 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/4/a/ii/solution

λ

is an eigenvalue of

T

T v = λ v

for some nonzero vector

v \in V

E_{λ} = {v \in V : T v = λ v} = ker (T - λ I) .

It is a vector subspace because it contains

0

and

T (a v + b w) = a T v + b Tw = λ (a v + b w)

whenever

v, w \in E_{λ}

and

a, b \in R

v \in E_{λ} \cap E_{μ}

, then

λ v = T v = μv

, so

(λ - μ) v = 0

. Since

λ \neq = μ

v = 0

. Therefore

E_{λ} \cap E_{μ} = {0} .

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