User:Eml5526.s11.team04.premchand/HW2

= Problem 2.5 - Proving equivalence =

Problem Statement
Given that

$$\textbf{b}_i.\textbf{P}(\textbf{v})=0 \qquad \left \{ i=1,2...n \right.\left. \right \}$$ .....equation 1

$$\textbf{W}.\textbf{P}(\textbf{v})=0  \qquad \forall \textbf{W}\in \mathbb{R}^{nd}$$......equation 2

such that $$\textbf{W}= \sum \alpha _i\textbf{b}_i$$

Prove

equation 1=equation 2

Solution
Equation 1 can be rewritten as multiple equations

such that

$$\textbf{b}_1.\textbf{P}(\textbf{v})=0$$    .... equation a

$$\textbf{b}_2.\textbf{P}(\textbf{v})=0$$    .... equation b

$$\textbf{b}_3.\textbf{P}(\textbf{v})=0$$    .... equation c

upto

$$\textbf{b}_n.\textbf{P}(\textbf{v})=0$$    .... equation n

Here $$\textbf{b}_1,\textbf{b}_2,\textbf{b}_3..\textbf{b}_n$$ are basis vectors in the given coordinate.Each of the equations a,b,c...n can be considered as particular cases in the same coordinate.

From the given fact that $$\textbf{W}= \sum \alpha _i\textbf{b}_i$$

we can infer $$\textbf{W}= \alpha _1\textbf{b}_1+\alpha _2\textbf{b}_2+\alpha _3\textbf{b}_3+...\alpha _n\textbf{b}_n$$ where $$\alpha _1,\alpha _2,\alpha _3...\alpha _n$$ are components of the vector $$\textbf{W}$$

This implies that $$(\alpha _1\textbf{b}_1+\alpha _2\textbf{b}_2+\alpha _3\textbf{b}_3+...\alpha _n\textbf{b}_n).\textbf{P}(\textbf{v})=0$$

This is simply, another way of writing

$$\textbf{W}.\textbf{P}(\textbf{v})=0$$

Hence the equivalence is proved