Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 7G (Linear Mathematics)

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- Solution 7G

Solution

For

x \in ker (α - λ I)

, commutation gives

(α - λ I) β x = β (α - λ I) x = 0.

Thus each eigenspace of

α

is an invariant subspace for

β

.

If

α = β^{2}

, the two endomorphisms commute. The

n

distinct real eigenvalues of

α

have one-dimensional real eigenspaces, since their direct sum is the

n

-dimensional vector space. Take a nonzero real eigenvector

x_{i}

. Its eigenspace is invariant under

β

, so

β x_{i} = c_{i} x_{i}

for a real scalar

c_{i}

. Squaring gives

λ_{i} x_{i} = α x_{i} = β^{2} x_{i} = c_{i}^{2} x_{i}, λ_{i} = c_{i}^{2} \geq 0.

The simple-spectrum hypothesis is essential: the real

90

-degree rotation matrix squares to

- I

in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

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