 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/digital-systems-linear-algebra-and-introduction-to-formal-proof/6/e/solution

A

x^{⊤} A x

equals its own transpose, so

x^{⊤} A x = x^{⊤} A^{⊤} x = - x^{⊤} A x,

and it is zero over the real numbers.

Conversely, suppose the quadratic form

x^{⊤} A x

vanishes for every real vector. Substituting

x = e_{i}

gives

a_{ii} = 0

. Substituting

x = e_{i} + e_{j}

gives

a_{ij} + a_{ji} = 0

for

i \neq = j

. Therefore

A^{⊤} = - A

x^{⊤} A x = 0 for every x ⟺ A^{⊤} = - A .

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