Condition number of a transposed matrix

Question

Does the condition number of matrix $A^T$ equal to the condition number of matrix $A$?

The condition number of a matrix is $||A|| \cdot ||A^{-1}||$.

Answer 1

This is true. Note that for every square matrix $A$ we have $\|A\| = \|A^\intercal\|$. Hence,

\begin{align} \kappa(A^\intercal) &= \| A^\intercal \| \| (A^\intercal)^{-1} \| \\ &= \| A^\intercal \| \| (A^{-1})^\intercal \| \\ &= \| A \| \| A^{-1} \| \\ &= \kappa(A). \end{align}

Answer 2

A general proof that works even for the case when $A$ is a rectangular matrix is stated below for interest.

Suppose that $A$ is any real $m \times n$ matrix.

When the rank of $A$ is full, then we can define the condition number of $A$ (in $2$-norm or spectral norm) as $$ \kappa(A) = {\sigma_{\max}(A) \over \sigma_{\min}(A)} $$

Suppose that the singular value decomposition of $A$ is given as $$ A = U S V^T $$ where $U$ and $V$ are orthogonal matrices.

Then it follows that $$ A^T = V S^T U^T $$

Thus, it follows that the singular values of $A$ and $A^T$ are the same.

Consequently, it follows that $$ \sigma_{\max}(A^T) = \sigma_{\max}(A) $$ and $$ \sigma_{\min}(A^T) = \sigma_{\min}(A) $$

Hence, it is immediate that $$ \kappa(A^T) = \kappa(A) $$