 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2010/1/a/solution

Every vector decomposes as

v = E v + (v - E v)

, where

E v \in im E

and

E (v - E v) = 0

. If

w

lies in both the image and kernel, write

w = E u

to obtain

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

. Thus

V = im E \oplus ker E,

and

E

is the projection onto its image along its kernel. It is the identity on its image and zero on its kernel. Let

N = dim V

and

r = rank E

. A basis adapted to this direct sum gives matrix

diag (I_{r}, 0_{N - r})

. Since

E

is not the identity,

N - r > 0

, and therefore

det E = 0.

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