Unformatted text preview: 1.3 Vector Equations Vectors have important mathematical and physical meanings, but for now, just consider them as lists of numbers. column vectors u, v ∈ R2 : u = 1 2 ,v = −3
1 8 u = v if and only if corresponding entries are equal u+v = −2
1 28 ← add corresponding entries 1 −3v = 1
1 − 24 1 ← multiply − 3 times each entry −2u + 4v =? Properties: u, v, w ∈ Rn , scalars c, d
(1) u + v = v + u (2) (u + v ) + w = u + (v + w ) (3) u + 0 = u, 0 has all entries 0 (4) u + (−u) = 0, −u = (−1)u (5) c(u + v ) = cu + cv (6) (c + d)u = cu + du (7) c(du) = (cd)u (8) 1 · u = u
1 Representing vectors interpretation. EX: For w = 4 4
y a b 3 1
v by (a, b), we get a geometrical = + 1 3 = u + v , we get
w u x Parallelogram Rule: 4th vertex of parallelogram is
u+v
x2 u+v u v 0 x1 2 Picture of vectors in R3 :
x3 2u u x1 x2 Note: u is a 3 × 1 matrix · represent u in R3 by a point (x, y , z ) = (1, 2, 3) · where is 2u, 3u, −u, −2u? Pictures in Rn?
u 1 u 2 · point (u1 , u2 , . . . un ) ↔ u = . . . un 3 EX: Sums of multiples of two vectors u = 2 1 and v = −3 1 3v 2v v 0 −2u + v −2u −u w 2u u 3u u−v −v −2v Describe the vertices of the intersections of the solid lines: au + bv for integers a, b (e.g., see vector −2u + 2v above) Q: By taking all combinations au + bv a, b ∈ R,
can we get all vectors w ∈ R2 ? A: Yes. As an example, for the vector w labelled above,
w= 3 2
4 u+v Deﬁnition:
Let v1 , v2, · · · , vp ∈ Rn . We say that y ∈ Rn is a linear combination of v1 , v2, · · · , vp if y = c1 v1 + c2 v2 + · · · + cp vp for some socalled weights c1 , c2 , · · · , cp EX: example above:
x2 w 2v v 0 u 2u x1 For w = 2u + 2v , the weights are 2, 2 Given another vector, say b = so that 10 −4 , ﬁnd weights x1 , x2 b = x1 u + x2 v Write x1 u + x2v = b, or
5 x1 2 1 2x1 x1 + x2 −3 1 = 10 −4 10 −4 10 , or + −3x2 x2 = , or 2x1 − 3x2 x1 + x2 2x1 − 3x2 = 10 x 1 + x 2 = −4 = −4 , or Therefore, solve system by taking augmented matrix 2 −3 1 10 , 1 −4 converting to RE form, and then ﬁnding weights x2 , x1 by back substitution. Note: Can solve this for any vector b = b1 b2 Now we go directly from this special case to the general situation: Given vectors v1 , v2, · · · , vp ∈ Rn , H = Span {v1 , v2, · · · , vp} is the set of all vectors which are linear combinations of v1, v2 , · · · , vp , so any vector w ∈ Rn in this set satisﬁes w = c1 v1 + c2 v2 + · · · + cp vp
6 EX: previous example: u = v1 = 2 and v = v2 = −3 1 and H = Span {u, v} 1 2 is all of R . EX: 1 1 0 2 ∈ Span 1 , 1 ? Q: Is 0 1 3 1 0 1 Solution: Yes, if c1 1 + c2 1 = 2 , so solve 0 1 3 system by considering augmented matrix for =1 c2 = 3 101 R2 ← R2 − R1 A=1 1 2 −→ 013 101 R3 ← R3 − R2 0 1 1 −→ 002 101 0 1 1 013 c1 c1 +c2 = 2 inconsistent, so NO. 7 EX: 1 0 1 Q: Does Span 1 , 1 0 = R3? 0 1 1 Solution: Yes, if 1 c1 1 + c2 1 + c3 0 = y 0 1 1 z is solvable x for all y ∈ R3. z 10 1 x 101x 0 1 −1 y − x 1 1 0 y −→ 01 1 z 011z 10 1 x 0 1 −1 y−x 00 2 z−y+x 0 1 x −→ consistent for all values of x, y , z , so YES. Remark: Notice that to answer the last question, we
don’t need the last column. We only need to check if every row has a leading entry (pivot position). 8 Observation about a change of notation: A vector eqn
x 1 a1 + x 2 a2 + · · · + x n an = b has the same soln set as the linear system whose augmented matrix is a1  a2  · · · an b . So b ∈ Span {a1 , a2, · · · , an} iﬀ linear system has a soln. EX: 1 1 Q: What is Span 1 , 0 0 −1 ? A: it is a plane in R3 . EX: Practical example (Problem 27 in book) In Production of goods, we have {#of days of production}×{output per day} = {total production} Problem Statement: A mining company has 2 mines.
Mine # 1 produces ore with 20 metric tons of copper, 550 kg of silver per day Mine # 2 produces or with 30 metric tons of copper, 500
9 kg of silver per day so output per day is v1 = 30 500 20 550 for mine #1, v2 = for mine #2. (a) What is the physical interpretation of 5v2? A: It is 5 days production in mine #2. (b) If mine #1 operates for x1 days and #2 operates for x2 days, write a vector eqn whose soln gives # of days needed to produce 150 metric tons of copper and 2825 kg of silver. A: x1v1 + x2 v2 = 150 2825 (c) Solve for x1 , x2 . Solution: 20 30 150 → 20 30 150 , so 550 500 2825 x2 = 4, x1 = 3 (check!) 2 0 −325 −1300 10 ...
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 Spring '10
 Russel
 Linear Algebra, Algebra, Equations, Vectors, Vector Space, #, ex

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