 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/11/g/solution

A B = B diag (4, - 2, - 2)

. Multiplying on the left by

B^{- 1}

therefore gives

B^{- 1} A B = 400 0 - 2 0 00 - 2 .

This is an orthogonal diagonalization: the linear map acts independently by multiplication by its eigenvalues in the new orthonormal basis. The repeated eigenvalue explains the freedom to rotate the last two basis directions.

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