06-240/Classnotes For Tuesday October 10 - Revision history

Wongpak: /* Scan of Lecture Notes */

2006-10-25T02:07:19Z

Scan of Lecture Notes

← Older revision		Revision as of 22:07, 24 October 2006
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	* PDF file by [[User:Alla]]: [[Media:MAT_Lect009.pdf\|Week 5 Lecture 1 notes]]		* PDF file by [[User:Alla]]: [[Media:MAT_Lect009.pdf\|Week 5 Lecture 1 notes]]
			* PDF file by [[User:Gokmen]]: [[Media:06-240-lecture-10-october.pdf\|Week 5 Lecture 1 notes]]

	==A Quick Summary by {{Dror}}==		==A Quick Summary by {{Dror}}==

Adamgolding: /* A Quick Summary by {{Dror}} */

2006-10-22T08:34:39Z

A Quick Summary by {{Dror}}

← Older revision		Revision as of 04:34, 22 October 2006
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	'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.		'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.

	'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One used in expressing the newcomer in terms of the vectors already in ~~the set~~. Such a vector must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.		'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One used in expressing the newcomer in terms of the vectors already in <math>G</math>. Such a vector must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.

	'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".		'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".

Adamgolding: /* A Quick Summary by {{Dror}} */

2006-10-22T08:31:20Z

A Quick Summary by {{Dror}}

← Older revision		Revision as of 04:31, 22 October 2006
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	'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.		'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.

	'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One used in expressing the newcomer in terms of the vectors already in the set. Such ~~one~~ must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.		'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One used in expressing the newcomer in terms of the vectors already in the set. Such a vector must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.

	'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".		'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".

70.49.220.158: /* A Quick Summary by {{Dror}} */

2006-10-22T08:28:09Z

A Quick Summary by {{Dror}}

← Older revision		Revision as of 04:28, 22 October 2006
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	'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.		'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.

	'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One ~~that is~~ used in expressing the newcomer in terms of the ~~vector~~ ~~currently~~ ~~being~~ ~~inserted~~. Such one must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.		'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One used in expressing the newcomer in terms of the vectors already in the set. Such one must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.

	'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".		'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".

Alla: Change heading title

2006-10-13T00:27:30Z

Change heading title

← Older revision		Revision as of 20:27, 12 October 2006
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	{{06-240/Navigation}}		{{06-240/Navigation}}

	===~~Links~~ to ~~Classnotes~~===		====Scan of Lecture Notes====

	* PDF file by [[User:Alla]]: [[Media:MAT_Lect009.pdf\|Week 5 Lecture 1 notes]]		* PDF file by [[User:Alla]]: [[Media:MAT_Lect009.pdf\|Week 5 Lecture 1 notes]]

Alla: Posted Week 5 Lecture 1 notes

2006-10-13T00:21:36Z

Posted Week 5 Lecture 1 notes

← Older revision		Revision as of 20:21, 12 October 2006
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	{{06-240/Navigation}}		{{06-240/Navigation}}

			===Links to Classnotes===
			* PDF file by [[User:Alla]]: [[Media:MAT_Lect009.pdf\|Week 5 Lecture 1 notes]]

	==A Quick Summary by {{Dror}}==		==A Quick Summary by {{Dror}}==

Drorbn at 13:03, 12 October 2006

2006-10-12T13:03:43Z

← Older revision		Revision as of 09:03, 12 October 2006
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	==A Quick Summary by {{Dror}}==		==A Quick Summary by {{Dror}}==
	(Intentionally terse. A sea of details appears in the book and already appeared on the blackboard. But these are useless without some '''organizing principles'''; in some sense, "understanding" is precisely being able to see those principles ~~with~~ the sea of details. Yet don't fool yourself into thinking that the principles are enough even without the details!)		(Intentionally terse. A sea of details appears in the book and already appeared on the blackboard. But these are useless without some '''organizing principles'''; in some sense, "understanding" is precisely being able to see those principles within the sea of details. Yet don't fool yourself into thinking that the principles are enough even without the details!)

	'''Theorem.''' A finite generating set <math>G</math> has a subset which is a basis.		'''Theorem.''' A finite generating set <math>G</math> has a subset which is a basis.
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	'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.		'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>\|L\|\leq\|G\|</math> and you can replace <math>\|L\|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.

	'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One that is used in expressing the newcomer in terms of the vector currently ~~active~~. Such one must exist or else the newcomer is a linear combination of the elements of <math>L</math>, but <math>L</math> is linearly independent.		'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One that is used in expressing the newcomer in terms of the vector currently being inserted. Such one must exist or else the newcomer is a linear combination of some of the elements of <math>L</math>, but <math>L</math> is linearly independent.

	'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".		'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".

Drorbn: /* A Quick Summary by {{Dror}} */

2006-10-12T01:44:33Z

A Quick Summary by {{Dror}}

← Older revision		Revision as of 21:44, 11 October 2006
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	'''Proof Sketch.'''		'''Proof Sketch.'''
	# Find a basis within <math>G</math>; it has <math>n</math> elements.		# Find a basis within <math>G</math>; it has <math>n</math> elements.
	# Use replacement to place the elements of <math>L</math> ~~with~~ some ~~other~~ basis.		# Use replacement to place the elements of <math>L</math> within some basis.

Drorbn at 01:41, 12 October 2006

2006-10-12T01:41:29Z

New page

{{06-240/Navigation}}

==A Quick Summary by {{Dror}}==
(Intentionally terse. A sea of details appears in the book and already appeared on the blackboard. But these are useless without some '''organizing principles'''; in some sense, "understanding" is precisely being able to see those principles with the sea of details. Yet don't fool yourself into thinking that the principles are enough even without the details!)

'''Theorem.''' A finite generating set <math>G</math> has a subset which is a basis.

'''Proof Sketch.''' Grab more and more elements of <math>G</math> so long as they are linearly independent. When you can't any more, you have a basis.

'''Lemma.''' (The Replacement Lemma) If <math>G</math> generates and <math>L</math> is linearly independent, then <math>|L|\leq|G|</math> and you can replace <math>|L|</math> of the elements of <math>G</math> by the elements of <math>L</math>, and still have a generating set.

'''Proof Sketch.''' Insert the elements of <math>L</math> one by one, and for each one that comes in, take one out of <math>G</math>. Which one? One that is used in expressing the newcomer in terms of the vector currently active. Such one must exist or else the newcomer is a linear combination of the elements of <math>L</math>, but <math>L</math> is linearly independent.

'''Theorem.''' If a vector space <math>V</math> has a finite basis, all bases thereof are finite and have the same number of elements, the "dimension of <math>V</math>".

'''Proof Sketch.''' By replacement, <math>|\alpha|\leq|\beta|</math> and <math>|\beta|\leq|\alpha|</math>.

'''Theorem.''' Assume <math>\dim V=n</math>.
# If <math>G</math> generates, <math>|G|\geq n</math>. In case of equality, <math>G</math> is a basis.
# If <math>L</math> is linearly independent, <math>|L|\leq n</math>. In case of equality, <math>L</math> is a basis.

'''Proof Sketch.'''
# Find a basis within <math>G</math>; it has <math>n</math> elements.
# Use replacement to place the elements of <math>L</math> with some other basis.