14-240/Tutorial-November4: Difference between revisions
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We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> |
We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is both one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>. |
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'''Approach 2: Use the Rank-Nullity Theorem''' |
'''Approach 2: Use the Rank-Nullity Theorem''' |
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Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>. |
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'''Approach 3: Find a Basis with the Decomposed Polynomial''' |
'''Approach 3: Find a Basis with the Decomposed Polynomial''' |
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This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>. |
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'''Approach 4: Find a Basis without the Decomposed Polynomial''' |
'''Approach 4: Find a Basis without the Decomposed Polynomial''' |
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This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>. |
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====Cite Carefully==== |
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'''Boris's Section Only''' |
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If you use in your proof '''Corollary 1 of the Fundamental Theorem of Algebra''', then please cite it as "Corollary 1 of the Fundamental Theorem of Algebra". Do not cite it as the "Fundamental Theorem of Algebra" since that means you are citing the fundamental theorem instead of its corollary. |
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==Nikita== |
Latest revision as of 17:49, 30 November 2014
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Boris
Question 26 on Page 57 in Homework 5
Let and be a subspace of . Find .
First, let . Then we can decompose since there is a such that . From here, there are several approaches:
Approach 1: Use Isomorphisms
We show that is isomorphic to . Let be the standard ordered basis of and be a subset of . Then there is a unique linear transformation such that where . Show that is both one-to-one and onto and conclude that .
Approach 2: Use the Rank-Nullity Theorem
Let be the standard ordered basis of and . Then where and . Define by . Then it is easy to show that is both well-defined and linear. Afterwards, show that and use the rank-nullity theorem to conclude that .
Approach 3: Find a Basis with the Decomposed Polynomial
This approach is straightforward. Show that is a basis of .
Approach 4: Find a Basis without the Decomposed Polynomial
This approach requires a little more cleverness when constructing the basis: .
Cite Carefully
Boris's Section Only
If you use in your proof Corollary 1 of the Fundamental Theorem of Algebra, then please cite it as "Corollary 1 of the Fundamental Theorem of Algebra". Do not cite it as the "Fundamental Theorem of Algebra" since that means you are citing the fundamental theorem instead of its corollary.