When is the operator norm inequality for matrix product an equality? Let $\lVert \cdot \rVert$ denote the operator 2-norm of a matrix. It is well-known that for any two matrices $A \in \mathbb{R}^{n \times p}$ and $B \in \mathbb{R}^{p \times n}$ we have
$$\lVert AB \rVert \leq \lVert A \rVert \lVert B \rVert.$$
I am interested to know when the above inequality is an equality. One obvious scenario is when either of $A$ or $B$ is an identity matrix (upto a scalar multiple). However, is this the only possible case? Does there exist a characterization of $A$ and $B$ when equality holds? Any references to relevant research papers are welcome.
 A: The matrix $A \in \mathbb R^{n \times p}$ represents a linear map from $V = \mathbb R^n $ to  $W = \mathbb R^p $ which without ambiguity we can denote as $A: V \to W$
Similarly $B \in \mathbb R^{p \times n}$  represents $B: W \to V$, and $A \circ B: W \to W$
The 2-norm of a matrix is the same as operator norm of the linear map that it represents (see, e.g. https://math.stackexchange.com/q/482191 ), where the operator norm of, for example,  $A$ is defined as  $||A|| = sup _{0 \ne v \in V} (||Av||_W/||v||_V)$.
Since the spaces here are finite dimensional, it follows that any operator attains its norm, e.g. for $A$, there is at least one $v \in V$ with $||Av|| = ||A||.||v||$
(Proof: use equivalent definition of $||A|| = sup_{||v|| = 1} (||Av||) $then since the unit ball is compact in finite dimensions and $A$ is continuous then ||Av|| attains its maximum on the unit ball.)
Then for any $A$ we can define a set $S_A = \{v \in V: ||A v|| = ||A||.||v||\}$ and similarly, $S_B$.
If we now want $||A B|| = ||A||.||B||$ then we require a vector $w \in W$ with $||w|| = ||ABw|| = ||A(B w)||$ which can happen iff $w \in S_B$ and $Bw \in S_A$, i.e. $S_{AB} = S_A \cap B(S_B) \ne 0$.

I don't think that this makes for a very interesting property of $A, B$.
For example let $\{v_i\}_{i=1, n}$ be a basis of unit length vectors for $V$ and $\{w_i\}_{i=1, p}$ for $W$.
Define $B$ by $B(w_1) = v_2$ and $B(w_i) = 0 $ otherwise.
Define $A$ by $A(v_2) = w_3$ and $A(v_i) = 0 $ otherwise.
Then $||A|| = ||B|| = ||A \circ B|| = 1$
