The Bordered Hessian will always be a
Square matrix
In constrained optimisation (maximise or minimise f(x1, …, xn) subject to g(x) = c), the second-order condition is checked with the bordered Hessian. It is the Hessian of the Lagrangian (the matrix of second-order partial derivatives) bordered by a row and a column of the first-order partials of the constraint, with zero in the corner.
For n variables and one constraint the bordered Hessian has n + 1 rows and n + 1 columns, e.g. for two variables:
\(\bar{H} = \begin{vmatrix} 0 & g_1 & g_2 \\ g_1 & Z_{11} & Z_{12} \\ g_2 & Z_{21} & Z_{22} \end{vmatrix}\).
With m constraints it is (n + m) × (n + m). Its border-preserving principal minors are evaluated as determinants, and a determinant exists only for a matrix with equal numbers of rows and columns, so the bordered Hessian is always a square matrix.
It is not rectangular, it is not an identity matrix (its elements are derivatives, not 1s on the diagonal and 0s elsewhere), and it is not a null matrix (only the corner element is zero).
Hence, the answer is Square matrix.
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