06-240/Classnotes For Thursday, September 28: Difference between revisions
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''Example'' |
''Example'' |
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1. Let P<sub>3</sub>(<math>\Re</math>)={ax<sup>3</sup>+bx<sup>2</sup>+cx+d}<math>\subseteq</math>P(<math>\Re</math>), ''a'', ''b'', ''c'', ''d'', <math>\in \Re</math>. |
1. Let P<sub>3</sub>(<math>\Re</math>)={ax<sup>3</sup>+bx<sup>2</sup>+cx+d}<math>\subseteq</math>P(<math>\Re</math>), ''a'', ''b'', ''c'', ''d'', <math>\in \Re</math>.<BR> |
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''u''<sub>1</sub>=''x''<sup>3</sup>-2''x''<sup>2</sup>-5''x''-3 |
''u''<sub>1</sub>=''x''<sup>3</sup>-2''x''<sup>2</sup>-5''x''-3<BR> |
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''u''<sub>2</sub>=3''x''<sup>3</sup>-5''x''<sup>2</sup>-4''x''-9 |
''u''<sub>2</sub>=3''x''<sup>3</sup>-5''x''<sup>2</sup>-4''x''-9<BR> |
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''v''=2''x''<sup>3</sup>-2''x''<sup>2</sup>+12''x''-6 |
''v''=2''x''<sup>3</sup>-2''x''<sup>2</sup>+12''x''-6<BR> |
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Let W=spab(''u''<sub>1</sub>, ''u''<sub>2</sub>), <BR> |
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Does ''v'' <math> \in </math> W?<BR> |
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''v'' is in W if ''v''=''a''<sub>1</sub>''u''<sub>1</sub>+''a''<sub>1</sub>''u''<sub>2</sub><br> for some ''a''<sub>1</sub>, ''a''<sub>2</sub> <math> \in \Re </math>. |
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If <math>\exists</math> ''a''<sub>1</sub>, ''a''<sub>2</sub> <math>\in \Re</math>, <br> |
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|2''x''<sup>3</sup>-2''x''<sup>2</sup>+12''x''-6 |
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|= ''a''<sub>1</sub>(''x''-2''x''<sup>2</sup>-5''x''-3) + ''a''<sub>2</sub>(3''x''<sup>3</sup>-5''x''<sup>2</sup>-4''x''-9) |
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|=(''a''<sub>1</sub>+3''a''<sub>2</sub>)''x''<sup>3</sup> + (-2''a''<sub>1</sub> -5''a''<sub>2</sub>)''x''<sup>2</sup> + (-5''a''<sub>1</sub>-4''a''<sub>2</sub>)''x'' + (-3''a''<sub>1</sub>-9''a''<sub>2</sub>) |
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|<div align="right"><math>\Leftrightarrow</math>2</div> |
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|=''a''<sub>1</sub>+3''a''<sub>2</sub> |
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|<div align="right">-2</div> |
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|=-2''a''<sub>1</sub>-5''a''<sub>2</sub> |
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|<div align="right">12</div> |
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|=-5''a''<sub>1</sub>-4''a''<sub>2</sub> |
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|<div align="right">-6</div> |
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|=-3''a''<sub>1</sub>-9''a''<sub>2</sub> |
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Solve the four equations above and we will get ''a''<sub>1</sub>=-4 and ''a''<sub>2</sub>=2.<br> |
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Check if ''a''<sub>1</sub>=-4 and ''a''<sub>2</sub>=2 hold for all the 4 equations.<br> |
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Since it's hold, <math>\Rightarrow</math> ''v'' <math>\in</math> W. |
Revision as of 19:59, 29 September 2006
Linear Combination
Definition: Let (ui) = (u1, u2, ..., un) be a sequence of vectors in V. A sum of the form
- ai F, aiui = a1u1 + a2u2+ ... +anun
is called a "Linear Combination" of the ui.
Span
span(ui):= The set of all possible linear combinations of the ui's.
If V is any subset,
span | := The set of all linear combination of vectors in |
= |
even if is empty.
Theorem: For any V, span is a subspace of V.
Proof:
1. 0 span .
2. Let x span , Let x span ,
x = aiui, ui , y = bivi, vi .
x+y = aiui + bivi = ciwi where ci=(a1, a2,...,an, b1, b2,...,bm) and wi=ci=(u1, u2,...,un, v1, v2,...,vm).
3. cx= c aiui= (cai)ui span .
Example
1. Let P3()={ax3+bx2+cx+d}P(), a, b, c, d, .
u1=x3-2x2-5x-3
u2=3x3-5x2-4x-9
v=2x3-2x2+12x-6
Let W=spab(u1, u2),
Does v W?
v is in W if v=a1u1+a1u2
for some a1, a2 .
If a1, a2 ,
2x3-2x2+12x-6 | = a1(x-2x2-5x-3) + a2(3x3-5x2-4x-9) | |
=(a1+3a2)x3 + (-2a1 -5a2)x2 + (-5a1-4a2)x + (-3a1-9a2) | ||
2
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=a1+3a2 | |
-2
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=-2a1-5a2 | |
12
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=-5a1-4a2 | |
-6
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=-3a1-9a2 |
Solve the four equations above and we will get a1=-4 and a2=2.
Check if a1=-4 and a2=2 hold for all the 4 equations.
Since it's hold, v W.