14-240/Tutorial-November4

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1	Sep 8	About This Class, What is this class about? (PDF, HTML), Monday, Wednesday
2	Sep 15	HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf
3	Sep 22	HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf
4	Sep 29	HW3, Wednesday, Tutorial, HW3_solutions.pdf
5	Oct 6	HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf
6	Oct 13	No Monday class (Thanksgiving), Wednesday, Tutorial
7	Oct 20	HW5, Term Test at tutorials on Tuesday, Wednesday
8	Oct 27	HW6, Monday, Why LinAlg?, Wednesday, Tutorial
9	Nov 3	Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial
10	Nov 10	HW8, Monday, Tutorial
11	Nov 17	Monday-Tuesday is UofT November break
12	Nov 24	HW9
13	Dec 1	Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial
F	Dec 8	The Final Exam
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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 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 $f(x)=\displaystyle \sum _{i=1}^{n}c_{i}g_{i}(x)$ where $c_{i}\in R$ and $g_{i}(x)\in K$ . Define $T:P_{n}(R)\to R$ by $T(\displaystyle \sum _{i=1}^{n}c_{i}g_{i}(x))=\displaystyle \sum _{i=1}^{n}c_{i}g_{i}(a)$ . 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 $dim(W)=n$ .