Is this a homomorphism? Matrix homomorphism $\phi:\Bbb R\to GL_2(\Bbb R)$

Question

My book claims that the following is a homomorphism:

$$\phi:\Bbb R\to GL_2(\Bbb R),\phi(a)=\begin{bmatrix}1&0\\a&1\end{bmatrix}$$

But it doesn't seem to be:

$$\phi(ab)=\begin{bmatrix}1&0\\ab&1\end{bmatrix}$$ $$\phi(a)\phi(b)=\begin{bmatrix}1&0\\a&1\end{bmatrix}\begin{bmatrix}1&0\\b&1\end{bmatrix}=\begin{bmatrix}1&0\\a+b&1\end{bmatrix}$$

Right? What am I missing?

Answer 1

Your computation is correct, $\mathbb{R}$ is to be taken as the additive group (it is implied since $\mathbb{R}$ is not a group for multiplication because of $0$). So $\phi$ is indeed an homomorphism from $(\mathbb{R}, +)$ to $(\operatorname{GL}_2(\mathbb{R}), \times)$ in the sense that you have

$$\phi(a+b) = \phi(a) \phi(b)$$

Answer 2

The group operation on the domain (that is, $\mathbb{R}$) is addition, not multiplication, so your calculation exactly shows that this is a homomorphism, that is, $\phi(a+b) = \phi(a)\phi(b)$.

Answer 3

It depends on the type of homomprphism your book claims it to be. If you want this to be a ring homomprphism, then as you pointed out, it certainly is not, but as a group homomprphism between the additive reals and $GL_2(\mathbb R)$, whose operation is matrix multiplication, then yes, it is a homomprphism because $$\phi(a+b)=\phi(a)\phi(b).$$ The notation is a little confusing because the binary operation in one group is denoted by $+$, while it is denoted with multiplication in the other, but the rules of a group homomprphism still hold.