invertible if and only if a 6= −1.



Vector Spaces

Determinants II

Example 4.8

Given the matrices

X =

−1 2 10 2 −20 0 −3

Y =

1 0 0−1 3 02 −2 1

Find the Following determinants

Solution

1) det(X ) = (−1)(2)(3) = 6 det(Y ) = det(Y T ) = 3

2) det(XY ) = (6)(3) = 18 det(YX ) = (3)(6) = 18

3) det(Y−1) = 1/3 det(XY−1) = 6/3 = 2



Vector Spaces

Determinants II

4) det(XYX−1) = det(Y ) = 3 det(X−1Y−1XY ) = 1

5) det(2X ) = 23det(X ) = 23(6) = 48

6) det(−3XTXY−4) = (−3)3(6)(6)(3)−4 = −12



Vector Spaces

Determinants II

Let us try to find a solution of a general system of 2 linearequations in 2 variables

{a11x1 + a12x2 = b1a21x1 + a22x2 = b2

Solve the 1st equation for

x = (b1−a12)ya11

Substitute into the 2nd equation

a21(b1−a12)ya11

+ a22y = b2

Solve for y

y = a11b2−a21b1a11a22−a12a21



Vector Spaces

Determinants II

Back substitution:

x = (b1−a12y)a11

= a22b1−a12b2a11a22−a12a21

Thus,

x =

∣∣∣∣ b1 a12b2 a22

∣∣∣∣∣∣∣∣ a11 a12a21 a22

∣∣∣∣ y =

∣∣∣∣ a11 b1a21 b2

∣∣∣∣∣∣∣∣ a11 a12a21 a22

∣∣∣∣



Vector Spaces

Determinants II

Cramer’s Rule

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

...an1x1 + an2x2 + · · ·+ annxn = bn

⇒ Ax = b

Theorem

Assume that the matrix A is invertible. Then the only solution ofthe system is given by



Vector Spaces

Determinants II

xi =detAi

detA, i = 1, 2, ..., n.

where the matrix Ai is obtained by substituting the vector b for theith column of A

Determinants and the inverse matrix

Definition

Let A be an n × n matrix. The cofactor matrix, denoted byCof(A) = (αij), is an n × n matrix, defined by

Cof(A)ij = (−1)i+jMij



Vector Spaces

Determinants II

where Mij is the ij-minor of the Matrix A.

Theorem

Cof(A)TA = ACof(A)T = (detA)I

Corollary

If det(A) 6= 0 then A−1 = (detA)−1Cof(A)T



Vector Spaces

Determinants II

Definition

Let A be an n × n matrix. The adjoint matrix, denoted byAdj(A), is an n × n matrix, defined by

Adj(A) = [Cof(A)]T

In this way we have that for a nonsingular matrix A, then A−1 isgiven by

A−1 =1

det(A)Adj(A)



Vector Spaces

Determinants II

Approach 3 (Original)

Definition

A permutation π of n elements is a one-to-one and onto functionhaving the set {1, 2, ..., n} as both its domain and image.

In other words, a permutation is a function

π : {1, 2, ..., n} → {1, 2, ..., n}such that for every integer i ∈ {1, 2, ..., n} there exists exactly oneinteger j ∈ {1, 2, ..., n} for which π(i) = j

Definition

The set of all permutations of n elements is denoted by Sn and iscommonly called the Symmetric Group



Vector Spaces

Determinants II

Given a permutation π ∈ Sn there are several common notationsused for specifying how π permutes the integers {1, 2, ..., n}, oneof those notations is given by

Definition

Given a permutation π ∈ Sn denote πi = π(i) for eachi ∈ {1, 2, ..., n}. Then the two-line notation for π is given by the2× n matrix

π =

(1 2 · · · nπ1 π2 · · · πn

)DefinitionLet π ∈ Sn be a permutation. Then an inversion pair (i , j) of π is apair of positive integers i , j ∈ {1, ..., n} for which i < j butπ(i) > π(j).



Vector Spaces

Determinants II

Example . We classify all inversion pairs for elements in S3:

id =

(1 2 31 2 3

)Has no inversion pairs since no elements are “out of order”

π =

(1 2 31 3 2

)Has the single pair (2, 3) since π(2) = 3 > 2 = π(3).



Vector Spaces

Determinants II

Definition

Let π ∈ Sn be a permutation . Then the sign of π, denoted bysign(π), is defined by

sign(π) = (−1)α{

1 if the number of inversions in π is even−1 if the number of inversions in π is odd

α = #of inversion pairs in π

Moreover, we call π an even permutation if sign(π) = 1 and wecall π an odd permutation if sign(π) = −1



Vector Spaces

Determinants II

Theorem

The number of elements in the symmetric group Sn is given by

|Sn| = n(n − 1)(n − 2) · · · 1 = n!

Definition

Let n be a positive integer and A be an n × n matrix with reals orcomplex entries. Then

det(A) =∑π∈Sn

sign(π)aπ1,1, aπ2,2, · · · , aπn,n,



Vector Spaces

Applications

Processing ImageDr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Applications

A Simple Model for Marital Status.

In a certain town, 30 percent of the married women get divorcedeach year and 20 percent of the single women get married eachyear. There are 8000 married women and 2000 single women.Assuming that the total population of women remains constant,how many married women and how many single women will therebe after 1 year?

Solution

Form a matrix A as follows:

1) The entries, in the first row of A will be the percent of marriedand single women,respectively, that are married after 1 year.



Vector Spaces

Applications

2) The entries in the second row will be the percent of women whoare single after 1 year.

Thus, we obtain the matrix

A =

(0.70 0.200.30 0.80

)If we let

x0 =

(80002000

)

the number of married and single women after 1 year can becomputed by multiplying A times x0.



Vector Spaces

Applications

x1 = Ax0 =

(0.70 0.200.30 0.80

)(80002000

)=

(60004000

)After 1 year there will be 6000 married women and 4000 singlewomen.

Excersise 1. Find the number of married and single women after 2years.

Excersise 2. Find the number of married and single women after 3years.



Vector Spaces

Applications

Excersise 3. Write a formula (that is, an equation or expression)for how to calculate the number of single and married women aftern years. (Do not attempt an explicit calculation, just write downthe formula.)

This is a classical problem of “population dynamics” we canapproach this example in a more advanced setting, namely,Markov Chains



Vector Spaces

Applications

Production Costs.

A company manufactures three products. Its production expensesare divided into three categories. In each category, an estimate isgiven for the cost of producing a single item of each product. Anestimate is also made of the amount of each product to beproduced per quarter. These estimates are given in Tables 1 and 2:

ProductExpenses A B C

Raw Materials 0.10 0.30 0.15Labor 0.30 0.40 0.25

Overhead 0.10 0.20 0.15

Table 1. Production Costs Per Item (dollars)



Vector Spaces

Applications

SeassonProduct Summer Fall Winter Spring

A 4000 4500 4500 4000B 2000 2600 2400 2200C 5800 6200 6000 6000

Table 2. Amount Produced Per Quarter

The company would like to present at their stockholders’ meetinga single table showing the total costs for each quarter in each ofthe three categories: raw materials, labor, and overhead.



Vector Spaces

Applications

Solution

The costs for the summer quarter, can be obtained from theTables:

Raw materials: (0.10)(4000) + (0.30)(2000) + (0.15)(5800) = 1870

Labor:(0.30)(4000) + (0.40)(2000) + (0.25)(5800) = 3450

Overhead:(0.10)(4000) + (0.20)(2000) + (0.15)(5800) = 1670



Vector Spaces

Applications

The second column will represent the costs for the fall quarter:

Raw materials: (0.10)(4500) + (0.30)(2600) + (0.15)(6200) = 2160

Labor:(0.30)(4500) + (0.40)(2600) + (0.25)(6200) = 3940

Overhead:(0.10)(4500) + (0.20)(2600) + (0.15)(6200) = 1900

Following in this way, we can calculate the costs for the winter :

Raw materials: = 2070

Labor: = 3810

Overhead: = 1830Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Applications

and spring quarter:

Raw materials: = 1960

Labor: = 3580

Overhead: = 1740

Now, consider the tables, given at the very beginning, asrepresented by the matrices

M =

0.10 0.30 0.150.30 0.40 0.250.10 0.20 0.15

P =

4000 4500 4500 40002000 2600 2400 22005800 6200 6000 6000



Vector Spaces

Applications

If we form the product MP, then every column represents the costsfor the summer, fall, winter, and spring quarters

MP =

1870 2160 2070 19603450 3940 3810 35801670 1900 1830 1740

On the other hand, the entries in the first row of MP represent thetotal cost of raw materials for each of the four quarters. Theentries in rows 2 and 3 represent the total cost for labor andoverhead, respectively, for each of the four quarters. The yearlyexpenses in each of the columns may be added to obtain the totalproduction costs for each quarter. Table 3 summarizes the totalproduction costs.



Vector Spaces

Applications

SeassonSummer Fall Winter Spring

Raw Materials 1870 2160 2070 1960Labor 3450 3940 3810 3580

Overhead 1670 1900 1830 1740

Total Costs 6990 8000 7710 29980

Exercise 1 . A toy manufacturer makes toy airplanes, boats, andcars. Each toy is fabricated in a factory F1 in Taiwan and thenassembled in factory F2 in the U.S. The total cost of each productconsists of the manufacturing cost and the shipping cost. Then thecosts at each factory (in $ US) can be described as



Vector Spaces

Applications

Manuf. Ship.Costs Costs

F1 =

0.32 0.400.50 0.800.70 0.20

AirplanesBoatsCars

F2 =

0.40 0.600.50 0.501.30 0.20

AirplanesBoatsCars

Find a matrix that gives the total manufacturing and shippingcosts for each product.

( Warning: you need to use some practical reasoning here; do notmimic the previous example exactly !!!! )



Vector Spaces

Applications

Economic Models for Exchange of Goods.

Suppose that in a primitive society, the members of a communityare engaged in three occupations: farming, manufacturing of toolsand utensils, and the weaving and sewing of clothing.

Assume that initially the community has no monetary system andthat all goods and services are bartered.

Let the three groups be denoted by F ,M, and C , respectively.Suppose that this directed graph indicates how the barteringsystem works in practice:



Vector Spaces

Applications

This figure indicates that the farmers keep half of their produceand give one quarter of their produce to the manufacturers and onquarter to the clothing producers. The manufacturers divide thegoods evenly among three groups, one third goes to each(including themselves). The group producing clothes gives half ofthe clothes to the farmers and divides the other half evenlybetween themselves and the manufacturers. The results aresummarized in the above table:



Vector Spaces

Applications

The first column of the table indicates the distribution of thegoods produced by the farmers, the second column indicates thedistribution of the manufactured goods, and the third columnindicates the distribution of the clothing.

As the size of the community grows, the system of barteringbecomes too cumbersome and, consequently, the communitydecides to institute a monetary system of exchange. For thissimple economic system, we assume that there will be noaccumulation of capital or debt and that the prices for each of thethree types of goods will reflect the values of the existing barteringsystem. The question is how to assign values to the three types ofgoods that fairly represent the current bartering system.



Vector Spaces

Applications

Solution

This problem can be turned into a linear system using a modelthat was originally developed by the Nobel prize winning economistWassily Leontief.

For now, let x1 be the monetary value of the goods produced bythe farmers, x2 be the value of the manufactured goods, and x3 bethe value of the clothing produced.

According to the first row of the table, the value of the goodsreceived by the farmers amounts to half the value of the farmgoods produced, plus one-third the value of the manufacturedproducts, and half the value of the clothing goods. Thus the totalvalue of the goods received by the farmers is 1

2x1 + 13x2 + 1

2x3. Ifthe system is fair, the total value of goods received by the farmersshould equal x1 the total value of the farm goods produced.



Vector Spaces

Applications

Thus, we have the linear equation

1

2x1 +

1

3x2 +

1

2x3 = x1

Using the second row of the table and equating the value of thegoods produced and received by the manufacturers, we obtain asecond equation

1

4x1 +

1

3x2 +

1

4x3 = x2

Finally, the third row of the table yields:

1

4x1 +

1

3x2 +

1

4x3 = x3



Vector Spaces

Applications

These equations can be rewritten as a homogeneous system:12x1 + 1

3x2 + 12x3 = x1

14x1 + 1

3x2 + 14x3 = x2

14x1 + 1

3x2 + 14x3 = x3

The reduced row-echelon form of the augmented matrix for thissystem is 1 0 −5

3 00 −1 −1 00 0 0 0

We can find a parametric representation for the solution set of thissystem by letting x3 = t. Then, we immediately getx1 = 5

3 t, x2 = −t and x3 = t by back-substitution. It follows thatthe variables x1, x2, x3 should be assigned values in the ratio.

x1 : x2 : x3 = 5 : 3 : 3



Vector Spaces

ApplicationsConsider the parallel system with modern currency systems; itdoesn’t matter what the actual units of money are, just that onecan find a fair system of exchange. This simple system is anexample of the closed Leontief input-output model. Leontief’smodels are fundamental to our understanding of economic systems.

Excersise 1 Determine the relative values of x1, x2, x3 if thedistribution of goods is as described in this table:

F M C

Raw Materials 13

13

13

Labor 13

12

16

Overhead 13

16

12



Vector Spaces

Applications

Chemistry. Application 1

It takes three different ingredients A,B, and C , to produce acertain chemical substance. A,B, and C have to be dissolved inwater separately before they interact to form the chemical.Suppose that the solution containing A at 1.5g/cm3 combinedwith the solution containing A at 3.6g/cm3 combined with thesolution containing C at 5.3g/cm3 makes 25.07g of the chemical.If the proportion for A,B,C in these solutions are changed to2.5, 4.3, and 2.4g/cm3, respectively (while the volumes remain thesame), then 22.36g of the chemical is produced. Finally, if theproportions are 2.7, 5.5, and 3.2g/cm3, respectively, then 28.14 gof the chemical is produced. What are the volumes (in cubiccentimeters) of the solutions containing A,B, and C



Vector Spaces

Applications

Solution

Let x, y, z be the corresponding volumes (in cubic centimeters) ofthe solutions containing A,B, and C . Then 1.5x is the mass of Ain the first case, 3.6y is the mass of B, and 5.3z is the mass of C .Added together, the three masses should give 25.07g . So

1.5x + 3.6y + 5.3z = 25.07

The same reasoning applies to the other two cases. This gives thelinear system

1.5x + 3.6y + 5.3z = 25.072.5x + 4.3y + 2.4z = 22.362.7x + 5.5y + 3.2z = 28.14



Vector Spaces

Applications

The augmented matrix of this system is 1.5 3.6 5.3 25.072.5 4.3 2.4 22.362.7 5.5 3.2 28.14

The reduced row-echelon form of the augmented matrix for thissystem is 1 0 0 1.5

0 1 0 3.10 0 1 2.2

and the solution is

x = 1.5, y = 3.2, z = 2.2



Vector Spaces

Applications

Chemistry. Application 2

Another typical application of linear systems to chemistry isbalancing a chemical equation. The rationale behind this is theLaw of conservation of mass, which states the following:

“mass is neither created nor destroyed in any chemical reaction.Therefore balancing of equations requires the same number ofatoms on both sides of a chemical reaction. The mass of all thereactants (the substances going into a reaction) must equal themass of the products (the substances produced by the reaction).”

As an example consider the following chemical equation

C2H6 + O2 → CO2 + H2O

Balance this chemical reaction.Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces

Applications

Solution

Balancing this chemical reaction means finding values of x , y , zand t so that the number of atoms of each element is the same onboth sides of the equation:

xC2H6 + yO2 → zCO2 + tH2OThis gives the following linear system:

2x = z6x = 2t

2y = 2z + t

The augmented matrix of this system is 2 0 −1 0 06 0 0 −2 00 2 −2 −1 0



Vector Spaces

Applications

The reduced row-echelon form of the augmented matrix for thissystem is 1 0 0 −12/36 0

0 1 0 −7/6 00 0 1 −4/6 0

x =

2

6α, y =

7

6α, z =

4

6α, t = α

Since we are looking for whole values of the variables x , y , z , andt, choose t = 6 and get x = 2, y = 7, and z = 4. The balancedequation is then:

2C2H6 + 7O2 → 4CO2 + 6H2O



Vector Spaces

IntroductionSubspaces

Vector Spaces

Vector Spaces

Subspaces of a Vector Space (R3)



Vector Spaces


Vector Spaces

Addition and Scalar multiplication on vectors

Let x = (x1, x2, · · · , xn) and y = (y1, y2, · · · , yn) be n-dimensionalvectors, and r ∈ R be a scalar

Vector Sum: x + y = (x1 + y1, x2 + y2, · · · , xn + yn)

Scalar Multiplication: rx = (rx1, rx2, · · · , rxn)

Zero Vector: 0 = (0, 0, · · · , 0)

Negative of a vector: −x = (−x1,−x2, · · · ,−xn)

Vector Difference: x− y = (x1 − y1, x2 − y2, · · · , xn − yn)



Vector Spaces


Vector Spaces

Properties of Addition and Scalar multiplication on vectors

x + y = y + x

(x + y) + z = x + (y) + z)

x + 0 = 0 + x = x

x + (−x) = (−x) + x = 0

r(x + y) = ry + rx

(r + s)x = rx + sx

(rs)x = r(sx)

1x = x

0x = 0

(−1)x = −xDr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces


Vector Spaces

Let A = (aij) and B = (bij) be m × n matrices, and r ∈ R be ascalar

Addition and Scalar multiplication on matrices

Matrix Sum: A + B = (aij + bij)

Scalar Multiplication: rA = (raij)

Zero Matrix: All entries are zer aij = 0 for 1 ≤ i ≤ m, 1 ≤ j ≤ nj

Negative of a Matrix: −A = −aij for 1 ≤ i ≤ m, 1 ≤ j ≤ nj

Matrix Difference: A− B = aij − bij for 1 ≤ i ≤ m, 1 ≤ j ≤ nj

Note

The m × n matrices have the same properties as mn-dimensionalvectors ... !!!!



Vector Spaces


Vector Spaces

Vector space: An introduction

As we saw in the previous two examples, we can conclude thatthey share the same algebraic properties, even though, they arecompletely different entities.

Thus, we can say that both form part of a more general structurethat contains them.

This bigger structure that we are talking about here is going to becalled a Vector Space. Inside of this concept we will have somany different sets sharing the same algebraic propertis for the twobasic operations defined on it



Vector Spaces


Vector Spaces

Roughly speaking, a Vector Space is a structured set of elementscalled vectors together with two operations vector addition andmultiplication by a scalar (real numbers). These operations ( andtheir rules ) must produce vectors inside of the set and mustsatisfy some properties.



Vector Spaces


Vector Spaces

Vector Space

Definition

A vector Space is a set V equipped with two operationsα : V × V → V and µ : R× V → V that satisfy certain properties

The operation α is called addition. For any u, v, the elementα(u, v) is denoted by u + v

The operation µ is called scalar multiplication . For any u ∈ V andany r ∈ R, the element µ(r ,u) is denoted by ru



Vector Spaces


Vector Spaces

Properties of Addition and Scalar Multiplication for a VectorSpace V

A1 x + y = y + x for all x, y ∈ V

A2 (x + y) + z = x + (y + z) for all x, y, and z ∈ V

A3 There exists an element of V, called zero and denoted by 0such that x + 0 = 0 + x = x for all x ∈ V

A4 For any x ∈ V there exists an element of V denoted by −xsuch that x + (−x) = (−x) + x = 0

A5 r(x + y) = rx + ry for all x, y ∈ V and r ∈ R

A6 (r + s)x = rx + sx for all x ∈ V and r , s ∈ R

A7 (rs)x = r(sx) for all x ∈ V and r , s ∈ R

A8 1x = x for all x ∈ VDr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces


Vector Spaces

Consequences of Axioms

1) Associativity of addition implies that a multiple sumu1 + u2 + · · ·+ un is well defined for any u1,u2, · · · ,un

2) Subtraction in V is defined as follows:

x− y = x + (−y)

3) Addition and scalar multiplication are called linear operations.Given u1,u2, · · · ,uk ∈ V and r1, r2, · · · , rk

r1u1 + r2u2 + · · ·+ rkun

is called a linear combination of u1,u2, · · · ,unDr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces


Vector Spaces

Examples of Vector Spaces

Rn : n-dimensional coordinate vectors.

Mm×n : m × n matrices with real entries.

R∞ : infinite sequences (x1, x2, · · · ), xi ∈ RFor any x = (x1, x2, · · · ), y = (y1, y2, · · · ) ∈ R∞ and r ∈ Rx + y = (x1 + y1, x2 + y2, · · · )rx = (rx1, rx2, · · · )Then 0 = (0, 0, · · · ) and −0 = (0, 0, · · · , ){0}: The trivial Vector Space0 + 0 = 0 r0 = 0



Vector Spaces


Vector Spaces

Functional Vector Spaces

F (R) : the set of all functions f : R→ R.Given functions f , g ,∈ F (R) and a scalar r ∈ R, let(f + g)(x) = f (x) + g(x) and (rf )(x) = rf (x) for all x ∈ RZero vector: o(x) = 0. Negative (−f )(x) = −f (x).

C (R) : all continuous functions f : R→ R.Linear operations are inherited from F (R). We only need tocheck that f , g ,∈ C (R)⇒ f + g , rf ∈ C (R),the zero function is continuous, and f ∈ C (R)⇒ −f ∈ C (R)

C 1(R) : all continuously differentiable functions f : R→ R.



Vector Spaces


Vector Spaces

Functional Vector Spaces

C∞(R) : all smooth ( infinitely differentiable ) functionsf : R→ R.

P : all polynomials p(x) = a0 + a1x1 + a2x

2 + · · ·+ akxk

V : {x : x is formal Power Series around 0 }.Let x =

∑∞n=1 anx

n, y =∑∞

n=1 bnxn, and r ∈ R, then

x + y :=∑∞

n=1(an + bn)xn rx :=∑∞

n=1(ran)xn

The zero vector is given by x = 0 =∑∞

n=1 anxn an = 0 for

all n



Vector Spaces


Vector Spaces

Special Example 1

Consider the set V = R+ with the nonstandard, addition (⊕) andscalar multiplication (�):

x⊕ y = xy for any x, y ∈ R+

r � x = xr for any x ∈ R+ and r ∈ R

A1. x⊕ y = y ⊕ x⇔ xy = yx

A2. (x⊕ y)⊕ z = x⊕ (y ⊕ z)⇔ (xy)z = x(yz)

A3. x⊕ η = η ⊕ x = x⇔ xη = ηx = x ( holds forη = 1 [η = 0])



Vector Spaces


Vector Spaces

A4. x⊕ θ = θ ⊕ x = 0 = η ⇔ xθ = θx = 1 ( holds forθ = x−1 [θ = −x])

A5. r � (x⊕ y) = (r � x)⊕ (r � x)⇔ (xy)r = xryr

A6. (r + s)� x = (r � x)⊕ (s � x)⇔ xr+s = xrxs

A7. (rs)� x = r � (s � x)⇔ xrs = (xs)r

A8. 1� (x) = x⇔ x1 = x

All the properties are satisfied therefore, this example respresent aVector Space ... !!!!



Vector Spaces


Vector Spaces

Special Example 2

V : { The set of all symbols of the form

∣∣∣∣ ab∣∣∣∣}, where a can be

any real number but b must be a positive number.

Define the nonstandard addition and scalar multiplication:

∣∣∣∣ a1b1∣∣∣∣⊕ ∣∣∣∣ a2b2

∣∣∣∣ =

∣∣∣∣ a1 + b1a2 · b2

∣∣∣∣r �

∣∣∣∣ ab∣∣∣∣ =

∣∣∣∣ rabr

∣∣∣∣As we can check that the set with the two operations abovedefined, {V;⊕,�}, form a Vector Space.



Vector Spaces


Vector Spaces

Special Example 3

V = R

Define the nonstandard addition and scalar multiplication:

x ⊕ y = x + y + 7 for any x , y ∈ R

r � x = rx + 7(r − 1)for any x ∈ R and r ∈ R

As we can check that the set with the two operations abovedefined, {V;⊕,�}, form a Vector Space.



Vector Spaces


Vector Spaces

Counterexample 1.

Consider the set V = Rn with the standard addition and anonstandard scalar multiplication:

r � x = x for any x ∈ Rn and r ∈ R

Properties A1− A4 hold because they do not involve scalarmultiplication.

A5. r � (x + y) = r � x + r � x⇔ x + y = x + y

A6. (r + s)� x = r � x + s � x⇔ x = x + x

A7. (rs)� x = r � (s � x)⇔ x = x

A8. 1� (x) = x⇔ x = x

The only property that fails is A6.Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces


Vector Spaces

Counterexample 2

Consider the set V = Rn with the standard addition and anonstandard scalar multiplication:

r � x = 0 for any x ∈ Rn and r ∈ RProperties A1− A4 hold because they do not involve scalarmultiplication.

A5. r � (x + y) = r � x + r � x⇔ 0 = 0 + 0

A6. (r + s)� x = r � x + s � x⇔ 0 = 0 + 0

A7. (rs)� x = r � (s � x)⇔ 0 = 0

A8. 1� (x) = x⇔ 0 = x

The only property that fails is A8.Dr. Marco A Roque Sol Linear Algebra. Session 4


Vector Spaces


Vector Spaces

Some general observations

The zero vector is unique.Suppose z1 and z2 are zero vectors.Then z1 + z2 = z2 sincez1 is a zero vector and z1 + z2 = z1 sincez2 is a zero vector.Hence z1 = z2.

For any x ∈ V, the negative −x is unique.Suppose y and y′ are both negatives of x.Let us compute the sum y′ + x + y in two ways:

(y′ + x) + y = 0 + yy′ + (x + y) = y′ + 0By associativity of the vector addition, y′ = y .



Vector Spaces


Vector Spaces

Some general observations

(cancellation law) x + y = x′ + y implies x = x′ for anyx, x′ ∈ V. If x + y = x′ + y then

(x + y) + (−y) = (x′ + y) + (−y)By associativity, (x + y) + (−y) = x + (y + (−y)) =x+ 0 = x and (x′+ y) + (−y) = x′+ (y + (−y)) = x′+ 0 = x′

Hence x = x′

0x = 0 for any x ∈ V.

Indeed, 0x + x = 0x + 1x = (0 + 1)x = 1x = x = 0 + x.By the cancellation law, 0x = 0.



Vector Spaces


Vector Spaces

(−1)x = −x for any x ∈ V.

Indeed, x + (−1)x = (−1)x + x =(−1)x + (1)x = (−1 + 1)x = (0)x = 0

therefore

(−1)x = −x



Vector Spaces


Subspaces

Subspaces of vector spaces

Definition.If V0, is any non-empty subset of V such that

1) For each u and v are in V0, u⊕ v is in V0. (In this case we sayV0 is closed under vector addition.)

2) For each r ∈ R and v ∈ V0, r � u is in V0. (In this case we sayV0 is closed under vector scalar multiplication.)

then V0 is called a Subspace of V0.



Vector Spaces


Subspaces

Examples

C (R) : is a Subspace of F (R) :

Pn : Polynomials of degree less than n is a subset of P

P∗n : Polynomials of degree n is not a subset of P[ −xn + (xn + 1) = 1] /∈ P∗n ⇒ P∗n is not closed under vectoraddition ).

R+ is not a subset of R

The line x − y = 0 is a subspace of V = R2



Vector Spaces


Subspaces

The parabola y − x2 = 0 is a subspace of V = R2

The plane z = 0 is a subspace of V = R3

The plane z = 1 is not a subspace of V = R3

The line t(1, 1, 0), t ∈ R is a subspace of V = R3 and asubspace of the plane z = 0.

The line (1, 1, 1) + t(1,−1, 0), t ∈ R is not a subspace ofV = R3 as it lies in the plane x + y + z = 3, which does notcontains 0.

In general, a straight line or a plane in R3 is a subspace if andonly if it passes through the origin.


Linear Algebra. Session 4 - Texas A&M Universityroquesol/Math_304_Fall_2019_Session… · Dr. Marco A Roque Sol Linear Algebra. Session 4. Determinants II Applications Vector Spaces

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