System of $4$ matrix equations in $4$ unkowns

Question

$\begin{pmatrix} a \\ b \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}^T \begin{pmatrix} C & D \\ D^T & E \end{pmatrix}= \begin{pmatrix} I_m & 0\\ 0 & I_n \end{pmatrix} $

$a$ and $b$ are vectors with length $m$ and $n$ respectively. C has dimension $m$ by $m$ and E $n$ by $n$.

From above, I get $4$ matrix equations in $4$ unknowns:

$$ a a^T C + a b^T D^T = I_n \tag{1}$$

$$ b a^T C + b b^T D^T = 0 \tag{2}$$

$$ a a^T D + a b^T E^T = 0 \tag{3}$$

$$ b a^T D + b b^T E^T = I_m \tag{4}$$

where $I$ is an identity matrix and $a$ is known. Is it possible to get $b, C, D, E$?

The matrix $\begin{pmatrix} C & D \\ D^T & E \end{pmatrix}$ is symmetric and the sum of a single row(or column) of this matrix is zero.

I know $C$, $aa^T$, and $bb^T$ are symmetric, but I'm not sure if this information would help. Thank you.

Answer 1

There are no solutions in the case where $A$ and $B$ are square. By 1 and 4, $A^{-1}=A^TC+B^TD^T$ and $B^{-1}=A^TD+B^TE^T$. This implies that $A$, $B$, and their inverses all have full rank, as do any product of these. But 2 and 3 report $BA^{-1}=AB^{-1}=0$, a contradiction.

Answer 2

There is no solution because of the followingobvious reason. The rank of the right-hand side is $n+m$, while that of the left hand side is bounded by the rank of $\begin{pmatrix} a \\ b\end{pmatrix}$, that is $m$.