How can you calculate the rank of an nxn matrix with the given conditions? Let $A=(a_{i,j})$ a square matrix whose elements are:

*

*$0$ if $i=j$.

*$1$ if $j>i$.

*$-1$ if $j<i$.

Is there a simple way to find its rank?

that was neat. Same rank as the matrix with entry $1$ when $j=i+1,$ then $-1$ when $j=i-1,$ otherwise $0$
Congruence:
$$ \tiny
\left(
\begin{array}{rrrrr}
1&0&0&0&0 \\
0&1&0&0&0 \\
0&-1&1&0&0 \\
0&0&-1&1&0 \\
0&0&0&-1&1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
0&1&1&1&1\\
-1&0&1&1&1 \\
-1&-1&0&1&1 \\
-1&-1&-1&0&1 \\
-1&-1&-1&-1&0 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1&0&0&0&0 \\
0&1&-1&0&0 \\
0&0&1&-1&0 \\
0&0&0&1&-1 \\
0&0&0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
0&1&0&0&0 \\
-1&0&1&0&0 \\
0&-1&0&1&0 \\
0&0&-1&0&1 \\
0&0&0&-1&0 \\
\end{array}
\right)
$$
The final matrix is the Skew Normal Form in Newman, especially pages 56-60.

The process is now repeated with $E$ until the canonical form
described by the theorem is obtained, and the procedure makes it clear
that the rank of $A$ must be even.

Once the matrix is in this normal form, the new null  (zero eigenvector)vector(s) becomes quickly  apparent: none when $n$ even, exactly one when $n$ is odd, namely $(1,0,1,0,1)^T$ and multiplied by any constant.
