Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 7B / a

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- Solution a

Solution

Expanding the characteristic polynomial gives

det (M - t I) = (2 - t) [(1 - t) (3 - t) + 2] - 2 (2 - t) = (2 - t) (1 - t) (3 - t) .

Thus the eigenvalues are

1, 2, 3

. Solving the corresponding homogeneous linear systems gives

E_{1} = span {(- 1, 0, 1)^{T}}, E_{2} = span {(1, 1, 0)^{T}}, E_{3} = span {(1, 1, 1)^{T}} .

Every nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of

R^{3}

.

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