Infinity matrix norm example

I have a brief question regarding the infinity matrix norm.

The subordinate matrix infinity norm is defined as:

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

This is derived from the general definition of a subordinate matrix norm which is defined as:

$$\|A\| = \max \left\{\frac{\|Ax\|}{\|x\|} : x \in K^{n}, x \neq 0\right\}.$$

I wanted to try this out in an example. So say we define the matrix:

$$A = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 2 \\ 4 & 4 & 3 \end{bmatrix}$$

and

$$x = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$

Now if we use the first definition, it is easy to see that $\|A\|_{\infty} = 11$

But if we use the general definition, we get:

$$\|A\|_{\infty} =\max \left\{\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} : x \in K^{n}, x \neq 0\right\}.$$

Now, we have:

$$Ax = \begin{bmatrix} 15 \\ 11 \\ 21 \end{bmatrix}.$$

Since the infinity vector norm is defined as:

$$\|x\|_{\infty} =\max_{1 \leq i \leq n} |x_i|$$

it follows that:

$$\|Ax\|_{\infty} = 21$$

and:

$$\|x\|_{\infty} = 3$$

But then we have:

$$\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} = \frac{21}3 = 7$$

that does not correlate with the fact that we previously found that $\|A\|_{\infty} = 11$.

If anyone can explain to me what is wrong with my reasoning here, I would appreciate it!

Solution 1:

The supremum occurs at $$x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$