# Condition Numbers

I'm working my way through a basic textbook on Numerical Linear Algebra, and one of the problems asks if the fact that an arbitrary invertible matrix $$A \in \mathbb{R}^{m \times m}$$ is perfectly conditioned in the 1-norm implies that it is perfectly conditioned in the $$\infty$$-norm. Formally:

cond$$_{1}(A) = 1 \Longrightarrow \text{cond}_{\infty}(A) = 1$$ ?

I can prove that this holds for symmetric matrices (i.e when $$A = A^{T}$$), but not for the general case. My intuition was to try to disprove this by contradiction, but I cannot think of/compute any non-symmetric matrix $$B$$ such that $$det(B) \neq 0$$ and $$\text{cond}_{1}(B) = 1$$. I've also been trying to come up with something utilizing $$\text{cond}_{1}(A^{T}A)$$, $$\text{cond}_{1}(AA^{T})$$, or $$\text{cond}_{1}(A^{T})$$, but so far have only succeeded in showing that the lower bound $$1 \leq \text{cond}_{\infty}(A)$$.

If anyone could offer any help with finding a suitable counterexample or give a hint on how to approach this, I would greatly appreciate it.

$$B = \begin{pmatrix}0 & \sqrt{10} \\ -\sqrt{10} & 0 \end{pmatrix}$$

• $$B$$ is antisymmetric.
• $$\det(B) = 10$$.
• $$\kappa_1(B) = \sqrt{10} \cdot \frac{1}{\sqrt{10}} = 1$$.

This is the example you requested. It is not an example with $$\kappa_\infty \neq 1$$.

There is no example with $$\kappa_\infty = 1$$.

$$||B||_1$$ is the maximum absolute column sum of $$B = \left( b_{ij} \right)_{i,j=1}^n$$ and $$||B||_\infty$$ is the maximum absolute row sum of $$B$$. Then \begin{align*} ||B||_1 &= \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}| \\ ||B||_\infty &= \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}| \\ \end{align*} Observe that $$||B^{-1}||_1$$ is the quotient of the maximum absolute row sum of $$B$$ and the determinant of $$B$$. Observe that $$||B^{-1}||_\infty$$ is the quotient of the maximum absolute column sum of $$B$$ and the determinant of $$B$$. \begin{align*} ||B^{-1}||_1 &= \frac{1}{\det B} \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}| \\ ||B^{-1}||_\infty &= \frac{1}{\det B} \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}| \\ \end{align*}

Then \begin{align*} \kappa_1(B) &= ||B||_1 \, ||B^{-1}||_1 \\ &= \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}| \cdot \frac{1}{\det B} \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}| \\ &= \max_{1 \leq i \leq n} \sum_{j=1}^n|b_{ij}| \cdot \frac{1}{\det B} \max_{1 \leq j \leq n} \sum_{i=1}^n|b_{ij}| \\ &= ||B||_\infty \, ||B^{-1}||_\infty \\ &= \kappa_\infty(B) \text{.} \end{align*}

(Since you only ask for a hint, I won't explain why the formulas for the norms of the inverse matrices are correct.)

• I checked this with Python, and got that cond(A,1) = 5, not 1. Commented Mar 18, 2021 at 23:10
• @victor__von__doom : Did you include the $1/5$ in front? Commented Mar 18, 2021 at 23:11
• Yeah, this is what I ran: A = 1/5 * np.array([[2,1], [3,0]], dtype=float); print(A,cond(A,1)) --- it outputted 5 for me. Lmk if I'm just doing something dumb Commented Mar 18, 2021 at 23:13
• @victor__von__doom : You write "perfectly conditioned in the 1-norm implies that it is perfectly conditioned in the $\infty$-norm. Formally: $$\kappa_1(A) = 1 \implies \kappa)\infty(A) = 1$$". That wording asserts $\kappa_1$ is the $1$-norm. Did you intend it to be something that is not declared in the Question? Commented Mar 18, 2021 at 23:25
• @victor__von__doom : Updated. Commented Mar 19, 2021 at 12:36