For non-square $n \times m$ matrices, when will $\| A \hat x\|$ be constant for all unit vectors $\hat x$?

Question

For non-square $n \times m$ real matrices, when will $\| A \hat x\|$ be constant for all unit vectors $\hat x$?

If $A$ is square, the answer is: If and only if $$A = \lambda I Q$$ where $\lambda$ is a scalar and $Q$ an orthogonal matrix. What about non-square matrices?

Background & context: This question came up while working on this question from Arnold's Mathematical Methods of Classical Mechanics, I noticed that:

Let $A$ be a $2^n \times n$ real matrix where each of the $2^n$ combinations of $\begin{bmatrix}\pm 1 & \pm 1 & \pm 1 & ... \end{bmatrix}$ is a row of $A$. Then $$\|A \hat x \| = 2^n$$ for all $x$.

Proof: Let $\sum_\pm$ denote summation over all combinations of $\pm$. Then $$\begin{align*} \|A \hat x \|&= \sum_{0 < i < 2^n} (a_i \cdot x)^2 \\ &= \sum_{\pm} (\pm x_1 \pm x_2 \pm ... \pm x_n)^2 \\ &= \sum_{0 < i \leq n} 2^nx_i^2 + \sum_{0 < i < j \leq n}2^n x_i x_j - \sum_{0 < i < j \leq n} 2 ^nx_i x_j \\ &= 2^n \|\hat x\| \\ &= 2^n. \end{align*}$$

This is intriguing, and prompted this question. Does this generalize? Is it a special case of a known result?

Answer 1

When $c\ge0$, the condition that $\|Au\|=c$ for every unit vector $u$ is equivalent to $u^T(A^TA-c^2I_m)u=\|Au\|^2-c^2\|u\|^2=0$ for every unit vector $u$. By the polarisation identity or the parallelogram law, it is also equivalent to $v^T(A^TA-c^2I_m)u=0$ for every pair of vectors $u$ and $v$. In turn, it is equivalent to $A^TA=c^2I_m$, which means $A$ is $c$ times a matrix with orthonormal columns (i.e., the columns of $A$ are mutually orthogonal and each of them has norm $c$).

Answer 2

The answer which you know for square matrices generalizes to $m\times n$ matrices under the following form: $$A=\lambda Q$$ where $\lambda\in\Bbb R$ and $Q$ is an $m\times n$ matrix with orthonormal columns (i.e. whose columns are pairwise orthogonal and of norm $1$). $\lambda$ may even be chosen non-negative.

To see that this condition is necessary, assume $\|A(x)\|=\lambda$ for each unit vector $x$. If $\lambda=0$ there is nothing to prove. If $\lambda>0$, consider $Q:=A/\lambda$. Since $Q$ is isometrical, it columns $q_1,\dots,q_m$ are of norm $1$, as images of the vectors of the canonical orthonormal basis $(e_1,\dots,e_m)$ of $\Bbb R^m$. They are also pairwise orthogonal, as for $i\ne j$, $2=\|e_i+e_j\|^2=\|q_i+q_j\|^2=2+2\langle q_i,q_j\rangle$.

Conversely, this condition is sufficient, i.e. any such matrix $Q$ is isometrical, since it satisfies $\|Q(x)\|^2=\|\sum x_iq_i\|^2=\sum x_i^2=\|x\|^2$ ($\forall x\in\Bbb R^m$).