Linear Algebra Done Right · Problem 13

Operators on Real Vector Spaces — Sheldon Axler

Mathematics — Linear Algebra · nilpotent-jordan-null-space

Problem

Suppose T∈L(V)T \in \mathcal{L}(V) is nilpotent. Prove that if v1,…,vnv_1, \ldots, v_n are vectors in VV and m1,…,mnm_1, \ldots, m_n are nonnegative integers such that Tm1v1,…,Tv1,v1,…,Tmnvn,…,Tvn,vn is a basis of VT^{m_1}v_1, \ldots, Tv_1, v_1, \ldots, T^{m_n}v_n, \ldots, Tv_n, v_n \text{ is a basis of } V and Tm1+1v1=⋯=Tmn+1vn=0,T^{m_1+1}v_1 = \cdots = T^{m_n+1}v_n = 0, then Tm1v1,…,TmnvnT^{m_1}v_1, \ldots, T^{m_n}v_n is a basis of null TT. This exercise shows that n=dim⁡null Tn = \dim \text{null}\, T. Thus the positive integer nn that appears above depends only on TT and not on the specific Jordan basis chosen for TT.

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