Suppose V is finite-dimensional with dim V≥2. Prove that there exist S,T in L(V,V) such that ST≠TS.

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Jim Hunt · Accepted Answer

According to question given that  C is finite dimensional with dim V≥2  Let dim V=2  ∴S,T∈L(V)  S(x,y)=(x,x)  T(x,y)=(o,x+y)  Therefore (ST)(x,y)=S{T(x,y)}=S(o,x+y)=(0,0)  (TS)(n,y)=T{S(n,y)}=T(x,x,)=(0,2x)  ∴ST≠TS

lovagwb · Accepted Answer

I’m going to assume you meant finite dimensional, otherwise  dim⁡(V)=2 would make no sense. Here’s quite an elementary example.   Take two linearly independent vectors, say e1,e2, to form a basis of  V. Then every element  v∈V  can be written as a linear combination of  e1  and  e2 , i.e.  v=ae1+be2, where  a  and  b  are coefficients from some underlying field  F .   Now say  T  is the linear map that projects any vector to its first coordinate, i.e. T(ae1+be2)=ae1. Suppose  S  is another linear map that switches coefficients: S(ae1+be2)=be1+ae2. Convince yourself these maps are indeed linear.   The composition  S∘T maps ae1+be2 → ae2, but  T∘S  maps it to be1

Don Sumner · Accepted Answer

Here&#039;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,v2} for V, we define in that basis the operators N1,N2∈L(V,V) as follows: N1(v1)=0,N1(v2)=v1; N2(v1)=v2,N2(v2)=0 Then for any vector w=av1+bv2 we have N2N1(w)=aN2N1(v1)+bN2N1(v2)=bv2 but N1N2(w)=aN1N2(v1)+bN1N2(v2)=av1 We see from (3) and (4) and the linear independence of v1,v2 that N1N2(w)≠N2N1(w) unless a=b=0, that is, unless w=0. Thus N1N2≠N2N1 as operators in L(V,V). In the event that dimV=n&amp;gt;2, we may build upon the construction of N1,N2 as follows: choosing a basis {v1,v2,...,vn} for V, we now define N1,N2 on v1,v2 as above, and set N1(vi)=N2(vi)=0 for 3≤i≤n. Then for any w=∑aivi∈V we have as above N1N2(w)≠N2N1(w) provided at least one of a1,a2≠0. Thus N1N2≠N2N1 We have thus shown that for any finite dimensional vector space V over any field F, dimV&amp;gt;1 implies the existence of a noncommutating pair of operators S,T∈L(V,V):TS≠ST

Suppose V is finite-dimensional with dim V \geq 2. Prove

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