Which of the following real quadratic forms on \(\mathbb{R}\)² is positive definite?

Concept:

(i) Q ∶ R2 → R is said to be positive definite if Q(x, y) > 0 ∀ (x, y) ≠ (0, 0)

(ii) A symmetric matrix is positive definite ⇔ It's all eigenvalues are positive

Explanation:

(1) Q(1, -1) = -1 \(\ngtr\) 0, so no positive definite.

(3) Q(x, -x) = x² - 2x² + x² = 0 \(\ngtr\) 0, so no positive definite.

(4) Q(x, -x) = x² - x² = 0 \(\ngtr\) 0, so no positive definite. 

So, option (1), (3), (4) are false and option (2) is true.

Alternate Method

(1). Matrix Representation for Q(x, y) = xy is,

\(A=\left[\begin{array}{cc} 0 & 1 / 2 \\ 1 / 2 & 0 \end{array}\right]\) then det (A) = -¼ < 0

So, It's all eigenvalues can not be positive.

∵ det (A) = λ₁ λ2)

option (1) is false

(2) \(A=\left[\begin{array}{cc} 1 & -1 / 2 \\ -1 / 2 & 1 \end{array}\right]\) then chA(x) = x2 - 2x + \(\frac{3}{4}\)

So, Eigen values of A are, x2 - 2x + \(\frac{3}{4}\) = 0

\(\Rightarrow x =\frac{2 \pm \sqrt{4-4 \times 3 / 4}}{2}=\frac{2 \pm 1}{2}\)

\(=\frac{3}{2}, \frac{1}{2}\)

⇒ all eigen values of A are positive.

⇒ quadratic form is positive definite. 

option (2) is true

(3) \(A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\) then det (A) = 0

⇒ one eigenvalue of A is zero and other is 2.

⇒ all eigenvalues are not positive.

⇒ Q(x, y) is not positive definite.

option (3) is false.

(4) \(A=\left[\begin{array}{cc} 1 & +1 / 2 \\ +1 / 2 & 0 \end{array}\right] \) then det (A) = -1/4

⇒ All eigenvalues can not be positive.

⇒ Q(x, y) is not positive definite 

option (4) is false.