User:Eml5526.s11.team3.ushnish12/Homework 2

=--Problem 2.3--=

It is given that $$\underline{\omega} = \sum_{i=1}^{n} \beta _{i}\underline{a_{i}}$$, where {$$ \underline{a_{i}} $$} is an orthonormal basis vector.

Then (1) p. 8-1 is equivalent to: $$ \underline{w} * {P} (\underline{v}) = 0 $$ $$ \forall $$ $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{i} $$ $$\in \mathbb{R}^n$$

We can also write

$$ \underline{w} *\underline {P} (\underline{v}) = 0$$ $$ \forall \left[ \beta_{1}, ..., \beta_{n} \right] \in \mathbb{R}^n$$

This can be represented as:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \underline{\omega} = \beta _{1}\underline{a_1}+\beta _{2}\underline{a_2}+\beta _{3}\underline{a_3}+....\beta _{n}\underline{a_n}$$
 * 
 * 

$$\displaystyle (Eq. 1) $$

Choice 1
 * }
 * }

Let $$\beta_1 = 1$$ and  $$\beta _2 = \beta _3=...\beta _n=0$$

Therefore, equation (1) can be written as-

$$\underline{a_1}.\underline{P}(\underline{v})=0$$

Choice 2

Let $$\beta_2 = 1$$ and  $$\beta _1 = \beta _3=...\beta _n=0$$

Therefore, equation (1) can be written as-

$$\underline{a_2}.\underline{P}(\underline{v})=0$$

.

.

Choice n

Let $$\beta_n = 1$$ and  $$\beta _1 = \beta _2=...\beta _{n-1}=0$$

Therefore, equation (1) can be written as-

$$\underline{a_n}.\underline{P}(\underline{v})=0$$

Hence we can state that :

$$ \underline{a}_{i} * \underline{P} (\underline{v}) = 0 $$        for $$ \forall_{i} = 1, ..., n $$

=--Problem 2.5--=

Problem Statement
To show that

$$ \underline{b}_{i} * \underline{P} (\underline{v}) = 0 $$ where i=1, ..., n

is equivalent to

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ where $$ \underline{w} = \Sigma_{i} \alpha_{i} \underline{b}_{i} $$

and $$ \forall \left[ \alpha_{1}, ..., \alpha_{n} \right] \in \mathbb{R}^n$$

Solution
It is given that-
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \underline{b}_{i} * \underline{P} (\underline{v}) = 0 $$  where   i = 1, ..., n $$
 * $$\displaystyle (Eq. 1)


 * }
 * }

Choice 1

Taking i=1, we get

$$ \underline{b}_{1} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{1}$$ , we get-


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \alpha _{1}\underline{b}_{1} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 2)


 * }
 * }

Choice 2

Taking i=2, we get

$$ \underline{b}_{2} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{2}$$ , we get-


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \alpha _{2}\underline{b}_{2} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 3)


 * }
 * }

......

......

......

Choice n

Taking i=n, we get

$$ \underline{b}_{n} * \underline{P} (\underline{v}) = 0 $$

Multiplying both sides by $$\alpha _{n}$$ , we get-


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \alpha _{n}\underline{b}_{n} * \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 4)


 * }
 * }

Adding Equations 2 to 4, we get


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle(\alpha _{1}\underline{b}_{1} + \alpha _{2}\underline{b}_{2} +\alpha _{n}\underline{b}_{n}$$) $$ *   \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 5)


 * }
 * }

or we can write Eq(5) as -


 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \sum_{i=1}^{n} \alpha _{i}\underline{b_{i}}$$* $$ \underline{P} (\underline{v}) = 0 $$ $$
 * $$\displaystyle (Eq. 6)


 * }
 * }

Now by definition,

$$\underline{\omega} = \sum_{i=1}^{n} \alpha _{i}\underline{b_{i}}$$ where $$ \forall \left[ \alpha_{1}, ..., \alpha_{n} \right] \in \mathbb{R}^n$$ and {$$\underline{b_{i}}$$} is an orthonormal basis vector.

So Eq.(6) can be written as -

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$