14-240/Tutorial-October28: Difference between revisions
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==Boris== |
==Boris== |
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====Be Efficient==== |
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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: |
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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: |
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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 |
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<math>S</math> cannot be linearly |
<math>S</math> cannot be linearly independent. Hence, the problem can be solved without solving any linear equations. |
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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: |
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'''Boris's tip (for concrete sets and vector spaces only)''': |
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more efficient strategy. |
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'''Consider the following strategy only if you are working with concrete sets and vector spaces:''' |
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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 |
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standard ordered basis. Here is an example: |
standard ordered basis. Here is an example: |
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vectors from <math>\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}</math>. The only question is which vectors should we add? |
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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 |
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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>. |
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==Nikita== |
==Nikita== |
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Latest revision as of 10:37, 29 November 2014
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Boris
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:
Q1: Determine if [math]\displaystyle{ S = \{(1, 4, -6), (1, 5, 8), (2, 1, 1), (0, 1, 0)\} }[/math] is linearly independent in [math]\displaystyle{ R^3 }[/math].
We can solve this linear equation to find the answer:
- [math]\displaystyle{ c_1(1, 4, -6) + c_2(1, 5, 8) + c_3 (2, 1, 1) + c_4(0, 1, 0) = (0, 0, 0) }[/math] where [math]\displaystyle{ c_i \in R }[/math].
Yet there is a less time-consuming approach that relies on two observations:
- (1) The dimension of [math]\displaystyle{ R^3 }[/math] is [math]\displaystyle{ 3 }[/math] so the size of a basis is also [math]\displaystyle{ 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]\displaystyle{ S }[/math] is [math]\displaystyle{ 4 }[/math], then the Replacement Theorem tells us that [math]\displaystyle{ S }[/math] cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.
Q2: Determine if the polynomials [math]\displaystyle{ x^3 - 2x ^2 + 1, 4x^2-x+3, 3x-2 }[/math] generate [math]\displaystyle{ P_3(R) }[/math].
Once again, we can solve a linear equation but we do not have to. Observe:
- (1) The dimension [math]\displaystyle{ P_3(R) }[/math] is [math]\displaystyle{ 4 }[/math] so the size of a basis is also [math]\displaystyle{ 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]\displaystyle{ 3 }[/math] polynomials, then the Corollary tells us that it cannot generate [math]\displaystyle{ P_3(R) }[/math]. Once again, we used a more efficient strategy.
Extending a Linearly Independent Set to a Basis
Boris's tip (for concrete sets and vector spaces only):
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:
Let [math]\displaystyle{ S = \{(-3, -6, 0), (0, 7, 0)\} }[/math] be a linearly independent subset of [math]\displaystyle{ R^3 }[/math]. To extend [math]\displaystyle{ S }[/math] to a basis, add vectors from [math]\displaystyle{ \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\} }[/math]. The only question is which vector(s) should we add?
We see that both vectors in [math]\displaystyle{ S }[/math] have a [math]\displaystyle{ 0 }[/math] as the third component so a safe choice is to add [math]\displaystyle{ (0, 0, 1) }[/math]. Since [math]\displaystyle{ R^3 }[/math] has a dimension of [math]\displaystyle{ 3 }[/math], then [math]\displaystyle{ \{(-3, -6, 0), (0, 7, 0), (0, 0, 1)\} }[/math] is a basis of [math]\displaystyle{ R^3 }[/math].
