Show that $\varphi:\mathbb{R}→Gl_2 (\mathbb{R})$ defined by $\varphi(a)=\begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix}$ is not an isomorphism

Question

$\varphi:\mathbb{R}→Gl_2 (\mathbb{R})$ defined by the matrix $\varphi(a)=\begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix}$

An isomorphism is a homomorphism that is also bijective. $\varphi(a)$ is a homomorphism so in order to show it is not an isomorphism I must show it is not 1 to 1 or onto or that it does not have a 2 sided inverse. I am not sure how to show these things. Any help?

Answer 1

Take $A:=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$; then $A\in GL_2(\Bbb R)$ but $\varphi(a)\neq A$ for every $a\in\Bbb R$, thus $\varphi$ is NOT surjective, and in particular it cannot be an isomorphism.

Answer 2

Hint: To show $\varphi$ is not onto, you just have to find a single element of $GL_2(\mathbb{R})$ that does not have the form $\begin{pmatrix}1 & a \\ 0 & 1\end{pmatrix}$ for any $a\in\mathbb{R}$. Can you find an invertible $2\times 2$ matrix that doesn't look like that?