# Question on Euclidean norm of a non-square matrix

Let $\;f:\mathbb R^n \rightarrow \mathbb R^m\;$ and consider the matrix $\; \nabla \cdot f=\begin{pmatrix} \frac{\partial f_1}{\partial x_1} \dots \frac{\partial f_1}{\partial x_n} \\ \dots \\ \frac{\partial f_m}{\partial x_1} \dots \frac{\partial f_m}{\partial x_n}\\ \end{pmatrix}\;$.

I want to compute this: $\;\frac{1}{2} {\vert \nabla f \vert }^2\;$ where $\;\vert \cdot \vert\;$ is the Euclidean norm of the matrix.

NOTE: $\;n\;$ is not necessary equal to $\;m\;$

Searching on google about "Euclidean norm of a non-square matrix", all the results I found, were about the Frobenius norm. So I thought it would be a good idea to compute the Frobenious norm of $\;\nabla f \;$ .

But then, I came across with this post What is the difference between the Frobenius norm and the 2-norm of a matrix? which confused me completely.

I haven't had any experience in norms of matrices until I was assigned to compute the above one. This is why I apologize in advance if my questions below are quite trivial or silly.

1. What is the Euclidean norm of the above matrix?
2. How should I proceed in order to compute $\;\frac{1}{2} {\vert \nabla f \vert }^2\;$ ?

Any help or hints would be valuable.

Thanks in advance!

• There are many norms on matrices, several of them (including Frobenius) being Euclidean norms: en.wikipedia.org/wiki/Matrix_norm
– J.G.
Commented Aug 19, 2017 at 18:23
• @J.G. Ok, but which one should I use? It has to be a restriction... Commented Aug 19, 2017 at 18:28

## 2 Answers

Usually we are concerned with the maximum expansion in the length of a (nonzero) vector. We have a matrix $A$ that may not be square, given a column vector $v,$ we want to know the maximum $m$ of $|Av|$ divided by $|v|.$ Well, $$m^2 = \max_{v \neq 0} \frac{(Av)\cdot (Av)}{v \cdot v} = \max \frac{v^T A^T Av}{ v^T v} = \max \frac{v^T Bv}{ v^T v},$$ where $B = A^T A$ is square and symmetric, also positive semidefinite, and square of the same dimension as that of $v.$ Therefore $B$ has a maximum real eigenvalue, and that is $m^2.$ So, $m$ is the square root of the largest eigenvalue of $B.$

There is probably a name for this; I wouldn't know.

The matrix you have written as $\nabla \cdot f$ is actually the Jacobian : https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

This is a common matrix in any differential field (such as differential geometry). It would be helpful to know why you need the norm to direct you in the right direction but assuming you need the determinant of a non square Jacobian I would direct you towards geometric algebra: https://en.wikipedia.org/wiki/Geometric_algebra

Here i would recommend using the Gram determinant as it is defined trough the exterior product (aka wedge product $\wedge$) of vectors (making them bivectors, trivectors, ect) as it has very deep geometric interpretation and can thus be coupled with almost every field of physics: https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant