 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2011/algebra/1/b/i/solution

To prove the sum is direct, suppose

\sum_{j = 1}^{r} v_{j} = 0

with

v_{j} \in V_{λ_{j}}

. Fix

i

P_{i} (T) = \prod_{j \neq = i} (T - λ_{j} I) .

Every

v_{j}

with

j \neq = i

is killed by one factor. On

v_{i}

, each factor is multiplication by

λ_{i} - λ_{j}

0 = P_{i} (T) \sum_{j} v_{j} = (\prod_{j \neq = i} (λ_{i} - λ_{j})) v_{i} .

The scalar coefficient is nonzero because the eigenvalues are distinct. Thus

v_{i} = 0

for every

i

, proving

V_{λ_{1}} + \dots + V_{λ_{r}} = V_{λ_{1}} \oplus \dots \oplus V_{λ_{r}} .

In particular, choosing a basis in each eigenspace yields a linearly independent combined list.

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