Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 14E / d

Solution

Use the standard basis

e_{1}, \dots, e_{n}

. For the smallest possible rank, set

α e_{j} = e_{j}

for

j \leq r

and zero otherwise, and set

β e_{j} = - e_{j}

for

j \leq s

and zero otherwise. Then the maps have ranks

r, s

, while their sum has nonzero diagonal entries only for

r < j \leq s

. Hence

rank (α + β) = s - r .

For full rank when

r + s \geq n

, instead take

α e_{j} = e_{j}

for

j \leq r

and zero otherwise, and

β e_{j} = e_{j}

for

j \geq n - s + 1

and zero otherwise. Their supports cover every basis index because

n - s + 1 \leq r + 1

. The sum has diagonal entries one where just one map is nonzero and two where both are nonzero. None vanish over

R

, so these linear maps have ranks

r, s

and

rank (α + β) = n .

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