Uniqueness of Ring Homomorphisms from $\mathbb{R}$ to $M_2(\mathbb{R})$

Question

Let $M_2(\mathbb{R})$ denote the set of all $2 \times 2$ real matrices. I am interested in investigating whether there exists a unique map $f: \mathbb{R} \to M_2(\mathbb{R})$ that satisfies the following conditions:

i) $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$;

ii) $f(xy) = f(x)f(y)$ for all $x, y \in \mathbb{R}$;

iii) $f(1) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. (This condition was added after edit, and I apologize for the omission due to being a bit sleepy when first writing this question.)

It's evident that $f(x) = x \cdot I$ (where $I$ is the identity matrix) is a solution. I feel that this might be the only solution, but I am unsure how to prove this.

A related question here deals with similar functional equations, though the codomain in that problem is $\mathbb{R}$.

Using a method similar to the one in the linked question, I've arrived at some preliminary conclusions. Here’s my attempt:

It's straightforward to show that $f(0) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. Let's consider the case where $x > 0$.

When $x \in \mathbb{Q}^+$, we can easily find that $f(x) = x \cdot I$.

Let $\det(\cdot)$ denote the determinant of a matrix, $\text{perm}(\cdot)$ the permanent, and $\text{tr}(\cdot)$ the trace. Then we have:

$$ \det(f(x)) = \det(f(\sqrt{x}))^2 > 0 \quad \text{and} \quad \text{perm}(f(x)) = \text{perm}(f(\sqrt{x}))^2 > 0. $$

They can't be zero; otherwise, matrices like $f(\frac{1}{x})$ cannot be defined. This implies:

$$ \text{tr}(f(x)) = \text{tr}(f(\sqrt{x})f(\sqrt{x})) = \text{tr}\left(\begin{pmatrix} f_{11}^2(\sqrt{x}) + f_{12}(\sqrt{x})f_{21}(\sqrt{x}) & \quad \\ \quad & f_{22}^2(\sqrt{x}) + f_{12}(\sqrt{x})f_{21}(\sqrt{x}) \end{pmatrix}\right) = (f_{11}(\sqrt{x}) - f_{22}(\sqrt{x}))^2 + 2 \cdot \text{perm}(f(\sqrt{x})) \geq 0.$$

Assuming $y > x > 0$, we then have:

$$ \det(f(y)) = \det(f(y-x) + f(x)) = \det(f(y-x)) + \det(f(x)) + \det(f(x)) \cdot \text{tr}(f^{-1}(x)f(y-x)) = \det(f(y-x)) + \det(f(x)) + \det(f(x)) \cdot \text{tr}\left(f\left(\frac{y-x}{x}\right)\right) \geq \det(f(x)), $$ $$ \text{tr}(f(y)) = \text{tr}(f(y-x)) + \text{tr}(f(x)) \geq \text{tr}(f(x)). $$

Therefore, $\det(f(\cdot))$ and $\text{tr}(f(\cdot))$ are monotonic. Given the density of rational numbers, we can always sandwich $x$ between two arbitrarily close rational numbers. By monotonicity, we can conclude that:

$$ \det(f(x)) = x^2 \quad \text{and} \quad \text{tr}(f(x)) = 2x. $$

At this point, I’m stuck. These conditions don’t seem sufficient to conclude the result I am aiming for.

Any help or hints would be greatly appreciated!

Edit: I apologize for initially omitting the condition $f(1) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. This condition is indeed crucial.

Answer 1

Whoops, I misspoke in the comments; even $f(1) = I$ is not enough to guarantee uniqueness if you don't require that $f$ is $\mathbb{R}$-linear. The reason is that $M_2(\mathbb{R})$ contains $\mathbb{C}$ as a subring, and there are uncountably many distinct homomorphisms $\mathbb{R} \to \mathbb{C}$, many of which even have infinite index. $\mathbb{C}$ is abstractly isomorphic to every other algebraically closed field of characteristic $0$ of the same cardinality $\mathfrak{c}$, so for example it is isomorphic to $\overline{\mathbb{R}(x)}$, and this produces an embedding $\mathbb{R} \to \mathbb{C}$ of infinite index. We can also apply arbitrary automorphisms of $\mathbb{C}$ to any such embedding which will usually produce a new one, and there are many of these.

Edit: These exotic embeddings $\mathbb{R} \to \mathbb{C}$ are also a counterexample to your claim about the determinant and trace being monotonic. I don't understand your argument for this and in particular I think you implicitly used that the permanent is multiplicative, which is not true.

Answer 2

You have edited your question to assume that $f(1)=I_2$. The question is actually easy if $f(1)$ is singular instead.

The three properties imply that $f(1)$ is nonzero and idempotent. If $f(1)$ is singular, we must have $f(1)=P\operatorname{diag}(1,0)P^{-1}$ for some invertible matrix $P$. Since $f(x)f(1)=f(x)=f(1)f(x)$, we in turn obtain $f(x)=P\operatorname{diag}(g(x),0)P^{-1}$ for some function $g$ that inherits the three analogous properties to (i)-(iii), so that $g$ is a ring homomorphism on $\mathbb R$. By an answer from Hagen von Eitzen or an answer from Yiorgos S. Smyrlis (without assuming the axiom of choice, continuity or linearity), $g(x)$ must be equal to $xg(1)$. Therefore $f(x)=xf(1)$. As $f(1)$ is nonzero, singular and idempotent, $f$ is unique up to similarity.

Answer 3

$\quad$ From uniqueness we are far away.

Choose a non-zero idempotent $P\in M_2(\mathbb{R})$, then define $f:\mathbb{R}\longrightarrow M_2(\mathbb{R})\,$ via $f(x)=xP\,$.
Concrete examples are $$P \:=\: \frac12\begin{pmatrix} 1& 1 \\ 1& 1 \end{pmatrix} \quad\text{or}\quad P\:=\:\begin{pmatrix}1&1\\0&0\end{pmatrix}.$$