Is there any inequality between 2-norm condition number and Frobenius norm condition number for rectangular matrix? What I have found in [1]
Condition number inequality between Frobenius norm and 2-norm for square matrix,
Consider a full rank matrix $X \in \mathbb{C}^{n \times m}$, $m=n$, then we can have,
$$n - 2 + \frac{1}{\kappa_2(X)} + \kappa_2(X) \le \kappa_F(X).$$
Question: Does this inequality works for the case when $m \neq n$?
If not, have you seen other relationship between them?
Update:
The following paper [2] mentioned that it is a natural extension to non-rectangular case, I don't get why. read the two equations below equation (3.7) in [2]. 
Ref:
[1] Smith, Russell A. "The condition numbers of the matrix eigenvalue problem." Numerische Mathematik 10.3 (1967): 232-240.
[2] Bazán, F. S. V. (2000). Conditioning of Rectangular Vandermonde Matrices with Nodes in the Unit Disk. SIAM Journal on Matrix Analysis and Applications, 21(2), 679–693. https://doi.org/10.1137/S0895479898336021
 A: Claim:
Define $\kappa(A) = \Vert A \Vert \cdot \Vert A^\dagger \Vert$ and suppose $X$ is rank $k$. Then,
$$
k - 2 + \frac{1}{\kappa_2(X)} + \kappa_2(X) \le \kappa_F(X)
$$
Moreover, this bound is tight since a diagonal matrix with $k$ identical values followed by zeros attains the bound.
Proof:
Suppose $X \in \mathbb{C}^{n\times m}$ with rank $k$ and singular values 
$$
\sigma_1 \geq \cdots \geq \sigma_k \geq 0 \geq \cdots \geq 0
$$
Then $X^{\dagger}$ has singular values,
$$
1/\sigma_k \geq \cdots \geq 1/\sigma_1 \geq 0 \geq \cdots \geq 0
$$
Therefore,
$$
\kappa_2(X) = \Vert X\Vert_2 \Vert X^\dagger \Vert_2 = \sigma_1/\sigma_k
$$
and
$$
\kappa_F(X) = \Vert X\Vert_F \Vert X^\dagger \Vert_F = \sqrt{\left(\sum_{i=1}^{k} \sigma_i^2\right)\left(\sum_{i=1}^{k} \frac{1}{\sigma_i^2}\right)}
$$
Define $X'$ to be the $k\times k$ matrix with $\sigma_1,\ldots, \sigma_k$ on the diagonal. Then $X'$ is full rank and square so the result holds by observing that $\kappa_2(X) = \kappa_2(X')$ and $\kappa_F(X) = \kappa_F(X')$.
