Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get 
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a similar pattern for 5 x 5, 6 x 6, ..., n x n matricies. 
How would we prove that we would achieve this pattern for all inverses of n x n matricies? 
 A: See https://en.wikipedia.org/wiki/Matrix_function   and the reference Higham (2008)
Also https://en.wikipedia.org/wiki/Jordan_normal_form#Matrix_functions
The matrix you call $A^{-1}$ is already in Jordan normal form. It is already written as $I + N,$ where $I$ is the identity and $N$ is the set of $1's$ off the diagonal. The importance of $N$ as a matrix is that it is nilpotent, in particular $N^n = 0$
So, we have the series
$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + x^4 - x^5 ...$
which converges when $-1 < x < 1.$
In comparison, the fact that $N$ is nilpotent, and $IN=NI,$ means
$$ (I+N)^{-1} = I - N + N^2 - N^3 + N^4 - + \cdots + (-1)^{(n-1)} N^{n-1} $$
and there is no need to worry about convergence. 
This is a basic fact of life. Given a real analytic function, we can evaluate that function on a square matrix if we know how to find the Jordan normal form of the matrix. There are whole books on this.
A: To elaborate on the other answer, which is correct, though presented with no concrete examples, note that your matrix $A$ is a sum $A = I + N$, where $N$ is the nilpotent matrix
$$N = \begin{bmatrix} 0 & 1 & & \\ & 0& 1 & \\ & & 0& 1 \\ & & &0 \end{bmatrix}$$
The matrix $N$ has powers of a simple form, for example
$$N^2 = \begin{bmatrix} 0 & 0 &1 & \\ & 0& 0 &1 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^3 = \begin{bmatrix} 0 & 0 &0 & 1\\ & 0& 0 &0 \\ & & 0& 0 \\ & & &0 \end{bmatrix}, \quad N^4 = 0$$
It is a basic fact in commutative algebra that when you have a nilpotent element $N$, then $1 + N$ is invertible, since (I'll suppose here that $N^4 = 0$) we have
$$I = I^4 + N^4 = (I + N)(I - N + N^2 - N^3)$$
and so $(I - N + N^2 - N^3)$ must be the inverse of $(I + N)$.
A: Hmm, you can also prove it from basic understanding of inverses of triangular matrices, and a proof by induction.
$$ \left( \begin{array}{c | c}
A_{TL} & A_{TR} \\ \hline
0 & A_{BR}
\end{array} \right)^{-1} =
\left( \begin{array}{c | c}
A_{TL}^{-1} & -A_{TL}^{-1} A_{TR} A_{BR}^{-1}\\ \hline
0 & A_{BR}^{-1}
\end{array} \right)
$$
Now, let's use $ A_{n \times n} $ to equal a matrix of the given form, of size $ n \times n $,
and $ x_n = \left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
$, of size $ n $.
Notice that
$$ A_{n \times n} = 
\left( \begin{array}{c | c}
A_{k \times k} & (-1)^k x_k  x_{(n-k)}^T \\ \hline
0 & A_{(n-k) \times (n-k)}
\end{array} \right)
$$
where
$$
x_k x_{n-k}^T = \left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
\left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)^T
=
\left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
\left( \begin{array}{c c c c}
1 & -1 & 1 & \cdots
\end{array}
\right)
=
\left( \begin{array}{c c c c}
1 & -1 & 1 & \cdots \\
-1 & 1 & -1 & \cdots \\
1 & -1 & 1 & \cdots \\
\vdots & \vdots & \vdots & 
\end{array} \right)
$$
So...
$$ A_{n \times n}^{-1} =
\left( \begin{array}{c | c}
A_{k \times k}^{-1} & - (-1)^k A_{k \times k}^{-1} x_k  x_{(n-k)}^T  A_{(n-k) \times (n-k)}^{-1} \\ \hline
0 & A_{(n-k) \times (n-k)}^{-1}
\end{array} \right)
$$
Now, 
$$
A_{k \times k}^{-1} x_k = 
-(-1)^{k} e_L
$$
where $ e_L $ is a vector of all zeroes except the last entry, which equals one.
Similarly,
$$
x_{(n-k)}^T A_{k \times k}^{-1} = 
e_F
$$
where $ e_F $ is a vector of all zeroes except the first entry, which equals one.
Now, I am notorious for loosing negative signs and minor typos, but you get the idea.
