Applied Linear Algebra: Vector Equations and Matrix Equations, Study Guides, Projects, Research of Linear Algebra

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Math 3191Applied Linear Algebra Lecture 3: Vector Equations^ Stephen BillupsUniversity of Colorado at Denver

Math 3191Applied Linear Algebra – p.1/

Announcements Visit course web page and register for the Discussionpage. Hwk 2, due Tuesday, Sept. 6 at beginning of class. Sec 1.3: 8,14,18,23,24,26,29,31,33. Sec 1.4: 6, 10, 14, 18, 20, 23,24, 26, 32, 34, 36. Sec 1.5: 4, 6, 12, 16, 20, 22.

Math 3191Applied Linear Algebra – p.2/

Back Substitution^232 4 −^10 6 12 2867670 0 −^2 0 7 1267 From row echelon form:^67670 0 0 0 1 /^2 1450 0 0 0 0 0 Use last nonzero row (row 3) to solve for last non-free variable (

Section 1.3: Vector Equations Key Concepts: Euclidean vectors (vectors in

n ).IR

Arithmetic operations on vectors. Linear Combinations. Vector Equations. Span.

Math 3191Applied Linear Algebra – p.5/

Notation Space saver notation: To save space, we will also represent a vector as justan ordered list of numbers: e.g.^ u^ = (1,^2 ,^ 3), instead of the more cumbersome 2316767 column notation u =. 2453 Note the use of parentheses instead of square brackets. Thus,^231 h^ i^6767 (1, 2 ,^ 3) =^6 =^.^21 2 345 3 n Subscripts: If u is a vector in^ , we denote the^ ith component ofIR

u^ by^ u. Fori

example, suppose^ u^ = (5,^4 ,^ 2). Then

u= 5, u= 4,^ and^ u= 2.^1 2 3 Math 3191Applied Linear Algebra – p.7/

Operations on Vectors n Vector addition : Given two vectors^ u, v^ ∈^ , their sumIR

u^ +^ v^ is obtained by adding corresponding entries of^ u^ and^ v. That is^ u^ +^ v^ =

23 u+^ v^1 16767 u+^ v^2 26767.^. 67. 67. 45 u+^ vn^ n Example:^ u^ = (1,^2 ,^ 3),^ v^ = (4,^5 ,^ 6) =⇒

u^ +^ v^ = (5,^7 ,^ 9). Scalar multiplication : Given a vector^ u^ ∈^ IR

n^ and a scalar (real number)^ c, the scalar multiple^ cu^ is obtained by multiplying each entry of

u^ by^ c. That is 23 cu (^16767) cu (^26767) cu =^.. 67. 67. 45 cun Example:^ u^ = (1,^2 ,^ 3),^ c^ =^ −2 =⇒^ cu^ = (

−^2 ,^ −^4 ,^ −6).^ Math 3191Applied Linear Algebra – p.8/

Adding Vectors Geometrically To compute^ u^ +^ v, draw^ v^ as an arrow with its tail touching the head of

u.^ u^ +^ v^ is the arrow starting at the tail of^ u^ and ending at the head of

v. v u^ Math 3191Applied Linear Algebra – p.10/

Adding Vectors Geometrically To compute^ u^ +^ v, draw^ v^ as an arrow with its tail touching the head of

u.^ u^ +^ v^ is the arrow starting at the tail of^ u^ and ending at the head of

v. u+v v u^ Math 3191Applied Linear Algebra – p.10/

Algebraic Properties Vector addition and scalar multiplication behave just like you’re usedto (from “grade school” addition and multiplication). That is, for alln^ u, v, w ∈ and scalars^ c^ and^ d:IR Commutativity:^ u^ +^ v^ =^ v^ +^ u. Associativity:^ (u^ +^ v) +^ w^ =^ u^ + (v^ +^ w) &

c(du) = (cd)u. Distributive properties:^ c(u^ +^ v) =

cu^ +^ cv^ & (c^ +^ d)u^ =^ cu^ +^ du. Zero vector. There exists a special vector (denoted by

0 ) such that u^ + 0 = 0 +^ u^ =^ u.^ Question: what is the zero vector? Additive inverse. Let^ −u^ denote the vector

(−1)u. Then u^ + (−u) =^ −u^ +^ u^ = 0.^ (Notation: Henceforth,

u^ −^ v^ will be used as a shortcut to denote^ u^ + (−v)

). Multiplicative identity.^1 u^ =^ u.

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Linear Combinations Given vectors^ v,... vand scalars^ c,... , c^1 p^1

, the vector^ yp

defined by^ y^ =^ c

v+^ · · ·^ cv 11 pp

is called a^ linear combination

of^ v,... , vwith weights^1 p^

c,... , c.^1 p Examples:1. Let^ v= (−^1 ,^ 1)^ and^ v= (2^1

,^ 1). The vector^ u^ = (0,^ 3)^ is a

linear combination of^ vand^1

v. In particular^ u^ = 2v+^21

v.^2

2. The vector^ u^ = (1,^1 ,^ 1)^ is

not^ a linear combination of thevectors v= (1, 0 , 0) and v= (0,^1 ,^ 0).^ Why not? 1 2

Question: How would you determine whether a vector

u^ is

a linear combination of given vectors

v,... , v? And if it is,^1 p

what are the weights?

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Vector Equations To determine if u^ is a linear combination of vectors v,... , v, we need to find weights^ x,... , x 1 p^1

such thatp xv+... xv= u. 11 pp

This is called a^ vector equation

. It is equivalent to solving the

system of linear equations^ vx+^ vx^111

+^ · · ·^ +^ vx=^ u 1 pp^1 vx+ vx+^ · · ·^ +^ vx=^ u 211 222 2 pp^2 · · · vx+ vx+^ · · ·^ +^ vx=^ um 11 m 22 mpp^ m

where^ v= (v, v,... , vi^1 i^2 imi

).^ Math 3191Applied Linear Algebra – p.15/

• Outline Back substitution. Vector equations. Matrix equations. Math 3191Applied Linear Algebra – p.3/
• x):^51 / 2 x= 1 =⇒ x= 2.
• Plug this value into row above (row 2): −^2 x+ 7(^23 |{z} x
• ) = 12 =⇒ x= 1.
• Plug xand xinto next row up:^3 5 2 x+ 4x− 10(^1 ) + 6x+ 12(^1 2 4 |{z} x
• 2 ) = 28 =⇒ x= −^2 x−^3 x+ 7.^1 2 4 |{z} x^5 Math 3191Applied Linear Algebra – p.4/
• Exercise Consider the vectors u = (1, 2) and v = (2, −3). 1. Calculate u + v algebraically.2. Calculate u + v graphically. Math 3191Applied Linear Algebra – p.11/
• Example Let v= (0, 4 , −1), v= (1,^6 , 3), v= (5, −^1 , −8). Determine 1 2 3 if u = (− 3 , 9 , −3) can be written as a linear combination of v, vand v.
• Math 3191Applied Linear Algebra – p.16/
• Example Let v= (0, 4 , −1), v= (1,^6 , 3), v= (5, −^1 , −8). Determine 1 2 3 if u = (− 3 , 9 , −3) can be written as a linear combination of v, vand v. 12 3 Solution: We need to solve xv+ xv+ xv= u. This is equivalent to solving 11 22 33 the linear system x+ 5x= − 32 3 4 x+ 6x− x= 91 2 3 −x+ 3x− 8 x= −
• Math 3191Applied Linear Algebra – p.16/
• Example Let v= (0, 4 , −1), v= (1,^6 , 3), v= (5, −^1 , −8). Determine 1 2 3 if u = (− 3 , 9 , −3) can be written as a linear combination of v, vand v. 12 3 Solution: We need to solve xv+ xv+ xv= u. This is equivalent to solving 11 22 33 the linear system^230 1 5 −^3 x+ 5x= − 32 3 67 OR^67^4 6 −^1 945 4 x+ 6x− x= 91 2 3 −^1 3 −^8 −^3 −x+ 3x− 8 x= −
• Math 3191Applied Linear Algebra – p.16/