Representations of finite groups -- Equivalent representations

[LagrangeEuler]

I am confused. Look for instance cyclic $C_2$ group representation where
$$D(e)= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$
and
$$D(g)= \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$
and let's take invertible matrix
$$A= \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}.$$
Then
$$A^{-1}= \frac{1}{2}\begin{bmatrix} -4 & 2\\ 3 & -1 \end{bmatrix}$$
Then
$$\tilde{D}(g)=A^{-1}\cdot D(g) \cdot A= \frac{1}{2}\begin{bmatrix} -4 & 2\\ 3 & -1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}= \begin{bmatrix} -5 & -6\\ 5 & 5 \end{bmatrix}$$
end that is not second order element, i.e. $\tilde{D}(g)\cdot \tilde{D}(g)$ is not equal to $I$. Why is that the case if with this transform one should get equivalent representation of group $C_2$?

 

[martinbn]

It should be
\begin{bmatrix}
-5 & -6\\
4 & 5
\end{bmatrix}

 

[LagrangeEuler]

It should be
\begin{bmatrix}
-5 & -6\\
4 & 5
\end{bmatrix}

Problem is still there. My question is why that element is not of second order, or why I do not get $C_2$ group with similarity transform?

 

[martinbn]

Problem is still there. My question is why that element is not of second order, or why I do not get $C_2$ group with similarity transform?

It is! Did you try the corrected one?

 

[LagrangeEuler]

It is! Did you try the corrected one?

No sorry. Everything is fine.

 

[LagrangeEuler]

Just one more. I found somewhere that any matrix representation of finite group is unitary. Matrix
$\begin{bmatrix} -5 & -6\\ 4 & 5 \end{bmatrix}$
however is not unitary.

 

[martinbn]

Just one more. I found somewhere that any matrix representation of finite group is unitary.

This statement means that it can be unitarized. In other words you can define an inner product such that the representation is unitary with respect to. But it doesn't say that it is unitary with respect to the standard inner product.