14-240/Tutorial-November4

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Boris

Question 26 on Page 57 in Homework 5

Let a \in R and W = \{f \in P_n(R): f(a) = 0\} be a subspace of P_n(R). Find dim(W).


First, let f(x) \in W. Then we can decompose f(x) since there is a g(x) \in P_{n - 1}(R) such that f(x) = (x - a)g(x). From here, there are several approaches:


Approach 1: Use Isomorphisms


We show that W is isomorphic to P_{n - 1}(R). Let B = \{1, x, x^2, ..., x^{n - 1}\} be the standard ordered basis of P_{n - 1}(R) and S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\} be a subset of W. Then there is a unique linear transformation T:P_{n - 1} \to W such that T(f(x)) = (x - a)f(x) where f(x) \in B. Show that T is both one-to-one and onto and conclude that dim(P_{n - 1}) = dim(W).


Approach 2: Use the Rank-Nullity Theorem


Let K = \{1, x, x^2, ..., x^{n - 1}, x^n\} be the standard ordered basis of P_n and f(x) \in P_{n}(R). Then f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x) where c_i \in R and g_i(x) \in K. Define T: P_{n}(R) \to R by T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a). Then it is easy to show that T is both well-defined and linear. Afterwards, show that rank(T) = 1 and use the rank-nullity theorem to conclude that dim(W) = n.


Approach 3: Find a Basis with the Decomposed Polynomial


This approach is straightforward. Show that S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\} is a basis of W.


Approach 4: Find a Basis without the Decomposed Polynomial


This approach requires a little more cleverness when constructing the basis: S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}.


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 Do not cite it as the "Fundamental Theorem of Algebra" since that means you are citing the fundamental theorem instead of its corollary.

Nikita