 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/d/solution

A^{⊤} = - A

, then the dot product satisfies

(A x) \cdot y = x^{⊤} A^{⊤} y = - x^{⊤} A y = - (A y) \cdot x .

Conversely, the assumed dot product identity gives

x^{⊤} (A^{⊤} + A) y = 0

for every

x, y

. Take

x = e_{i}

and

y = e_{j}

from the standard basis: every entry of

A^{⊤} + A

is zero. Hence $A^{⊤} = - A$ , proving the equivalence.

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