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

• How are you defining the condition numbers for non-square matrices? – tch Jan 17 at 2:04
• @tch That's a good question. $\kappa(A) = \lVert A \rVert \cdot \lVert A^{\dagger} \rVert$ – ArtificiallyIntelligence Jan 17 at 4:26
• By reducing to the SVD, you can express the 2-norm condition number as the ratio of the largest and smallest nonzero singular values, and similarly the Frobenius condition number as the square root of the ratio of the sum of the squares of the singular values with the sum of the squares of their reciprocals. You might be able to manipulate these expressions to get some relationships. – Christopher A. Wong Jan 17 at 5:35

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 $$\kappa_F(X) = \kappa_F(X')$$.

• This is neat. Previously I go back to ref [1] and [2] in detail, and it turns out that their proof in detail can be extended to rectangular case without harm to the generality of their proof for square case. – ArtificiallyIntelligence Jan 17 at 18:06