Can you please help me for this problem?
Show that the map $x:M\rightarrow \mathbb{R}$, $\mathbb{R}$ has group structures under addition, is defined by $x\left( \left[ \begin{matrix} a& b\\ o& c\end{matrix} \right] \right)$ =$ \log \left( \dfrac {a} {c}\right) $ is a group homomorphism.
Also, is x surjective and what is ker(x)?
On
I'll assume that $M \subset \mathrm{GL}_2(\mathbb{R})$ , where all of the elements in $M$ are upper triangular matrices.
To show that $x$ is a homomorphism, we just need to show that it preserves the identity and the operations, i.e. $x\left( I \right)= 0$ and for any $a,b \in M$, $x(ab)=x(a)+x(b)$.
Alan Wang did a nice job showing the later, and it's clear that $x \left( \begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix} \right) = \log(1/1) = 0$, however, I disagree on his kernal.
$\ker(x)$ should be all the elements that get sent to 0, i.e. $$ \ker(x) = \Bigg \{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \mid a,b \in \mathbb{R} \Bigg \} $$
(since $\log \left(\dfrac{a}{a} \right)= 0$ ).
$x$ is also surjective, Alan did a great job showing that as well.
PS: I would have just added this as a comment to the previous answer, but it turns out I don't have enough reputation to do so :(
Edit: I had the kernal wrong
I assume that $M \subseteq GL(2,\Bbb{R}).$ \begin{align*} x\left( \begin{bmatrix} a& b\\ 0& c\end{bmatrix} \begin{bmatrix} p& q\\ 0& r\end{bmatrix}\right) &= x\left( \begin{bmatrix} ap& aq+br\\ 0& cr\end{bmatrix}\right)\\ &= \log \left( \dfrac {ap} {cr}\right) \\ &=\log\left(\frac{a}{c}\right)+\log\left(\frac{p}{r}\right)\\ &= x\left( \begin{bmatrix} a& b\\ 0& c\end{bmatrix}\right)+x\left( \begin{bmatrix} p& q\\ 0& r\end{bmatrix}\right) \end{align*}
Beware that the operation of $M$ is multiplication.
The kernel of $x$ is $\left\{\begin{bmatrix} a& b\\ 0& c\end{bmatrix}\; \bigg|\;\log(a/c)=0 \right\}=\left\{\begin{bmatrix} a& b\\ 0& a\end{bmatrix}\; \bigg|\;a,b\in \Bbb{R},a\neq0\right\}$.
For $k\in \Bbb{R}$, $$x\left( \begin{bmatrix} e^k& 0\\ 0& 1\end{bmatrix} \right) =\log(e^k)=k$$ So the map is surjective.