Let x = (x₁, …, x_n) and y = (y₁, …, y_n) denote vectors in ℝⁿ for a fixed n ≥ 2. Which of the following defines an inner product on ℝⁿ?

Concept:

(a) Let V be a vector space. A function β : V × V → ℝ, usually denoted β(x, y) = <x, y>, is called an inner product on V if it is positive, symmetric, and bilinear. That is, if

(i) <x, y> ≥ 0, <x, x> = 0 only for x = x

Explanation:

x = (x1, …, xn) and y = (y1, …, yn) ∈ ℝn 

For n = 2

(1) 〈x, y〉 = \(\rm\displaystyle\sum_{i, j=1}^2\) xiyj = x1y1 + x1y2 + x2y1 + x2y2 

So, A = \(\begin{bmatrix}1&1\\1&1\end{bmatrix}\) 

Here a > 0, d > 0 but ad - bc = 1 - 1 = 0

So it does not define inner product space.

Option (1) is false

So, A = \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\) 

Here a = 0, d = 0 but ad - bc = 0 - 1 = -1 < 0

So it does not define inner product space.

Option (4) is false

(2) 

〈x, y〉 = \(\rm\displaystyle\sum_{i, j=1}^n\left(x_i^2+y_j^2\right)\)

= \((x_1^2 + y_1^2)+(x_1^2+y_2^2) + (x_2^2+y_1^2) + (x_2^2+y_2^2) \)

= \(2x_1^2 + 2y_1^2+2x_2^2+2y_2^2\)

Consider  x = (1,1)   and y = (1,1)

then <2x,y> = 2(2)² + 2 (2)² + 2 (1)² + 2(1)² = 20

But 2 <x,y> = 2( 2 (1)2 + 2(1)2 +  2 (1)2 + 2(1)2 ) = 16

Thus <2x,y> ≠ 2 <x,y> and so <x,y> is not an inner product. 

Option (2) is false

Hence option (3) is true.