Suppose $V$ is finite dimensional with $\dim V \geq 2$. Prove that there exist $S,T \in L(V,V)$ such that $ST\neq TS$ Does this have anything to do with basis? If this is talking about the identity map, I don't understand why $TS$ and $ST$ would be different. Please help!
 A: By selecting a basis, it is enough to consider the case where $\dim V = 2$.  In other words, we are looking for two $2\times 2$ matrices that do not commute.
A: Here's a specific example which holds for $V$ a vector space over any field $\Bbb F$.  Suppose first that $\dim V = 2$; then picking any basis $\{v_1, v_2\}$ for $V$, we define in that basis the operators $N_1, N_2 \in L(V, V)$ as folllows:
$N_1(v_1) = 0, \; N_1(v_2) = v_1;  \tag{1}$
$N_2(v_1) = v_2, \; N _2(v_2) = 0. \tag{2}$
Then for any vector $w = av_1 + bv_2$ we have
$N_2N_1(w) = aN_2N_1(v_1) + bN_2N_1(v_2) = bv_2, \tag{3}$
but
$N_1N_2(w) = aN_1N_2(v_1) + bN_1N_2(v_2) = av_1; \tag{4}$
We see from (3) and (4) and the linear independence of $v_1, v_2$ that 
$N_1N_2(w) \ne N_2N_1(w) \tag{5}$
unless $a = b = 0$, that is, unless $w = 0$.  Thus
$N_1N_2 \ne N_2N_1 \tag{6}$
as operators in $L(V, V)$.  In the event that $\dim V = n > 2$, we may build upon the construction of $N_1, N_2$ as follows:  choosing a basis $\{v_1, v_2, \ldots, v_n\}$ for $V$, we now define $N_1$, $N_2$ on $v_1$, $v_2$ as above, and set
$N_1(v_i) = N_2(v_i) = 0 \tag{7}$
for $3 \le i \le n$.  Then for any $w = \sum a_i v_i \in V$ we have as above
$N_1N_2(w) \ne N_2N_1(w) \tag{8}$
provided at least one of $a_1, a_2 \ne 0$.   Thus
$N_1N_2 \ne N_2N_1. \tag{9}$
We have thus shown that for any finite dimensional vector space $V$ over any field $\Bbb F$, $\dim V > 1$ implies the esistence of a noncommuting pair of operators $S, T \in L(V, V)$:  $TS \ne ST$.  QED
Note:  I think it is worth pointing out that both $N_1N_2$ and $N_2N_1$ are idempotent operators; that is, $(N_1N_2)^2 = N_1N_2$ etc.  This may easily be seen by evaluating $(N_1N_2)^2$ on $w = \sum a_i v_i$.  In fact we have that $N_1N_2(v_1) = v_1$, and likewise $N_2N_1(v_2) = v_2$ with $N_1N_2(v_i) = N_2N_1(v_i) = 0$ for $i >2$; $N_2N_1$ and $N_1N_2$ are thus projections onto the subspaces $\langle v_1 \rangle$ and $\langle v_2 \rangle$, respectively.  Furthermore we see that $N_1$, $N_2$ are themselves nilpotent operators, i.e. $N_i^2 = 0$, $i = 1, 2$.  This in turn implies that the projectors $N_1N_2$ and $N_2N_1$ are orthogonal  in the sense that $(N_1N_2)(N_2N_1) = N_1N_2^2N_1 = N_1(0)N_1 = 0$ etc.  We also see that when $\dim V = 2$, $N_1N_2 + N_2N_1 = I_V$, the identity map on $V$.  All these properties have appropriate generalizations to the case $\dim V > 2$; I leave them for my readership to discover and prove; the task is not difficult.  And though these assertions take us somewhat beyond the scope of the original question, they come for free with the answer!  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
