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\centerline{\bf SOLUTION FOR OCTOBER 2016}
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a. Show that if $A$ and $B$ are linear transformations from ${\mathbb R}^N \to {\mathbb R}^N$ then it is impossible for:
$$ AB - BA = I$$ where $I$ is the identity map.
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b. On the other hand show that there are linear transformations $A$, $B$ (defined on some infinite dimensionsal space) so that: $$AB-BA=I.$$
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{\bf SOLUTION:} In finite dimensions if $AB-BA=I$ then taking the trace of both sides gives
$$ tr(AB-BA) = tr(I) = N.$$ 
Using the fact that
$tr(AB) = tr(BA)$ and that $tr(C-D) = tr(C) - tr(D)$ we see that $tr(AB-BA) = tr(AB)-tr(BA) = 0$ and thus
$0=N$  which is impossible.
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In infinite dimensions let us consider the vector space of polynomials with real coeeficients:
$$ P = \{ a_{0} + a_{1}x + \cdots + a_{n}x^{n} | \ a_{0}, a_{1}, \cdots , a_{n}  \textrm{ are real} \}.$$
Now define $(Ap)(x) = p'(x)$ and $(Bp)(x) = xp(x).$ These are both linear and:
$$ ((AB)(p))(x) = (A(xp))(x) = xp'(x) + p(x) \textrm{ by the product rule from calculus } $$ and:
$$ ((BA)(p))(x) = (Bp')(x) = xp'(x).  $$
Thus: $$ ((AB-BA)(p))(x) = xp'(x) + p(x) - xp'(x) = p(x).  $$ 
Therefore: $$AB-BA =I \textrm{ on }  P.$$


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