A Hadamard matrix $H$ of order $4m$ is a $4m$ by $4m$ matrix with entries $1$ and $-1$. It satisfies $HH^T = (4m) I$ where I is the identity matirx. A normalized Hadamard matrix is a Hadamard matrix whose first row and first column elements are all $1$s.
Let $X$ be a set of size $v$. A $(v,k,\lambda)$ design is a collection $\mathcal{D}$ of distinct subsets of $X$ (called $blocks$) s.t.
- each set in $\mathcal{D}$ contains exactly $k$ elements from $X$;
- every pair of distinct elements in $X$ is contained in exactly $\lambda$ blocks.
We can obtain a $(4m-1)\times (4m-1)$ 0-1 matrix by deleting the first column and first row in a normalized Hadamard matrix and changing $-1$s to $0$s. This is actually an incidence matrix to a $(4m-1, 2m-1, m-1)$ design.
But how can I obtain a $(2m-1, m-1, m-2)$ design or a $(2m, m, m-1)$ design? Seems impossible.
You start with your matrix $\overline{H}$, obtained from a normalized $4m$-by-$4m$ Hadamard matrix by eliminating the first row and column and replacing $-1$s by $0$s.
Now the matrix $\overline{H}_0$ obtained from $\overline{H}$ by deleting all rows in which the value of the first column is $0$ turns out to be an incidence matrix for a $2$-$(2m,\,m,\,m-1)$ design.
Similarly, the matrix $\overline{H}_1$ obtained from $\overline{H}$ by deleting all rows in which the value of the first column is $1$ is the incidence matrix for a $2$-$(2m-1,\,m-1,\,m-2)$ design.