Note that since A_{11} is nonsingular (in fact SPD), we can do the following factorisation: \begin{bmatrix}A_{11}&A_{12}\\A_{21}&X\end{bmatrix} =\begin{bmatrix}I&0\\A_{21}A_{11}^{-1}&I\end{bmatrix} \begin{bmatrix}A_{11}&0\\0&X-A_{21}A_{11}^{-1}A_{12}\end{bmatrix} \begin{bmatrix}I&A_{11}^{-1}A_{12}\\0&I\end{bmatrix}. ...

Assuming that you've row-reduced correctly: We conclude that the system will be consistent if k^2 - k - 6 = 0 (so k = 3,-2) and therefore, for these k, the system have infinitely many ...

Your interpretation is correct. Indeed, (i) is straightforward, (ii) is always true because the matrix is symmetric. As for the last point, what you're talking about is called Sylvester's criterion ...

Finally managed to do it correctly (I finally tracked down the 2 errors, one of which was a simple sign drop), so I'll answer it (and accept it when I can, if nobody else answers) in order to prevent ...

Yes, well apart from the fact that there is no thing such as -0, you are correct. Cheers!

Given a k-dimensional subspace X of V, consider the subspace X^{\ast} = \{ f \in V^{\ast} : f(x) = 0 \forall x \in X \} of V^{\ast}. This subspace is (n-k)-dimensional and this defines ...