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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}} $ .

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  • $\begingroup$ Of course that you can use the formula for $\|A\|_2$. This is in fact the definition. $\endgroup$
    – John B
    Feb 13, 2016 at 23:35
  • $\begingroup$ I mean, how can I use it if I'm not given the x vector? Like, I'm given A and asked to find the 2-norm. $\endgroup$
    – user269711
    Feb 13, 2016 at 23:39
  • $\begingroup$ You vary $x$ over the domain, that is, you compute for all $x$ and then take the maximum. $\endgroup$
    – John B
    Feb 13, 2016 at 23:40
  • $\begingroup$ Note that even for square matrices, there is no relation between singular and eigenvalues. A first example is the matrix $\begin{bmatrix}1&n\\0&1\end{bmatrix}$ where the singular values depend on $n$,. $\endgroup$ Feb 14, 2016 at 7:23

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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 $y = A(A^Tx)$.

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