 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2008/2/i/solution

E

satisfying

E^{2} = E

; it need not be an orthogonal projection without an inner product. Pairwise annihilation gives

P^{2} = (\sum_{i} E_{i})^{2} = \sum_{i} E_{i}^{2} + \sum_{i \neq = j} E_{i} E_{j} = \sum_{i} E_{i} = P .

Thus their sum is a projection.

The identities

E_{i} P = E_{i}

imply

T (I - P) = 0

. Since

T

is nonsingular,

P = I

. Also

T E_{i} = \sum_{j} λ_{j} E_{j} E_{i} = λ_{i} E_{i},

im E_{i} \subseteq V_{i}

. Conversely, if

T x = λ_{i} x

, applying

E_{j}

gives

(λ_{j} - λ_{i}) E_{j} x = 0

. For

j \neq = i

the eigenvalues are distinct, hence

E_{j} x = 0

. Since

P = I

x = E_{i} x

, proving

im E_{i} = V_{i}

Every vector is

\sum_{i} E_{i} x

. If

\sum_{i} v_{i} = 0

with

v_{i} \in V_{i}

, applying

E_{j}

gives

v_{j} = 0

. Thus

V = V_{1} \oplus \dots \oplus V_{r} .

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