Let $\text{M}_n({R})$ denote the $\mathbb{R}$-algebra of the $n$-by-$n$ matrices over $\mathbb{R}$ and let $\text{GL}_n(\mathbb{R})$ be the unit group of $\text{M}_n({R})$.
Given $k \ge 0$, let $m(k)$ denote the largest integer $n \ge 0$ for which there exists an $\mathbb{R}$-linear map
$L: \mathbb{R}^{n} \rightarrow \text{M}_k(\mathbb{R})$ such that $1$ is an eigenvalue of $L(v)$ for every
$v \in \mathbb{S^{n - 1}} \Doteq \left\{
(x_1, \dots,x_n) \in \mathbb{R}^n \,\vert\, x_1^2 + \cdots + x_n^2 = 1 \right\}$.

The latter condition on $L$ can be rephrased by saying that $X^2 - v^2$ divides the characteristic polynomial $\chi_{L(v)}(X)$ of $L(v)$ for every $v \in \mathbb{R}^n$. This shows in particular that such an $L$ is injective and hence $m(k) \le k^2$.

It is easily checked that $m(0) = m(1) = 0$ and $m(k) \ge 1$ for every $k \ge 2$.

**Claim.**
We have $m(2) = 2$ and hence $m(k) \ge 2$ for every $k \ge 2$.

The proof will rely on

**Lemma 1.** Let $L: \mathbb{R}^2 \rightarrow \text{M}_2(\mathbb{R})$ be an $\mathbb{R}$-linear map such that $X^2 - v^2$ divides $\chi_{L(v)}(X)$ for every $v \in \mathbb{R}^2$. Then there is $C \in \text{GL}_2(\mathbb{R})$ and $\lambda \in \mathbb{R} \setminus \{0\}$, such that
$CL(x,y)C^{-1} = \begin{pmatrix} x & \lambda y \\ \lambda^{-1}y & -x \end{pmatrix}$ for every $(x, y) \in \mathbb{R}^2$.

*Proof*. Let $L$ be such an application. Then there are matrices $A, B \in \text{M}_2(\mathbb{R})$ lying in the conjugacy class of $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ such that $L(x, y) = x A + y B$. Conjugating $L(x, y)$ by an invertible matrix if needed, we can assume, without loss of generality, that $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Writing $B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we have $\text{det}(L(x, y)) = -x^2 - (a - d)xy -bcy^2$. The eigenvalues of $L(x, y)$ are $1$ and $-1$ if and only if $\text{det}(L(x, y)) = -1$, therefore $x^2 + (a - d)xy + bcy^2 = 1$ must be the equation of the unit circle $\mathbb{S}^1 \subset \mathbb{R}^2$. As a result, we have $a = d$ and $bc = 1$. Since the trace of $B$ is zero, it follows that $a = d = 0$, which completes the proof.

We are now in position to prove the claim.

*Proof of the claim.* Reasoning by contradiction, we assume the existence of $L: \mathbb{R}^3 \rightarrow \text{M}_2(\mathbb{R})$ with the desired property. Applying the above lemma to $L(x, y, 0)$ and $L(x, 0, z)$, we deduce that there is $C \in \text{GL}_2(\mathbb{R})$ and $\lambda, \mu \in \mathbb{R} \setminus \{0\}$, such that
$CL(x,y, z)C^{-1} = \begin{pmatrix} x & \lambda y + \mu z \\ \lambda^{-1}y + \mu^{-1} z& -x \end{pmatrix}$ for every $(x, y, z) \in \mathbb{R}^3$. Therefore $\det(L(x, y, z)) = -(x^2 + y^2 + z^2) - yz(\frac{\lambda^2 + \mu^2}{\lambda \mu})$. As this determinant equals $-1$ for every $(x, y, z) \in \mathbb{S}^2$, then we have $\lambda = \mu = 0$, a contradiction.

If $L: \mathbb{R}^3 \rightarrow \text{M}_3(\mathbb{R})$ be an $\mathbb{R}$-linear map such that $X^2 - v^2$ divides $\chi_{L(v)}(X)$ for every $v$, then the latter condition is equivalent to
$\det(L(v)) = -v^2 \text{tr}(L(v))$ for every $v \in \mathbb{R}^3$.

**Question.** What is the value of $m(3)$?