What's a sphere in the space of matrices? Given a normed vector space $V$ over a field $k$ it's sphere is the set of points of unit norm.
$$ S(V) = \{ v \in V \mid 1 = \lVert v \rVert^2 \}$$
Over the reals the n-spheres are just spheres of free modules.
$$ S^n = S(\mathbb{R}^{n + 1}) $$
Not sure about the terminology here for using fields other than the reals.
Anyhow suppose you want the sphere over matrices?
Assuming a Euclidean norm the sphere over m by n matrices ought to be the set of matrices with largest singular value equal to 1 (using the spectral norm.)
$$ S([\mathbb{R}^n, \mathbb{R}^m]) = \{ A \in [\mathbb{R}^n, \mathbb{R}^m] \mid 1 = \lVert A \rVert^2 \}$$
We want the largest (in magnitude) root of the characteristic polynomial to be 1. Plugging in 1 or -1 we get.
$$ \det(A^*A - \text{I}_n) = 0 $$
And
$$ \det(A^*A + \text{I}_n) = 0 $$
But this doesn't really make any sense to me and I can't visualize this. It just seems like weird symbols.
Is there a geometric picture of this?
I have a hunch spheres ought to be reducible to combinations of simpler n-spheres (might require axiom of choice) but I don't have the knowledge for such things.
 A: I don't know if the following picture is geometric per se, but perhaps you can get some intuition for it by considering what choices of the components of the singular value decomposition of $A$ make it such that $A$ is in the unit sphere of matrices.
If $A$ is a real $m \times n$ matrix, then the singular value decomposition will be of the form $$A = U\Sigma V$$
where $U$ is an $m \times m$ real orthogonal matrix, $\Sigma$ is a $m \times n$ real diagonal matrix whose entries are the singular values of $A$, and $V$ is a real $n \times n$ orthogonal matrix.
Notice that as long as we satisfy the singular value requirement in $\Sigma$, both $U$ and $V$ can be any orthogonal matrices of the appropriate size. Therefore, we would expect this sphere of matrices to be a direct product of the sets of all such orthogonal matrices, with whatever the set of possible $\Sigma$ matrices is. If we call the set of all orthogonal matrices of size $n$ to be $\mathbf{O}(n)$, then:
$$S([\mathbb{R}^n, \mathbb{R}^m]) = \mathbf{O}(m)\times\text{something}\times\mathbf{O}(n)$$
The number of independent parameters in an orthogonal matrix, i.e. the dimension of $\mathbf{O}(m)$, is $\frac{m^2-m}{2}$. So this gives you a sense that this sphere is at least $\frac{(m^2-m)(n^2-n)}{4}$ dimensional.
Now to $\Sigma$. The idea is that the largest value in $\Sigma$ must be $1$, and $\Sigma$ can have at most $p = \min(m,n)$ distinct singular values by definition. So the set of valid $\Sigma$ matrices appears to contain $p-1$ independently chosen numbers from $[0,1]$, one $1$ to satisfy the constraint, and some repetition of the singular values chosen previously in the remaining $m + n - 1 - p$ slots.
If we called this set $\mathbf{\Sigma_1}$, then $$S([\mathbb{R}^n, \mathbb{R}^m]) = \mathbf{O}(m)\times\mathbf{\Sigma_1}\times\mathbf{O}(n)$$
but this is basically the same thing I wrote above; just "something".
Under the caveat that I am being very imprecise, we can say that
$$S([\mathbb{R}^n, \mathbb{R}^m]) \approx \mathbf{O}(m)\times[0,1]^{\min(m,n) -1}\times \left\{1\right\}^\left({\min(m,n)\choose m+n-1-\min(m,n)}\right)\times\mathbf{O}(n)$$
The reason for this crude and imprecise statement is that we effectively have $p - 1 = \min(m,n) - 1$ independent values between $0$ and $1$ to pick for $\Sigma$, and $\left({\min(m,n)\choose m+n-1-\min(m,n)}\right)$ ways to distribute the $p-1$ numbers we picked, plus the required $1$, to the remaining $m+n-1-p$ slots on the diagonal of $\Sigma$.
The reason this is only approximate is because this is only valid when every selected number of those $p-1$ numbers is distinct; this combinatorial argument doesn't work when one of those selected numbers is repeated, and also doesn't consider other restrictions that may arise on the values in $\Sigma$ for specific choices of $U$ and $V$, but at least hopefully gives some sense of the approximate "geometry" or "structure" of the sphere.
