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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.

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  • $\begingroup$ What do you need the inner product for? $\endgroup$ – Federico Poloni Apr 12 at 21:05
  • $\begingroup$ @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 $v\in\mathbb{S}^{n-1}$. $\endgroup$ – Gianni del Fiore Apr 13 at 17:10
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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 $m(3)$?

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    $\begingroup$ 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 $m\left(k\right)=\max\left\{n\in\mathbb{Z}:nk^2-\left({n}\choose{k}\right)\geq0\right\}$? $\endgroup$ – Gianni del Fiore Apr 13 at 17:02
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    $\begingroup$ 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... $\endgroup$ – Pietro Majer Apr 13 at 18:02
  • $\begingroup$ @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? $\endgroup$ – Luc Guyot Apr 14 at 11:11
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    $\begingroup$ @PietroMajer Thanks for your comment, I edited my answer, stating that $m(1) = 0$ instead. $\endgroup$ – Luc Guyot Apr 14 at 11:12
  • $\begingroup$ @GiannidelFiore I just added a remark showing that $m(k) \le k^2$. $\endgroup$ – Luc Guyot Apr 14 at 11:54

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