2-Norm of Non-Square Matrices So, the 2-norm of an $m \times n$ matrix for $m\geq n$ is defined by the max singular value/square of the max eigenvalue. But, if it's not square, and you're only given a matrix A (no x-vector), what do you do if 1. m>n. Do I have to go through the whole SVD process, since I can't find an eigenvalue? or 2. if n>m, since I can't do SVD then. Please note that I'm talking about if I'm only given a matrix A, so I can't use that $||A||{_2} =max _{ \space \vec x \neq 0}  \frac{||A\vec x||{_2}}{||\vec x||_{2}}  $ .
 A: I prefer to use the equivalent definition
\begin{equation}
\|A\|_2 = \max \{ \|Ax\|_2 \: : \: \|x\|_2 \leq 1 \}
\end{equation}
because it reminds me that we are maximizing a real continuous function over the close unit ball. It is however, equally useless for practical computation!
In practice, we have to exploit that
\begin{equation}
\|A\|_2 = \sigma_1
\end{equation}
where $\sigma_1$ is the largest singular value of the matrix $A$. The singular value decomposition of a $A \in \mathbb{R}^{m \times n}$ is a factorization of the form
\begin{equation}
A = U \Sigma V^T
\end{equation}
where $U \in \mathbb{R}^{m \times m}$ and $V \in \mathbb{R}^{n \times n}$ are matrices with orthonormal columns, and $\Sigma \in \mathbb{R}^{m \times n}$ is a diagonal matrix with non-negative diagonal entries (the singular values). 
I want to stress the point that the singular value decomposition exists for all matrices regardless of the number of rows and columns. It is one of the more powerful tools in the field of matrix analysis.
In principle, you can obtain singular values by explicitly forming the matrices $AA^T$ or $A^T A$ and finding all eigenvalues as 
\begin{equation}
AA^T = U \Sigma \Sigma^T U^T, \quad \text{and} \quad A^TA = V \Sigma^T \Sigma V^T,
\end{equation}
but this is overkill, unless the matrix is of tiny dimension. In practical applications, the largest singular value is estimated by applying the power method to the problem, computing the necessary matrix vector product $y = AA^Tx$ with out explicitly forming the matrix $AA^T$, by exploiting the identity $y = A(A^Tx)$.
