 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/14/a/iii/solution

If every eigenvalue is nonzero, define

B = \sum_{n} λ_{n}^{- 1} P_{n}

, where

P_{n} = e_{n} e_{n}^{T}

. Using the mutually orthogonal projection identity

P_{n} P_{m} = δ_{nm} P_{n}

A B = \sum_{n, m} \frac{λ _{n}}{λ _{m}} P_{n} P_{m} = \sum_{n} P_{n} = I .

The same calculation gives

B A = I

A^{- 1} = n = 1 \sum N \frac{1}{λ _{n}} e_{n} e_{n}^{T} .

 New to topics? Read the docs here!