 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/17/c/i/solution

Q e_{i} = λ_{i} e_{i}

combine column by column into

QE = E D

. Since

∣ E ∣ \neq = 0

Q = E D E^{- 1} .

The nonzero determinant of

E

ensures that its columns form an eigenbasis.

 New to topics? Read the docs here!