 Solution

ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2021/1/a/iii/solution

Write

K_{j} = ker T^{j}

, where

T^{0} = I

and

K_{0} = {0}

. These kernels satisfy

K_{j} \subseteq K_{j + 1}

. Also

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

and

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

: for example,

T^{i} (T v) = T^{i + 1} v = 0

when

v \in K_{i + 1}

. Thus the proposed linear map between quotient vector spaces is well-defined and has the formula

\overline{T} : K_{i + 1} / K_{i} ⟶ K_{i} / K_{i - 1}, v + K_{i} ⟼ T v + K_{i - 1} .

v + K_{i}

belongs to its kernel, then

T v \in K_{i - 1}

. Therefore

T^{i} v = T^{i - 1} (T v) = 0

, so

v \in K_{i}

and

v + K_{i}

is zero. We have proved

\overline{T} is injective for every i \geq 1.

dim K_{i + 1} - dim K_{i} \leq dim K_{i} - dim K_{i - 1}

Surjectivity need not hold. Take a nonzero vector space

V

and the zero linear map

T = 0

. At

i = 1

, we have

K_{2} = K_{1} = V

and

K_{0} = 0

, so

\overline{T} : V / V ⟶ V /0 is the map 0 ⟶ V, which is not surjective .

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