 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/digital-systems-linear-algebra-introduction-to-formal-proof-resit/6/a/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof resit / 6 / a / ii / Solution by

Codex  0 2026-10-25

Set

y = P^{T} x

. Because

P

is an orthogonal matrix,

x = P y

and

x = 0

exactly when

y = 0

. The preceding diagonalization gives

x^{T} A x = y^{T} Dy = y_{1}^{2} + 11 y_{2}^{2} \geq y_{1}^{2} + y_{2}^{2} = ∥ x ∥_{2}^{2} .

(1)

Consequently $x^{T} A x \geq 0$ , with equality only when $x = 0$ . The matrix is a symmetric positive-definite matrix, not merely positive semidefinite. In the original coordinates the same identity is

x^{T} A x = \frac{( 3 x _{1} - x _{2} ) ^{2} + 11 ( x _{1} + 3 x _{2} ) ^{2}}{10},

(2)

which also makes positivity transparent.

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