# Condition number of a matrix-vector product

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?

• If you have questions you should put a comment and not edit my answer to reflect that. – Shogun May 26 at 15:46

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 $$\kappa$$

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

$$\$$