Past exam of the linear algebra course of the University of Oxford / 2023 / 2 / a / ii

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- Solution ii

Solution

Put

d_{i} = dim K_{i}

, including

d_{0} = 0

. The restriction of

T^{i}

to

K_{i + 1}

has kernel

K_{i}

, so the rank-nullity theorem gives

dim T^{i} (K_{i + 1}) = d_{i + 1} - d_{i} .

Likewise

T^{i - 1} ∣_{K_{i}}

has kernel

K_{i - 1}

and hence image dimension

d_{i} - d_{i - 1}

. But

T (K_{i + 1}) \subseteq K_{i}

, which implies

T^{i} (K_{i + 1}) = T^{i - 1} (T (K_{i + 1})) \subseteq T^{i - 1} (K_{i}) .

Comparing dimensions yields the required concavity of the kernel dimensions of powers of a nilpotent map:

d_{i + 1} - d_{i} \leq d_{i} - d_{i - 1}, 2 d_{i} \geq d_{i - 1} + d_{i + 1} (1 \leq i < m) .

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