What is the meaning of subtracting from the identity matrix? If I subtract the matrix $A$ from the identity matrix $I$, $I - A$, is there a meaning to the resulting matrix perhaps given some conditions like invertibility or symmetry? For example,
$$
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
- \begin{bmatrix} a & b \\ c & d \end{bmatrix}
= \begin{bmatrix} 1-a & -b \\ -c & 1-d \end{bmatrix}
$$
For future reference, is there a good reference to lookup answers to such questions when I can't derive the answer on my own?
 A: The difference $I - A$ is sometimes used with respect to projection matrices, where $A$ is the projection onto a subspace and $I - A$ is the projection onto the orthogonal complement of that subspace.
A: If there exists $n\in\mathbb{N}$ with $A^{n}=0$, then $I−A$ is invertible (try to prove this yourself before reading why!).
The reason: multiply out the brackets in the expression
$$(I−A)(A^{n−1}+A^{n−2}+\cdots+A+I)$$
and see that you get $I$. It follows that $A^{n−1}+A^{n−2}+\cdots+A+I$ is the inverse of $I−A$ in this case.
By the way, a matrix $A$ with $A^{n}=0$ for some $n\in\mathbb{N}$ is called nilpotent.

By the way, the fact presented above can be used to give a proof of the (basic) fact that all eigenvalues of a nilpotent matrix must equal zero. In particular, this means that the determinant and trace of a nilpotent matrix are always $0$.
If $A$ is nilpotent then so is $\frac{1}{\lambda}A$ for any $\lambda\neq0$. Hence $I-\frac{1}{\lambda}A=\frac{1}{\lambda}(\lambda I-A)$ has an inverse. It follows that $\det{(\lambda I-A)}\neq0$ whenever $\lambda\neq0$ and $A$ is nilpotent, so the characteristic polynomial only has $0$ as a root.
So considering $I-A$ can actually come in handy to prove things.
A: If A is invertible the resulting matrix B produced when you subtract $I$ from A may or may not be invertible. Can you think of examples?  If A is symmetric we have A = $A^t$, so, we have I - A = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
- \begin{bmatrix} a & b \\ b & d \end{bmatrix}
= \begin{bmatrix} 1-a & -b \\ -b & 1-d \end{bmatrix}$ is also symmetric. You see something similar to this when you are calculating the characteritic polynomial of a matrix to calculate eigenvectors and test for the diagbalizability of a matrix. but instead you are looking at Det ( A - cI)= 0, where c is in your field. 
