Show that $T= \lambda I$, for some $\lambda \in F$, where $I$ is the identity map.

Question

Let $T \in L(V,V)$. If $ST=TS$ $\forall S \in L(V,V)$. Show that $T= \lambda I$, for some $\lambda \in F$, where $I$ is the identity map.

Clearly $<T>$ is in the centre of $L(V,V)$ but how to prove my desired result without using any characteristic polynomial (Although I can't see anything using characteristic polynomial also). I am getting any clue also.

Consider matrices $S$ which have all but one entries equal to $0$. — Servaes, Sep 23 '15 at 17:27
I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) — Patrick Stevens, Sep 23 '15 at 17:31

score 3 · Accepted Answer · answered Sep 23 '15 at 17:41

I have got something, if I take $E_{ii}$ $i=1,...,n$ then for any matrix $A_{n \times n}$ we get $AE_{ii}$ the matrix with $i$ th column of $A$ only and $E_{ii} A$ the matrix with $i$ th row of $A$ only. So equating $AE_{ii}=E_{ii} A$ we will get all the elements except diagonal entries are all zero. After that take $E_{1j}A=AE_{1j}$ $\forall j=1,..,n$. We will get $A=\lambda I$.

Here I have taken the matrix of $T$ is $A$.

See :D

Show that $T= \lambda I$, for some $\lambda \in F$, where $I$ is the identity map.

1 Answers1