I am having difficulty with a rather simple proposition.
Let $\mathfrak{t}_{n}$ constitute the $n \times n$ upper triangular matrices over $\mathbb{C}$ with the usual matrix commutator, and $\mathfrak{n}_{n}$ those that are strictly upper triangular. A Lie algebra is reductive if it is the direct sum of its center and its simple ideals.
I would like to show that $\mathfrak{t}_{n}$ is not reductive for all $n>1$.
Intuitively, I would expect that because $\mathfrak{t}_{n}$ contains $\mathfrak{n}_{n}$ as an ideal, and $\mathfrak{t}_{n} / \mathfrak{n}_{n} \cong \{ n \times n $ diagonal matrices$\}$, which is not an ideal in $\mathfrak{t}_{n}$ (by considering the $n=2$ case and computing a commutator), that $\mathfrak{t}_{n}$ is not reductive; as the ideal $\mathfrak{n}_{n}$ has no complement. However, I am not convinced that this is the proper reasoning, nor rigorous.
In particular, how would this show that the direct sum of the center and simple ideals is not $\mathfrak{t}_{n}$ ($\mathfrak{n}_{n}$ is notably not semisimple by the characterization of finite dimensional semisimple Lie algebras)? Can one show that there are no non-Abelian simple proper ideals? Even for the $n=2$ case, I can think of no more elegant solution to show non-reductiveness other than brute ideal-checking with the three vector subspaces.
I would appreciate any direction towards the general case, or a nice way to do the $n=2$ case that is indicative towards the prior. (Preferably with as little machinery as possible, as the problem itself seems quite simple.)
This can be solved, for example, by using some properties of reductive Lie algebras.
Suppose $\mathfrak g$ is reductive. Then, as you mentioned, $\mathfrak g = Z_{\mathfrak g} \oplus \mathfrak g_{ss}$, where $Z_{\mathfrak g}$ is the center and $\mathfrak g_{ss}$ is the semisimple part, i.e. the part which splits as direct sum of the simple ideals: $\mathfrak g_{ss} = \mathfrak g_1 \oplus \dots \oplus \mathfrak g_k$.
Now, $[\mathfrak g_i, \mathfrak g_j]$ is zero if $i \neq j$ (this is what it means to have a direct sum) and is an ideal of $\mathfrak g_i$ if $i=j$. Because $\mathfrak g_i$ is simple it must be $[\mathfrak g_i, \mathfrak g_i] = \mathfrak g_i$.
Thus, $[\mathfrak g_{ss},\mathfrak g_{ss}] = \mathfrak g_{ss}$ and, since $Z_g$ is abelian, $[\mathfrak g,\mathfrak g] = \mathfrak g_{ss}$ so that $\mathfrak g = [\mathfrak g,\mathfrak g] \oplus Z_{\mathfrak g}$.
In your case, $Z_{\mathfrak t_n}$ is the scalar multiples of the identity and $[\mathfrak t_n,\mathfrak t_n] = \mathfrak n_n$, and the direct sum of those does not give the whole $\mathfrak t_n$ (unless $n=1$).