Is this block matrix invertible? Suppose that $A$ is full column rank matrix. Define 
$$
L=\begin{pmatrix}
A&0\\
0&A^{T}
\end{pmatrix}.
$$
Can we prove/disprove that $L$ is invertible?
 A: This is true only when $A$ is square.
Take 
$$
A = \begin{pmatrix} 
1 \\ 
0 \end{pmatrix}; \qquad 
L = \begin{pmatrix} 
1&0&0 \\
0&0&0 \\
0&1&0 \end{pmatrix}.$$
Our matrix $L$ is certainly not invertible. 
A: You multiply block matrices the same way you multiply two matrices. So, if $A$ is invertible, we get:
\begin{bmatrix}
    A^{-1}      & 0 \\
       0       & (A^{T})^{-1}
\end{bmatrix}
is the inverse of the original matrix.
A: Based on your comments, we have that $A$ is an $n \times m$ matrix, with rank $m$ and $n > m$. Thus $A^T$ is an $m \times n$ matrix. 
A quick aside: 

Suppose X, Y, and Z are matrices of dimension n × n, m × n, and m × m, respectively. Then $$\det \begin{pmatrix} X & 0 \\ Y & Z \end{pmatrix} = \det(X) \det(Z)$$
  (ripped from the wikipedia page on the determinant)

So in our case, the block matrix $\begin{pmatrix} A & 0 \\ 0 & A^T \end{pmatrix}$ isn't in the right form yet, since $A$ isn't square. By appending $n-m$ columns of $0$s onto $A$, and removing the first $n-m$ columns of $A^T$, we get a block matrix in the correct form.
For example, for a $3 \times 2$ matrix, $$\left ( \begin{array}{c c c | c c}
a & b & 0 & 0 & 0 \\
c & d & 0 & 0 & 0 \\
e & f & 0 & 0 & 0 \\
\hline
0 & 0 & a & c & e \\
0 & 0 & b & d & f \\
\end{array} \right )$$
However since we appended a column of zeros, the determinant of the upper square block is $0$,  which makes the total determinant $0$ as well. Therefore, the block matrix is not invertible.  
