Non-unitary representation How to prove $\pi :\mathbb R\to \mathbb C^2$,   defined by $t\mapsto \begin{pmatrix} 1 & t\\  0 & 1\end{pmatrix}$ is a non-unitary representation? 
Is the following correct? $\pi$ is a representation because 
$$\pi (t+s)=\begin{pmatrix}1 & t+s \\ 0 & 1\end{pmatrix}=\begin{pmatrix}1 & t\\  0  &1\end{pmatrix}\begin{pmatrix}1 & s \\ 0 & 1\end{pmatrix}  = \pi(t)\pi(s)$$
For non-unitary, I need to prove $U:\mathbb C^2\to \mathbb C^2$ is not unitary. 
 A: For $\pi$ to be a unitary representation, $\pi(x)$ must be a unitary matrix for every element $x$ of the group. Here, 
$$\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}$$ 
is not a unitary matrix unless $t=0$. Indeed,
$$\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}^* \begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix} 
=\begin{pmatrix}1 & 0 \\ \bar t & 1\end{pmatrix} \begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}   = \begin{pmatrix}1 & t \\ \bar t & 1+|t|^2\end{pmatrix} $$   which is not the identity matrix.
A: Note: this post largely composed before user147263 graciously corrected the equation in user147972's original question.  See below for details.  End of Note.
So we have that
$\pi(t) = \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}, \tag{1}$
and thus
$\pi(s + t) = \begin{bmatrix} 1 & s + t \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}, \tag{2}$
which shows that $\pi$ is a homomorphism from the additive group of real numbers to the multiplicative group of real $2 \times 2$ matrices of the form
$\begin{bmatrix} 1 & t\\ 0 & 1\end{bmatrix}. \tag{3}$
We note in passing that the OP apparently made an error in writing the equation
$\pi(ts) = \begin{bmatrix} 1 & ts \\ 0 & 1 \end{bmatrix} =  \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix} =  \begin{bmatrix} 1 & s + t \\ 0 & 1 \end{bmatrix}, \tag{4}$ 
since 
$\begin{bmatrix} 1 & ts \\ 0 & 1 \end{bmatrix} \ne \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}, \tag{5}$
and furthermore (4) implies $st = s + t$, clearly false for arbitrary $s, t \in \Bbb R$.  So, rather than bandy these points about in the comment stream, waiting until they were straightened out, I made the leap of faith as it were and tried to infer what was intended from my general mathematical background and common sense; hope I got it right!  Meanwhile I would like to suggest to user147972 that he/she carefully reconsider the equation for $\pi(ts)$ proferred in the question as stated, and to consider a judicious edit if he/she agrees witn my assessment.  In any event, we return to the (non-)unitarity of $\pi(t)$.  A unitary matrix $U$ may be defined by the fact that $U^\dagger U = UU^\dagger = I$, where $U^\dagger$  is the hermitian adjoint of $U$; as is well-known, $U^\dagger$ is given by taking the conjugate transpose of $U$; that is, writing
$U = [u_{ij}] \tag{6}$
we have
$(U^\dagger)_{ij} = \bar u_{ji}, \tag{7}$
or, in terms of (6), 
$U^\dagger = [\bar u_{ij}]^T = [\bar u_{ji}]. \tag{8}$
For any $t \in ,\Bbb C$ we thus may compute $\pi(t)^\dagger \pi(t)$ as follows:
$\pi(t)^\dagger \pi(t) = \begin{bmatrix} 1 & 0 \\ \bar t & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & t \\ \bar t & 1 + \bar t t \end{bmatrix} = I \Leftrightarrow t = 0; \tag{9}$
this shows that $\pi(t)$ is not unitary for $0 \ne t \in \Bbb C$, hence certainly not for $0 \ne t \in \Bbb R$.  QED.
And this is how we show that $\pi(t)$ is not in general unitary!
Hope this helps.  Cheerio,
and as always,  
Fiat Lux!!!
