14-240/Tutorial-November4: Difference between revisions
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'''Approach 1: Use Isomorphisms''' |
'''Approach 1: Use Isomorphisms''' |
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⚫ | Since <math>dim(P_{n - 1}(R)) = n</math>, then all we have to show is 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>. Let <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 one-to-one and onto to complete the proof. |
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Reminder: Finite dimensional vector spaces <math>V_1, V_2</math> over the same field are isomorphic to each other <math>\iff dim(V_1) = dim(V_2)</math>. |
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⚫ | Since <math>dim(P_{n - 1}(R)) = n</math>, then all we have to show is 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>. Let <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. |
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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 one-to-one and onto to complete the proof. |
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Revision as of 16:40, 29 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
Since , then all we have to show is that is isomorphic to . Let be the standard ordered basis of . Let be a subset of . Then there is a unique linear transformation such that where . Show that is one-to-one and onto to complete the proof.
Approach 2: Use the Rank-Nullity Theorem
Approach 3: Find a Basis with the Decomposed Polynomial
Approach 4: Find a Basis without the Decomposed Polynomial