Answer to Question #167049 in Linear Algebra for Sarita bartwal

Question

The function < , > : R^2× R^2→ R : <(x1), (y1)>

<(X2),(y2)>= x1^2+x2^2 define inner product over R^2

True or false with full explanation

Answer 1

Solution:

Given: $<x_1,y_1><x_2,y_2>=x_1^2+x_2^2$

To check whether it is defined as an inner product over $R^2$ or not.

For that we check the following properties:

(a) Linearity: $<au + bv, w> = a<u, w> + b<v, w>$ .

(b) Symmetric Property: $<u, v> = <v,u>$ .

(c) Positive Definite Property: For any $u ∈ V , <u,u> \ ≥ 0$ ; and $<u,u> = 0$ if and only if $u = 0$ .

Checking for linearity:

Consider $<ax_1,y_1><bx_2,y_2>$

$=ax_1^2 +bx_2^2$

Consider $a<x_1,y_1>b<x_2,y_2>$

$=ax_1^2+bx_2^2$

Since they are equal, linearity is satisfied.

Checking for Symmetric:

Consider $<x_1,y_1><x_2,y_2>=x_1^2+x_2^2$

Then $<x_2,y_2><x_1,y_1>=x_2^2+x_1^2$

Since they are equal, symmetric property is satisfied.

Checking for positive definite property:

$x_1^2+x_2^2\ge0$ for any value of $x_1,x_2$

Also, $x_1^2+x_2^2=0$ only when $x_1=x_2=0$

Hence, positive definite property is satisfied.

Thus, we can say that given relation is inner product.

Final answer - True.