14-240/Tutorial-November4: Difference between revisions

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:We that <math>W</math> is isomorphic to <math>P_{n - 1}</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 W. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)</math> where <math>f(x) \in P_{n - 1}</math>. Then show that <math>T</math> is one-to-one and onto.
:We 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 W. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)</math> where <math>f(x) \in P_{n - 1}</math>. Then show that <math>T</math> is one-to-one and onto.





Revision as of 16:22, 29 November 2014

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 that is isomorphic to . Let be the standard ordered basis of and be a subset of W. Then there is a unique linear transformation such that where . Then show that is one-to-one and onto.


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