Ytukyg

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?

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.

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.

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.