Determining if Upper Triangular Matrices is semisimple. I wanted to know if $U_3(\mathbb{R})$ that is the space of upper triangular matrices over $\mathbb{R}$ is semisimple or not.
One theorem that I am planning to use is the following proposition $4.7$ from Lang's Algebra, Chapter $XVII$.
It states that if $R$ is a $k-$sunalgebra of $End_k(V)$. Then $R$ is semisimple if and only if $V$ is semisimple $R$ module.
Now here I don't have any idea how to show if $\mathbb{R}^3$ is a semisimple $U_3$ algebra either.
 A: Note that the nil ideal consisting of matrices of the form
$$\begin{pmatrix}0&x_1&x_2\\0&0&x_3\\0&0&0\end{pmatrix},\quad x_1,x_2,x_3\in\mathbb{R}$$
is contained in the Jacobson radical $J(U_3(\mathbb{R}))$. So $U_3(\mathbb{R})$ is not semi-simple.
A: Since the ideal of strictly upper triangular matrices is nilpotent, it must be included in the Jacobson radical as @dromastyx already mentioned, hence it cannot be semi-simple.
To apply the theorem you mentioned, note that $V:=\{\begin{pmatrix} x \\ 0 \\ 0\end{pmatrix} | x\in\mathbb R\}$ is an invariant subspace of $\mathbb R$, but it doesn't have an complement. Given any $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$, if $a\not=0$, then $\begin{pmatrix} 1 & & \\ & 0 & \\ & & 0\end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} a \\ 0 \\ 0 \end{pmatrix}\in V$. Hence the complement if exists must be included in the $2$-dimensional subspace $\{\begin{pmatrix} 0 \\ x \\ y \end{pmatrix} | x, y\in\mathbb R\}$ which is $2$-dimensional and hence must be exactly the full complement, but it is not $U_3$-invariant: $\begin{pmatrix} 0 & a & b \\ & 0 & 0 \\ & & 0\end{pmatrix}\begin{pmatrix} 0 \\ x \\ y \end{pmatrix}=\begin{pmatrix} ax+by \\ 0 \\ 0\end{pmatrix}$ can be anything in $V$.
This is an overkill to see none of $U_n$ is semi-simple: By Wedderburn-Artin, if semi-simple, it would be isomorphic to $\oplus_i M_{n_i}(\mathbb R) \oplus_j M_{n_j}(\mathbb C)\oplus_k M_{n_k}(\mathbb H)$, as $\mathbb R, \mathbb C, \mathbb H$ are the only finite dimensional division rings over $\mathbb R$.
If it contains any copy of $M_n(\mathbb C)$ or $M_n(\mathbb H)$, it would have an element with imaginary numbers such as $i$ as its eigenvalues. But $U_n$ doesn't contain such an element: the eigenvalues of $x\in U_n$ are exactly its diagonal elements which must be all real. (Note that eigenvalues can be defined intrinsically without the canonical representation of acting on $\mathbb R^n$)
Therefore $U_n\simeq \oplus_i M_{n_i}(\mathbb R)$ and again since no element in $U_n$ has complex numbers as eigenvalues, we know it must be the case that $n_i=1$ for all $i$, but then $U_n$ must be commutative.
