Question #207246

Suppose V is finite-dimensional with dim V ≥ 2.

Prove that there exist S, T L(V; V ) such that ST ≠ T S.


please assist.


1
Expert's answer
2021-06-17T18:21:38-0400


Here's a specific example which holds for V a vector space over any field F. Suppose first that dimV=2; then picking any basis {v1,v2v_1,v_2 } for V, we define in that basis the operators N1,N2L(V,V)N_1,N_2∈L(V,V) as folllows:



N1(v1)=0,N1(v2)=v1;N2(v1)=v2,N2(v2)=0.N_1(v_1)=0,N_1(v_2)=v_1; \\[9pt] N_2(v_1)=v_2,N_2(v_2)=0.


Then for any vector w=av1+bv2w=av_1+bv_2 we have


N2N1(w)=aN2N1(v1)+bN2N1(v2)=bv2,         N_2N_1(w)=aN_2N_1(v_1)+bN_2N_1(v_2)=bv_2,~~~~~~~~~


but


N1N2(w)=aN1N2(v1)+bN1N2(v2)=av1;         N_1N_2(w)=aN_1N_2(v_1)+bN_1N_2(v_2)=av_1;~~~~~~~~~


We see from (3) and (4) and the linear independence of v1,v2v_1,v_2 that


N1N2(w)N2N1(w)         N_1N_2(w)≠N_2N_1(w)~~~~~~~~~


unless a=b=0, that is, unless w=0. Thus


N1N2N2N1        N_1N_2≠N_2N_1~~~~~~~~


as operators in L(V,V). In the event that dimV=n>2, we may build upon the construction of N1,N2N_1,N_2 as follows: choosing a basis {v1,v2,,vnv_1,v_2,…,v_n } for V, we now define N1,N2 on v1,v2N_1, N_2 \text{ on } v_1, v_2 as above, and set



N1(vi)=N2(vi)=0N_1(v_i)=N_2(v_i)=0


for 3≤i≤n. Then for any w=aiviVw=∑a_iv_i∈V we have as above


N1N2(w)≠N2N1(w)


provided at least one of a1,a2≠0. Thus


N1N2N2N1N_1N_2≠N_2N_1


We have thus shown that for any finite dimensional vector space V over any field F, dimV>1 implies the existence of a noncommutating pair of operators S,TL(V,V):TSSS,T∈L(V,V): TS≠S T



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS