Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 9F (Linear Algebra)

For matrices $A,B$ and scalars $r,s\in \mathbb{F}$, the definition of the trace givesHence the trace is a linear functional on the vector space of $n\times n$ matrices.

The definition of matrix multiplication givesThe middle equality uses commutativity of scalar multiplication in $\mathbb{R}$ or $\mathbb{C}$, together with interchange of finite sums. Thus the trace is unchanged by cyclically moving the two factors.

To prove that trace is the unique normalized cyclic linear functional, let $E_{ij}$ be the matrix units. For $i\ne j$, $E_{ii}{E}_{ij}=E_{ij}$ whereas $E_{ij}{E}_{ii}=0$, so the assumed identity for the linear functional gives $f(E_{ij})=0$. Also, $E_{ij}{E}_{ji}=E_{ii}$ and $E_{ji}{E}_{ij}=E_{jj}$, giving $f(E_{ii})=f(E_{jj})$. Write this common value as $c$. Since the matrix units form a basis,For $n=1$ the same conclusion follows immediately from the one-dimensional basis $E_{11}$. Finally, $f(I)=nc$; if this equals $n$, then $c=1$ over either specified field. The normalization makes $f$ exactly the trace.

Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.