 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2005/algebra/1/solution

Expanding gives

(I - E)^{2} - (I - E) = E^{2} - E

, so either operator is idempotent exactly when the other is. For an idempotent

E

v = (v - E v) + E v, E (v - E v) = 0.

w \in ker E \cap Im E

, write

w = E u

; then

0 = Ew = E^{2} u = E u = w

. Hence

V = ker E \oplus Im E

Next,

(E_{1} + E_{2})^{2} - (E_{1} + E_{2}) = E_{1} E_{2} + E_{2} E_{1} .

If the right side vanishes, left multiplication by

E_{1}

gives

E_{1} E_{2} + E_{1} E_{2} E_{1} = 0

, while right multiplication gives

E_{1} E_{2} E_{1} + E_{2} E_{1} = 0

E_{1} E_{2} = E_{2} E_{1}

, and their original sum now gives

2 E_{1} E_{2} = 0

. Over the real or complex numbers this implies $E_{1} E_{2} = E_{2} E_{1} = 0$ .

For three idempotents summing to

I

, each pair sum equals the complementary idempotent

I - E_{k}

. Thus every product of distinct operators vanishes. Every vector equals

E_{1} v + E_{2} v + E_{3} v

. If

v_{1} + v_{2} + v_{3} = 0

with

v_{i} \in Im E_{i}

, applying

E_{j}

gives

v_{j} = 0

. Therefore

V = Im E_{1} \oplus Im E_{2} \oplus Im E_{3} .

Take a basis of each summand and unite the three bases. A vector in the

j

th basis has eigenvalue one for

E_{j}

and zero for the other two, so this is the requested common eigenbasis. Zero summands simply contribute no basis vectors.

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