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It is clear that not every nonsingular square matrix $A$ can be factored as $A = LU$. For example, $A=\begin{bmatrix}1 & 0 &0\\0 & 0 & 2 \\0 & 1 & -1 \end{bmatrix}$ can not be factored. i.e.

$$\begin{bmatrix}1 & 0 &0\\0 & 0 & 2 \\0 & 1 & -1 \end{bmatrix}=\begin{bmatrix}1&0&0\\l_{21}&1&0\\l_{31}&l_{32}&1\end{bmatrix}\begin{bmatrix}u_{11}&u_{12}&u_{13}\\0&u_{22}&u_{23}\\0&0&u_{33}\end{bmatrix}$$ then by considering operations of first row of U with first column of L, and second row of U with second column of L:

$u_{22}=0,u_{23}=2, l_32.0=1?$. This implies $LU$ factorization doesn't exist.

However if it exists, it is unique. My confusion is the following: If $A = LU$ is unique then A is square nonsingular. Is it True? If so, how can we prove/disprove it? Thanks.

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    Yes, it's true. An easier example for your failure to have $LU$ decomposition is the matrix $\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. – 2017-01-31
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    @Ted Shifrin. Thanks. But, how can we prove it? – 2017-01-31
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    You know $L$ is invertible (nonsingular) by construction. So it seems I lied. How do you solve for $U$ in terms of $A$? – 2017-01-31

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If $A$ is $m\times n$, then $L$ is always a nonsingular $m\times m$ matrix. So regardless of shape or rank, $U$ is uniquely determined by $L$ and $A$. However, what you need to think about is this: If the rank of $A$ (and hence of $U$) is less than $m$, can you perhaps get more than one $L$ that will work? (Think about an example where $A$ is $3\times 3$ with rank $1$.)