I am working on an iPad Pro m4 11 inch with latest version of iOS and the tablet app.

My workflow with math books is as follows: I take a snapshot of the page, clip the text I am interested it, send to Mathpix and then paste the latexified data into a capacities Zettel. Normally this works great even in the tablet app. But today I pasted the text in and it didn't get rendered. I see a whole bunch of open Latex boxes. I had to tap on the screen as many times as there are Latex boxes on the page until it got rendered. This happens every time I open the Zettel or expand it.

it works fine on the web app on the iPad: when I open the exact same Zettel it renders fully and properly. Since I can't attach snapshots here, I can't show what it looks like, but I will share the Latex data. As you can see there is a lot of Latex so that means a lot of tapping going on till it renders. PS: the tablet app is great!

If $\mathbf{X}$ is a nonzero vector then the set of vectors $\{r \mathbf{X} \mid 0 \leqslant r \leqslant 1\}$ is the segment from 0 to the point $\mathbf{X}$. If $\mathbf{X}=\mathbf{0}$, then the collection $\{r \mathbf{X} \mid 0 \leqslant r \leqslant 1\}$ contains only the zero vector, and in this case we say that the segment degenerates to a point.

If $T$ is any linear transformation, then $T(r \mathbf{X})=r T(\mathbf{X})$, so the image of the segment $\{r \mathbf{X} \mid 0 \leqslant r \leqslant 1\}$ is the segment $\{r(T(\mathbf{X})) \mid 0 \leqslant r \leqslant 1\}$, possibly degenerate if $T(\mathbf{X})=\mathbf{0}$.

The set of points $\{\mathbf{U}+r \mathbf{X} \mid 0 \leqslant r \leqslant 1\}$ is also a segment, from $\mathbf{U}$ to $\mathbf{U}+\mathbf{X}$.

If $\mathbf{X}$ and $\mathbf{U}$ are linearly independent vectors, then the set of vectors $\{r \mathbf{X}+s \mathbf{U} \mid 0 \leqslant r \leqslant 1,0 \leqslant s \leqslant 1\}$ describes the parallelogram determined by $\mathbf{X}$ and $\mathbf{U}$ (see Fig. 2.27). The sets $\{r \mathbf{X} \mid 0 \leqslant r \leqslant 1\}$ and $\{s U \mid 0 \leqslant s \leqslant 1\}$ form two edges of the parallelogram and the other two edges are $\{r \mathbf{X}+$ $\mathbf{U} \mid 0 \leqslant r \leqslant 1\}$ and $\{\mathbf{X}+s \mathbf{U} \mid 0 \leqslant s \leqslant 1\}$. The four corners of the parallelogram are, in order: $\mathbf{U}, \mathbf{0}, \mathbf{X}, \mathbf{U}+\mathbf{X}$.

If $\mathbf{X}$ and $\mathbf{U}$ are linearly dependent, but not both $\mathbf{0}$, then the four points $\mathbf{U}, \mathbf{0}, \mathbf{X}$, and $\mathbf{U}+\mathbf{X}$ all lie on the same line and the set $\{r \mathbf{X}+s \mathbf{U} \mid 0 \leqslant r \leqslant 1$, $0 \leqslant s \leqslant 1\}$ is then a degenerate or collapsed parallelogram.