Question: If $A= \begin{bmatrix} 1&0&1\\1&1&-1 \end{bmatrix}$ and $\vec{x}$ is an element of $R^3$. It is given that $||\vec{x}|| =1$. I have to find the maximum value for $A\vec{x}$.
Solution:
I did this by maximizing $||A\vec{x}||$. I found $A^TA$ and since it was symmetric, so I found its eigenvalues. They are $0,2, 3$. So the maximum value should be $\sqrt{3}$ and the minimum is $0$.
Now, What would happen if instead of $||\vec{x}||=1$, we had $||\vec{x}||=2$
Observe that $$\max_{||x||=2} ||Ax|| = 2\max_{||x||=2} \frac{||Ax||}{2} = 2\max_{||x||=2} \frac{||Ax||}{||x||} = 2\max_{||x||=2} ||A\frac{x}{||x||}||=2\max_{||y||=1} ||Ay||.$$ So your maximum is $2\sqrt{3}$.