Question #167049

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


1
Expert's answer
2021-03-02T05:05:24-0500

Solution:

Given: <x1,y1><x2,y2>=x12+x22<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 R2R^2 or not.

For that we check the following properties:

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

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

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

Checking for linearity:

Consider <ax1,y1><bx2,y2><ax_1,y_1><bx_2,y_2>

=ax12+bx22=ax_1^2 +bx_2^2

Consider a<x1,y1>b<x2,y2>a<x_1,y_1>b<x_2,y_2>

=ax12+bx22=ax_1^2+bx_2^2

Since they are equal, linearity is satisfied.

Checking for Symmetric:

Consider <x1,y1><x2,y2>=x12+x22<x_1,y_1><x_2,y_2>=x_1^2+x_2^2

Then <x2,y2><x1,y1>=x22+x12<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:

x12+x220x_1^2+x_2^2\ge0 for any value of x1,x2x_1,x_2

Also, x12+x22=0x_1^2+x_2^2=0 only when x1=x2=0x_1=x_2=0

Hence, positive definite property is satisfied.

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

Final answer - True.



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