Let $$X_n = \left[ {\begin{array}{cc} 0 & I_m \\\ I_m & 0 \end{array} } \right]$$ when $n$ is even $(n=2m)$ and $$X_n = \left[ {\begin{array}{cc} 0 & I_m & 0\\\ I_m & 0 & 0\\\ 0 &0 &1 \end{array} } \right]$$ when n is odd (n=2m+1).
I want to find a matrix A such that $$A^TA= X_n$$ for the even case. I started with n=2 (m=1) and matrix $A$ to be $$ A = \left[ {\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\ -\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \end{array} } \right].$$ The professor said I could use the case for n=2 to find $A$ for a general even integer. I tried forming a 4 by 4 matrix A to see if I can quickly deduced the A for general even $n$ but I couldnt. Can anyone please give me hint to solce this? Thanks.