Showing that the set of normal operators is not a subspace of $L(V)$ if $\dim V\geq 2$ I proceeded as follows.
Let $U$ denote the set of normal operators over $V$. It is clear that $U$ is a subset of $L(V)$ and that $0\in U$ (since $0$ is self adjoint). Moreover, for all scalars $\lambda\in\mathbb{F}$,  we have $(\lambda S)(\lambda S)^*=(\lambda S)(\bar{\lambda }S^*)= \lambda \bar{\lambda }SS^*= \lambda \bar{\lambda }S^*S=(\bar{\lambda }S^*)(\lambda S)= (\lambda S)^*(\lambda S)$, so $U$ is closed under scalar multiplication. We can thus conclude from the test for a subspace that it is the additive condition that fails. Now using basic properties of operator addition and the adjoint, one is able to show $S+T$ is normal iff $ST^* +TS^*= S^*T +T^*S$. 
It is here that I am stuck. For a general f-d vector space with $\dim V\geq 2$, how do I construct two operators such that this condition does not hold? 
The question comes from section 7A of Axler's "Linear Algebra Done Right" (#10)
 A: The class of normal matrices contains all Hermitian matrices ($H^*=H$) and all anti-Hermitian matrices ($A^*=-A$). It is easy to see that for any complex square matrix $B$ one can write $B=H+A$ with $H=\frac12(B+B^*)$ Hermitian and $A=\frac12(B-B^*)$ anti-Hermitian. So it the set of normal matrices were a subspace, then all matrices would be normal, which is clearly false when $\dim V\geq2$.
For a concrete counterexample it suffices to take for $B$ any matrix that is not normal. The easiest is to take $B=(\begin{smallmatrix}0&2\\0&0\end{smallmatrix})$ which is not diagonalisable and therefore certainly not normal (this is also obvious from the definition of normal), and which is the sum of the Hermitian matrix $H=(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$ and the anti-Hermitian matrix $A=(\begin{smallmatrix}\phantom-0&1\\-1&0\end{smallmatrix})$. Extend $B$ with zeros to get a counterexample in higher dimensions.
But one could also take for $B$ a matrix that is diagonalisable but with non-orthogonal eigenvectors for different eigenvalues (this is easy to construct), and which is therefore not a normal matrix.
