What's the maximum vector length we can have from a linear transformation? Assume I have this linear transformation: 
$\ Ax = y $. What's the maximum magnitude(or length) of y we can obtain from this transformation? I feel somehow that the answer is related to eigenvalues. However, I'm not sure how to prove this. In other words, how to prove that the maximum length of $y$ would be the maximum eigenvalue of $A$ multiplied by its eigenvector. 
What about if $A$ is non square matrix? Is that related to singular values? if so, how can I prove that? 
 A: What you're looking for is the norm of the linear transformation. This is defined by
$$
\|A\| = \sup \{ \| Ax \| \mid \| x\| =1\}.
$$
Of course this doesn't really answer your question. In case $A$ is a normal matrix, we can say a bit more. In this case we have that the norm of $A$ is equal to the spectral radius $r(A)$ as you suspected. The spectral radius is defined by
$$
r(A) = \sup \{ \lvert \lambda \rvert \mid \lambda \text{ is an eigenvalue of } A\}.
$$
For a case where the spectral radius doesn't coincide with the norm, observe the following matrix
$$
A = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right).
$$
Note that it's only eigenvalue is $0$ but it's norm is equal to 1.
But we can still relate the singular values of $A$ to the norm of $A$. Note that a property of the norm is that
$$
\|A^\dagger A \| = \|A \|^2.
$$
So $\|A\|$ will always be equal to the square root of the largest eigenvalue of $A^\dagger A$ or in other words $\|A\|$ is equal to the largest singular value of $A$.
