Linear transformations $A$ for which $n^{*}A n$ is constant for all unit vectors $n$.

Question

Is there a name for linear transformations $A$ for which $n^{*} A n$ is constant for all unit vectors $n$?

The expression looks like the first fundamental form, if $A$ is symmetric, of course.

These are surprisingly non-obvious, the answer depending strongly on dimension. In $R^2$, they are just the rotations. In $R^3$, they are rotations with a scaling in the axial direction by the cosine of the angle of rotation.

Answer 1

In $R^3$ another class of solution consists of rotations of multiples of the case I mentioned in the original post:

$$ \begin{matrix} c & 0 & 0 \\ 0 & c & s \\ 0 & -s & c \\ \end{matrix} $$ where $c^2 + s^2 = 1$.

It was this example that first got my attention.

Answer 2

To close this, the last fact required is that any linear transform of a space can be decomposed into the sum of a symmetric transform and a skew-symmetric transform.

Let $A$ be a linear transform of a space into itself. Set $B = \frac{1}{2} (A + A^*)$ and $C = \frac{1}{2} (A - A^*)$. Thus $B$ is symmetric and $C$ is skew-symmetric, and $A = B + C$. As we have observed, $x^* C x = 0$ for any $x$. So

$x^* A x = x^* (B + C) x = x^* B x$

The general solution then is the sum of a symmetric solution of the problem with any skew symmetric matix. And we know already the solution for the symmetric case.

Answer 3

Here's a similar approach to classify some colleagues of mine came up with:

Having $\lambda = v^tAv$ is equivalent to having $v^t(A-\lambda I)v=0$

So it suffices to classify the $A$ for which $v^tAv=0$ these have to be skew symmetric because $0=v^tAv=\sum_{i,j}a_{ij}v_iv_j$ thus $a_{ij}+a_{ji}=0$ for all $i,j$. So $A^t=-A$.

Thus for any $\lambda$ we can write $A=X+\lambda I$ where $X$ is skew symmetric $X^t=-X$