14-240/Tutorial-October28 - Revision history

Bug at 15:37, 29 November 2014

2014-11-29T15:37:06Z

← Older revision		Revision as of 11:37, 29 November 2014
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	==Boris==		==Boris==

	====~~Avoid~~ ~~the Einstellung Effect~~====		====Be Efficient====

	By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:		By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:

Bug: /* Boris */

2014-11-23T01:37:48Z

Boris

← Older revision		Revision as of 21:37, 22 November 2014
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	==Boris==		==Boris==

	====~~Try to~~ Avoid the Einstellung Effect====		====Avoid the Einstellung Effect====

	By this point in the course, we become good at solving systems of linear equations. However, we should not use this same		By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:

	old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:


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	Yet there is a less time-consuming approach that relies on two observations:		Yet there is a less time-consuming approach that relies on two observations:
		⚫	:(1) The dimension of <math>R^3</math> is <math>3</math> so the size of a basis is also <math>3</math>.

		⚫	:(2) No linearly independent set can have more vectors than a generating set (by the Replacement Theorem).
⚫	::(1) The dimension of <math>R^3</math> is <math>3</math> so the size of a basis is also <math>3</math>.

⚫	::(2) No linearly independent set can have more vectors than a generating set (by the Replacement Theorem).

	Since a basis is a generating set and the size of <math>S</math> is <math>4</math>, then the Replacement Theorem tells us that		Since a basis is a generating set and the size of <math>S</math> is <math>4</math>, then the Replacement Theorem tells us that
	<math>S</math> cannot be linearly		<math>S</math> cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.

	independent. Hence, the problem can be solved without solving any linear equations.


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	Once again, we can solve a linear equation but we do not have to. Observe:		Once again, we can solve a linear equation but we do not have to. Observe:
		⚫	:(1) The dimension <math>P_3(R)</math> is <math>4</math> so the size of a basis is also <math>4</math>.
		⚫	:(2) No generating set can have fewer vectors than a basis (by a Corollary to the Replacement Theorem).
		⚫	Since there are only <math>3</math> polynomials, then the Corollary tells us that it cannot generate <math>P_3(R)</math>. Once again, we used a more efficient strategy.

⚫	::(1) The dimension <math>P_3(R)</math> is <math>4</math> so the size of a basis is also <math>4</math>.

		⚫	====Extending a Linearly Independent Set to a Basis====
⚫	::(2) No generating set can have fewer vectors than a basis (by a Corollary to the Replacement Theorem).

			'''Boris's tip (for concrete sets and vector spaces only)''':
⚫	Since there are only <math>3</math> polynomials, then the Corollary tells us that it cannot generate <math>P_3(R)</math>. Once again, we used a

	more efficient strategy.



⚫	====Extending a Linearly Independent Set to a Basis====

	'''Consider the following strategy only if you are working with concrete sets and vector spaces:'''

	If a problem requires us to extend a linearly independent set to a basis, then the easiest approach is to add vectors from the		If a problem requires us to extend a linearly independent set to a basis, then the easiest approach is to add vectors from the

	standard ordered basis. Here is an example:		standard ordered basis. Here is an example:


		⚫	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis, add vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math>. The only question is which vector(s) should we add?

⚫	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>~~R^3~~</math>, add

	vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math>. The only question is which vectors should we add?


	We see that both vectors in <math>S</math> have a <math>0</math> as the third component so a safe choice is to add <math>(0, 0, 1)</math>. Since <math>R^3</math> has a

	dimension of <math>3</math>, then <math>\{(-3, -6, 0), (0, 7, 0), (0, 0, 1)\}</math> is a basis of <math>R^3</math>.		We see that both vectors in <math>S</math> have a <math>0</math> as the third component so a safe choice is to add <math>(0, 0, 1)</math>. Since <math>R^3</math> has a dimension of <math>3</math>, then <math>\{(-3, -6, 0), (0, 7, 0), (0, 0, 1)\}</math> is a basis of <math>R^3</math>.

	==Nikita==		==Nikita==

Bug at 01:16, 23 November 2014

2014-11-23T01:16:53Z

← Older revision		Revision as of 21:16, 22 November 2014
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	Since there are only <math>3</math> polynomials, then the Corollary tells us that it cannot generate <math>P_3(R)</math>. Once again, we used a		Since there are only <math>3</math> polynomials, then the Corollary tells us that it cannot generate <math>P_3(R)</math>. Once again, we used a

	more efficient strategy ~~to solve a problem~~.		more efficient strategy.

Bug: /* Extending a Linearly Independent Set to a Basis */

2014-11-09T18:14:46Z

Extending a Linearly Independent Set to a Basis

← Older revision		Revision as of 14:14, 9 November 2014
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	'''Consider the following strategy only if you are working with concrete sets and vector spaces:'''		'''Consider the following strategy only if you are working with concrete sets and vector spaces:'''

	If a problem requires us to extend a linearly independent ~~subset of a vector space~~ to a basis, then the easiest approach is to		If a problem requires us to extend a linearly independent set to a basis, then the easiest approach is to add vectors from the

	~~add vectors from the~~ standard ordered basis. Here is an example:		standard ordered basis. Here is an example:

Bug: /* Boris */

2014-11-09T18:13:55Z

Boris

← Older revision		Revision as of 14:13, 9 November 2014
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	more efficient strategy to solve a problem.		more efficient strategy to solve a problem.

⚫	====Extending a Linearly Independent Set to a Basis====


		⚫	====Extending a Linearly Independent Set to a Basis====

	'''Consider the following strategy only if you are working with concrete sets and vector spaces:'''		'''Consider the following strategy only if you are working with concrete sets and vector spaces:'''

Bug: /* Try to Avoid the Einstellung Effect */

2014-11-08T00:26:27Z

Try to Avoid the Einstellung Effect

← Older revision		Revision as of 20:26, 7 November 2014
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	'''Q2''': Determine if the polynomials <math>x^3 - 2x ^2 + 1, 4x^2-x+3, 3x-2 </math> generate <math>P_3(R)</math>.		'''Q2''': Determine if the polynomials <math>x^3 - 2x ^2 + 1, 4x^2-x+3, 3x-2 </math> generate <math>P_3(R)</math>.

	Once again, we can solve a ~~linear~~ linear equation but we do not have to. Observe:		Once again, we can solve a linear equation but we do not have to. Observe:

	::(1) The dimension <math>P_3(R)</math> is <math>4</math> so the size of a basis is also <math>4</math>.		::(1) The dimension <math>P_3(R)</math> is <math>4</math> so the size of a basis is also <math>4</math>.
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	more efficient strategy to solve a problem.		more efficient strategy to solve a problem.



	====Extending a Linearly Independent Set to a Basis====		====Extending a Linearly Independent Set to a Basis====

Bug: /* Extending a Linearly Independent Set to a Basis */

2014-11-08T00:25:22Z

Extending a Linearly Independent Set to a Basis

← Older revision		Revision as of 20:25, 7 November 2014
Line 58:		Line 58:

	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add		Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add


	vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> ~~without~~ ~~disturbing~~ ~~the~~ ~~linear~~ ~~independence~~ of ~~the~~ ~~set.~~ ~~The only question~~		vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math>. The only question is which vectors should we add?

	is which vectors should we add?

Bug: /* Extending a Linearly Independent Set to a Basis */

2014-11-08T00:24:43Z

Extending a Linearly Independent Set to a Basis

← Older revision		Revision as of 20:24, 7 November 2014
Line 59:		Line 59:
	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add		Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add

	vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> without disturbing the linear independence of the set. The only		vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> without disturbing the linear independence of the set. The only question

	~~question~~ is which vectors should we add?		is which vectors should we add?

Bug: /* Extending a Linearly Independent Set to a Basis */

2014-11-08T00:24:29Z

Extending a Linearly Independent Set to a Basis

← Older revision		Revision as of 20:24, 7 November 2014
Line 59:		Line 59:
	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add		Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add

	vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> ~~that~~ ~~fulfill~~ the ~~criteria~~ of ~~linear~~ ~~independence~~. The only ~~question is which~~		vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> without disturbing the linear independence of the set. The only

	vectors should we add?		question is which vectors should we add?

Bug: /* Extending a Linearly Independent Set to a Basis */

2014-11-08T00:23:39Z

Extending a Linearly Independent Set to a Basis

← Older revision		Revision as of 20:23, 7 November 2014
Line 59:		Line 59:
	Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add		Let <math>S = \{(-3, -6, 0), (0, 7, 0)\}</math> be a linearly independent subset of <math>R^3</math>. To extend <math>S</math> to a basis of <math>R^3</math>, add

	vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> that fulfill the criteria of linear independence. The only question which		vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math> that fulfill the criteria of linear independence. The only question is which

	vectors should we add?		vectors should we add?