Past exam of the year 1 of the computer science course of the University of Oxford / 2003 / Calculus, probability, linear algebra, discrete mathematics, logic and proof, and models of computation / 3 / b / i

Solution

V

e_{1}, \dots, e_{m}

with

T e_{j} = λ_{j} e_{j}

. Arrange the basis so that

λ_{1} = \dots = λ_{r} = 0

and the remaining eigenvalues are nonzero. If there are no zero or no nonzero eigenvalues, the corresponding linear span below is simply the zero vector subspace.

For

v = \sum_{j} a_{j} e_{j}

T v = 0

holds precisely when

a_{j} = 0

for every

j > r

. Thus

ker T = span {e_{1}, \dots, e_{r}} .

Every output is in

span {e_{r + 1}, \dots, e_{m}}

, and conversely

e_{j} = T (λ_{j}^{- 1} e_{j})

for each

j > r

. Therefore

im T = span {e_{r + 1}, \dots, e_{m}} .

These two disjoint parts of a basis span

V

and their linear spans intersect only at zero. This is exactly the direct sum of vector spaces assertion

V = ker T \oplus im T .

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