How to determine singular or nonsingular matrix I have the following matrix:
$$G = \left[
\begin{array}{cc}
A & e \\
e^T & 0
\end{array}
\right]$$
Where $A = BB^T$ is symmetric, positive semi-definite, and $e = [1,1,...,1]^T$.
let says $A$ is $n \times n$ and $e$ is $n \times 1$. 
Is matrix $G$ a nonsingular matrix ? Why?
My question is from the stanford lecture: https://see.stanford.edu/materials/lsocoee364a/hw8sol.pdf
The prof says the KKT matrix is nonsingular, if it satisfies one of four conditions.
Is there any matrix $Q$ such that $A$ + $eQe^T > 0$  like the prof's fourth condition
Thanks.
 A: First of all, there is no way to tell if $G$ is nonsingular if we do not specify the matrix $A$. If we are free to choose any $A$ symmetric positive semi-definite, we can easily make G singular. 
Elaborating what was said in the comments, let $A$ be the zero matrix. Then, the vector $z = [x \ 0]^T$ is in the null space of $G$, for any $x$ in the null space of $e^T$.
$$Gz = \left[
\begin{array}{cc}
0 & e \\
e^T & 0
\end{array}
\right] 
\left[
\begin{array}{cc}
x \\
0
\end{array}
\right] =
\left[
\begin{array}{cc}
0 \\
e^Tx
\end{array}
\right] = 0,$$e.g. for $x = [-1 \ 1 \ 0 \ \cdots 0]$.
Regarding the condition for the nonsingularity of $G$: 
$$A + eQe^T > 0,$$
 for some $Q \geq 0$, and still considering $A$ to be the zero matrix, we can show that there is no $G$ that satisfies this condition.
First, note that in this case, $Q$ is actually a scalar, so the condition simplifies to
$$Q(ee^T) > 0.$$
We know that $ee^T$ has one eigenvalue equal to $n$, and $n-1$ eigenvalues equal to $0$, therefore, is positive semi-definite and there is no $Q$ that makes $Q(ee^T)$ positive definite.
