# Continuous group homomorphism

Let A be the set of matrix of type $$\begin{bmatrix} a&0\\ 0 & a^{-1} \end{bmatrix}$$ where a is a positive number. How to show that if $$\phi$$ is a Group homomorphism from A to $$\mathbb{R}_{+}$$ with multiplication, then there exists some x $$\in$$ $$\mathbb{R}$$ such that $$\phi(a_t)=e^{tx}$$, where $$a_{t}$$=$$\begin{bmatrix} e^{t}&0 \\ 0 & e^{-t}\\ \end{bmatrix}$$

I understand that every element of A can be represented in the form of $$a_t$$, but I am not getting why a continuous group homomorphism is of type this.

Assuming that you are using the subspace topology on $$A$$, you have mutually inverse group homeomorphisms $$f_A: (\mathbb{R}, +) \to A, \, t \mapsto a_t \text{ and } g_A: A \to (\mathbb{R},+), \, M \mapsto \ln(\pi_{1,1}(M))$$ where $$\pi_{1,1}: A \to \mathbb{R}_+$$ denotes the projection onto the top-left entry. Similarly there are mutually inverse group homeomorphisms $$f_+: (\mathbb{R}, +) \to (\mathbb{R}_+, \cdot), \, t \mapsto e^t \text{ and } g_+: (\mathbb{R}_+, \cdot) \to (\mathbb{R},+), \, x \mapsto \ln(x).$$

With this we can transform your question to a well-known problem: Let $$\phi : A \to (\mathbb{R}_+, \cdot)$$ be a continuous group homomorphism. Then $$\psi := g_+ \circ \phi \circ f_A$$ is a continuous group homomorphism $$(\mathbb{R}, +) \to (\mathbb{R}, +)$$, so it is given by $$\psi(t) = t \cdot \psi(1) \text{ for all } t \in \mathbb{R}.$$ This is immediate for $$t \in \mathbb{Z}$$ by the homomorphism property, then follows for $$t \in \mathbb{Q}$$ by clearing denominators and lastly follows for $$t \in \mathbb{R}$$ since $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ and limits in $$\mathbb{R}$$ are unique.

Now we have $$\phi = f_+ \circ \psi \circ g_A$$ and hence $$\phi(a_t) = \phi \circ f_A(t) = f_+ \circ \psi(t) = f_+(t \cdot \psi(1)) = e^{t \cdot \psi(1)} \text{ for all } t \in \mathbb{R}.$$