Condition number of a matrix-vector product

Question

If $A$ is an $m \times n$ matrix and $x$ is an $n \times 1$ vector then the linear transformation $y=Ax$ maps $\mathbb{R}^{n} $ to $\mathbb{R}^{m}$, so the linear transformation should have a condition number $\mbox{cond}_{Ax}(x)$. Assume that $\| \cdot \| $ is a subordinate norm.

1. Show that we can define $$\mbox{cond}_{Ax}(x) := \frac{\|A\| \|x\|}{\|Ax\|}$$ for every $x \neq 0$

2. Find the condition number of the linear transformation at $x = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}^{T}$ using the $\infty$-norm

$$ T = \begin{bmatrix} 1 & 7 & -1 \\ 3 & 2 & 1 \\ 5 & -9 & 3 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} $$

I think that finding the condition number using the $\infty$-norm of a vector is impossible because that vector can not invertible. Am I right? Can somebody explain what my mistake is?

Answer 1

The matrix doesn't need to be invertible. First note the following

$$ \| x \|_{\infty} = \max_{1 \leq i \leq m} |x_{i}| $$

The vector $\infty$ norm is the maximum absolute value of an entry in a vector

$$ \| A \|_{\infty} = \max_{1 \leq i \leq m} \sum_{j=1}^{n} |a_{ij}| $$

and the matrix $\infty$ norm is the maximum absolute row sum. If $ \textrm{Cond}_{Ax}(x)$ is defined as above then we would get

$$ \|x \|_{\infty} = 2 $$

$$ \|T \|_{\infty} = 17$$

Then we see that

$$ T(x) = \begin{bmatrix} 1 & 7 & -1 \\ 3 & 2 & 1 \\ 5 & -9 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} (1-7-2) \\ ( 3-2+2) \\ (5+9+6) \end{bmatrix} = \begin{bmatrix} -8 \\3 \\ 20 \end{bmatrix}$$

So the value of $\|T(x)\|_{\infty}$ is

$$ \|T(x)\|_{\infty} = 20 $$

I really don't like $\textrm{cond}_{Ax}(x)$ so I'm going to use $\kappa$

$$ \kappa(x) = \frac{17 \cdot 2}{20 } = \frac{17}{10} = 1.7$$

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