Is there a geometrical interpretation of subtraction of two matrices, with a special case of $I -A$ (subtraction of a matrix from identity matrix)?
Reference: If $A$ is an idempotent matrix, the range of $A$ and the range of $I-A$ are disjoint sets. Trying to understand it geometrically.
If anyone can explain the general case of matrix subtraction geometrically, it will be a good help.
I don't think there is a general answer for $A-B$, but in the case of $I-A$, more precisely in the case of $Q=I-P$ where $P$ is an orthogonal projection matrix (i.e., an idempotent matrix as you say) on a certain subspace $S$, then $Q=I-P$ is the orthogonal projection on the orthogonal complement $S^{\perp}$ of $S$.
For example, in 3D, consider line $S$ with equations $x=y=z$, with normed unit vector $v=\frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}$. The orthogonal projection matrix onto $S$ is the rank-one matrix (rank one because the range space is one-dimensional):
$$P=vv^T=\frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}\frac{1}{\sqrt{3}}\begin{pmatrix}1&1&1\end{pmatrix}=\frac13\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$$
and
$$I-P=\frac13\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}$$
is the orthogonal projection onto plane $S^{\perp}$ orthogonal to $S$ with equation $x+y+z=0$, (with a rank-2 matrix, because the range space is now 2-dimensional).