# Maximal dimension of a vector subspace of real matrices with a special spectral property

What is the maximal $$n$$ for which there exists a linear map $$L:\mathbb{R}^n\to\text{End}\left(\mathbb{R}^k\right)$$ such that 1 is always an eigenvalue of $$L_v$$ for every $$v\in\mathbb{S}^{n-1}$$?

Guess: would $$n=\max\left\{m\in\mathbb{Z}:mk^2-{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{m}{2k}\right)\kern-.3em\right)}\geq0\right\}$$?

Motivation for the guess: the assumption implies that $$v\cdot v$$ is an eigenvalue of $$L_v^2$$ for every $$v\in\mathbb{R}^n$$. In other words, $$\det\left(L_v^2-v\cdot v I\right)=0$$, which is a homogeneous polynomial equation of degree $$2k$$ in the components of $$v$$. Since the number of monomials in such a polynomial is $${\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n}{2k}\right)\kern-.3em\right)}$$ and their coefficients are expressions in the $$nk^2$$ entries of $$L_{e_i},\ 1\leq i\leq n$$, the condition $$nk^2-{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n}{2k}\right)\kern-.3em\right)}\geq0$$ says the system is not overdetermined.

• What do you need the inner product for? Commented Apr 12, 2019 at 21:05
• @Federico notice that I am asking 1 and -1 to be eigenvalues of $Lv$ for every unit vector $v\in V$, i.e., $\left\|v\right\|=1$. Of course we can assume that $V=\mathbb{R}^n$ with the usual dot product, so $v\in\mathbb{S}^{n-1}$. Notice also that, since $L\left(-v\right)=-Lv$, it is enough to assume that 1 is an eigenvalue of $L\left(v\right)$ for every $v\in\mathbb{S}^{n-1}$. Commented Apr 13, 2019 at 17:10

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 $$m(3)$$?

• thank you for your answer. I was aware of the statement in your claim. Notice that, since $L\left(v\right)=-L\left(v\right)$, we can actually ask only 1 to be an eigenvalue of $L\left(v\right)$ for every $v\in\mathbb{S}^{n-1}$. Is it possible that $m\left(k\right)=\max\left\{n\in\mathbb{Z}:nk^2-\left({n}\choose{k}\right)\geq0\right\}$? Commented Apr 13, 2019 at 17:02
• Maybe we can say that for $n=0$ such a map does exist-- after all, the unit sphere of $\mathbb{R}^0$ is empty, so the condition is vacuously satisfied by the zero map... Commented Apr 13, 2019 at 18:02
• @GiannidelFiore It is really good that you gave an heuristic explaining why $m(k)$ should be finite. But it is still unclear to me where the inconsistency of such an overdetermined system would come from. Did you also settle the case of $m(3)$? What exactly do you know/ignore about the question? Any numerical computations supporting your guess? Commented Apr 14, 2019 at 11:11
• @PietroMajer Thanks for your comment, I edited my answer, stating that $m(1) = 0$ instead. Commented Apr 14, 2019 at 11:12
• @GiannidelFiore I just added a remark showing that $m(k) \le k^2$. Commented Apr 14, 2019 at 11:54