Let $V$ be a finite-dimensional vector space and $T$ be a linear map from $V$ to itself (in other words, it's an operator on $V$).      I need to prove that $T$ is a scalar multiple of the identity (i.e. $T(v)=cv$ for all $v \in V$ and for some $c \in F$, the underlying field) if and only if $ST=TS$ for all operators $S$ on $V$.   Note that $ST=TS$ does not necessarily mean that they are both equal to $I$, the identity operator. The only if part is trivial. But I cannot see how to attack the if part. Any help would be greatly appreciated.
The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study.  We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in {\cal L}(V), TS = ST$. Here, ${\cal L}(V)$ denotes the set of all linear operators over $V$.  One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.  I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction,  ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress.   Thanks in advance!
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This causes browsers to complain about "unsecured content": Booting a Windows 7 installation on different hardware I'm in a situation where I could do with very quickly migrating a Windows 7 (RTM x64) installation from one machine to another.  What options are open to me in terms of getting W7 to boot after the drive is picked up and moved from one box to another? I thought it was supposed to be a little less sensitive to this kind of move than XP, but it doesn't work - it is stuck in a reboot loop and never reaches a GUI.  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