Answer to Question #209638 in Linear Algebra for anuj

Question

Suppose V is finite-dimensional with dim V greater or equal to 2. Prove that there exist S, T €L(V, V) such that ST is not equal to T S

Answer 1

solution

N₁,N₂ $\in$ L(V,V) as follows

N₁($\upsilon$₁) =0, N₁($\upsilon$₂ )=$\upsilon$₁ ; (1)

N₂($\upsilon$₁)=$\upsilon$₂ ,N₂($\upsilon$₂)=0 (2)

Then for any vector w=a$\upsilon$₁ + b$\upsilon$₂ we have

N₂N₁(w)= aN₂N₁($\upsilon$₁)+bN₂N₁($\upsilon$₂)=b$\upsilon$₂ (3)

but

N₁N₂(w)=aN₁N2($\upsilon$₁) + bN₁N₂($\upsilon$₂)=a$\upsilon$₁ (4)

we see from 3 and 4 the linear independence of $\upsilon$₁,$\upsilon$₂ that

N₁N₂(w)$\neq$ N₂N₁(w)

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

N₁N₂ $\neq$ N₂N₁

as operators in L(V,V). In the event that dim V=n>2, we may build upon a construction of N₁,N₂ follows : choosing a basis {$\upsilon$₁,$\upsilon$₂......,$\upsilon$_n } for V, we define N₁,N₂ on $\upsilon$₁,$\upsilon$₂

as the above set

N₁($\upsilon$_i)=N₂($\upsilon$_i)=0

for 3$\leq$i$\leq$n, then for any w=$\sum$ a_i$\upsilon$_i$\in$V we have as above

N₁N₂(w) $\neq$ N₂N₁(w)

providing at least one of a₁,a₂$\neq$ 0, thus

N₁N₂$\neq$ N₂N_1.