So I'm learning about matrix norms and I think I understand what they represent, but I'm confused with something...
To solve this problem, consider the effect that matrix $A$ has on the simplest unit vectors, $(1,0)^T$ and $(0,1)^T$: \begin{align} \begin{bmatrix}1 & -3 \\ 2 & 8\end{bmatrix} \cdot \color{red}{\begin{bmatrix}1 \\ 0\end{bmatrix}} &= \begin{bmatrix}1 \\ 2\end{bmatrix} & \left\| \begin{bmatrix}1 \\2 \end{bmatrix}\right\|_1 &= |1|+|2| = 3 \\ \begin{bmatrix}1 & -3 \\ 2 & 8\end{bmatrix} \cdot \color{red}{\begin{bmatrix}0 \\ 1\end{bmatrix}} &= \begin{bmatrix}-3 \\ 8\end{bmatrix} & \left\| \begin{bmatrix}-3 \\8 \end{bmatrix}\right\|_1 &= |{-}3|+|8| = 11 \end{align}
As you can see, the matrix is multiplied by two vectors, first by [1;0] and then by [0,1], in order to find it's norm
-Can I multiply it with [1;1] and see if it produces a "better" result?
Because later on in an example it multiplies a 2x3 matrix by [1;1;1] vector so... there can be all ones? or not?
It seems like you're talking about the operator matrix norm.
This can be defined as: $$ ||A||_p = \sup_{v\,:\,||v||_p\ne 0} \frac{||Av||_p}{||v||_p} $$ Or equivalently as: $$ ||A||_p = \sup_{v\,:\,||v||_p = 1} ||Av||_p $$ which I tend to prefer.
So, in this case, yes, you not only should consider other vectors, you have to do so. The norm is defined to take the maximum over all possible unit vectors. This means that such norms can be hard to directly calculate from its definition.
However, for some cases, there are easier shortcuts. For instance, when $p=1$, one can use: $$ ||A||_1 = \max_j \sum_i |a_{ij}| $$ or when $p=\infty$, it is $$ ||A||_\infty = \max_i \sum_j |a_{ij}| $$ or when $p=2$, one can compute $$ ||A||_2 = \max_i\sigma_i(A) = \max_i\sqrt{\lambda_i(A^TA)} $$ where $\sigma_j(M)$ and $\lambda_j(M)$ are the singular values and eigenvalues of $M$.