If $A$ is the upper triangular matrix of order $n$, given below, is it possible to work out, (or at least bound) its rank?
$A = \begin{bmatrix}
0 & 100 & \star & \dots & \star & \star \\
0 & 0 & \star & \dots & \star & \star \\
0 & 0 & 203 & \dots & \star & \star \\
\vdots & \vdots & \vdots & \ddots & \star & \star \\
0 & 0 & 0 & \dots & \star & \star \\
0 & 0 & 0 & \dots & 0 & \star
\end{bmatrix}
$
Where each $\star$ represents any number (not necessarily the same) from $\mathbb{C}$.
Best Answer
I assume that $\star$ is allowed to be zero.
We attain the minimal possible rank by setting each $\star = 0$. Any matrix in this pattern will necessarily have rank at least $2$ because we always have the rank $2$ submatrix $$ \pmatrix{100&\star \\0 & 203} $$ We attain the maximal possible rank by setting each $\star = 1$. Since the matrix is in row-echelon form, the rank is simply the number of leading non-zero entries, which is $n-1$. We cannot attain rank $n$ because the first column is always $0$.
It is possible to attain any rank in between by setting columns equal to $0$.