# 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.

• There are many norms on matrices, several of them (including Frobenius) being Euclidean norms: en.wikipedia.org/wiki/Matrix_norm – J.G. Aug 19 '17 at 18:23
• @J.G. Ok, but which one should I use? It has to be a restriction... – kaithkolesidou Aug 19 '17 at 18:28

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