Given a matrix $M$, denote by $M^+$ its Moore-Penrose inverse. Let $B \in \mathbb{R}^{n \times r}$ and $A_1,\dots,A_t \in \mathbb{R}^{m \times n}$.
Is there a way to estimate the Frobenius norm of the matrix $[A_1 B \vert \dots \vert A_t B]^+ \in \mathbb{R}^{r t \times m}$ in terms of the Frobenius norm of $B^+$ and the matrices $A_1,\dots,A_t$?
I initially thought that something like $$ \lVert [A_1 B \vert \dots \vert A_t B]^+ \rVert_F^2 \leq \det\begin{pmatrix} \mathrm{Tr}(A_1^T A_1) & \dots & \mathrm{Tr}(A_1^T A_t) \\ \vdots & \vdots & \vdots \\ \mathrm{Tr}(A_t^T A_1) & \dots & \mathrm{Tr}(A_t^T A_t)\end{pmatrix} \cdot \lVert B^+ \rVert_F^2 $$ would be true, but after some experiments I believe that this is false. Alternatively, I would be happy with something like $$ \lVert [A_1 B \vert \dots \vert A_t B]^+ \rVert_F^2 \leq \left\lVert \begin{pmatrix} A_1^+ A_1 & \dots & A_1^+ A_t \\ \vdots & \vdots & \vdots \\ A_t^+ A_1 & \dots & A_t^+ A_t \end{pmatrix} \right\rVert_F^2 \cdot \lVert B^+ \rVert_F^2 $$ or the analogous bound where $A_i^+ A_j$ is replaced by $A_i^T A_j$. In fact, any other bound that separates $B$ from $A_1,\dots,A_t$ would already be very interesting. Can such a bound exist? What about other bound that mix the two? What is the best bound in the case $t = 1$?
If this helps, assume that $B$ has rank $r$ and that $[A_1 B \vert \dots \vert A_t B]$ has rank $r t$.
