Let $A \in \mathbb{R}^{m \times n}$ be a random matrix with i.i.d. entries (the distribution is not important), where $m < n$ (i.e. $A$ is a "wide" matrix). I would like a lower bound on 
$$
\phi(A) \triangleq \min\_x \frac{\lVert Ax \rVert}{\lVert x \rVert}
$$
that holds with high probability (apologies if the notation $\phi(A)$ conflicts with any established usage).


When $m \geq n$, evidently $\phi(A) = \sigma\_{min}(A)$, the least singular value of $A$ (although I am not certain why this is true). Of course the distribution of the least singular value of a random matrix has been well-studied.


But when $m < n$, it seems that $\phi(A) \neq \sigma\_{min}(A)$ in general. For example, if $m = 1$ and $n > 1$, then $\phi(A) = 0$ (just choose $x$ to be orthogonal to the vector $A$), but $\sigma\_{min}(A)$ is the Euclidean norm of the vector $A$, which usually will not be $0$.



Question about "wide" random matrices
I spoke to someone locally, and we think the issue is which convention is used to define the singular values of a matrix. If one defines the singular values of a matrix $A$ to be the eigenvalues of the matrix
$$
\sqrt{A^TA}
$$
then if $A$ is $m \times n$ with $m < n$ we have $\sigma\_{\min}(A) = 0$ but $\sigma\_{\min}(A^T) \neq 0$ in general. This agrees with the identity $\phi(A) = \sigma\_{\min}(A)$.


However, if one defines the singular values of $A$ to be the diagonal entries of the matrix $\Sigma$ in the singular value decomposition
$$
A = U\Sigma V^T
$$


then $A$ and $A^T$ have exactly the same singular values, and $\phi(A) \neq \sigma\_{\min}(A)$ in general.


