Answer to Question #207246 in Linear Algebra for cayyy

Question

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.

Answer 1

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 {$v_1,v_2$ } for V, we define in that basis the operators $N_1,N_2∈L(V,V)$ as folllows:

$N_1(v_1)=0,N_1(v_2)=v_1; \\[9pt] N_2(v_1)=v_2,N_2(v_2)=0.$

Then for any vector $w=av_1+bv_2$ we have

$N_2N_1(w)=aN_2N_1(v_1)+bN_2N_1(v_2)=bv_2,~~~~~~~~~$

but

$N_1N_2(w)=aN_1N_2(v_1)+bN_1N_2(v_2)=av_1;~~~~~~~~~$

We see from (3) and (4) and the linear independence of $v_1,v_2$ that

$N_1N_2(w)≠N_2N_1(w)~~~~~~~~~$

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

$N_1N_2≠N_2N_1~~~~~~~~$

as operators in L(V,V). In the event that dimV=n>2, we may build upon the construction of $N_1,N_2$ as follows: choosing a basis {$v_1,v_2,…,v_n$ } for V, we now define $N_1, N_2 \text{ on } v_1, v_2$ as above, and set

$N_1(v_i)=N_2(v_i)=0$

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

N₁N₂(w)≠N₂N₁(w)

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

$N_1N_2≠N_2N_1$

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 $S,T∈L(V,V): TS≠S$ T