# Is “$:$” so-called Frobenius inner product?

I see some notation like \begin{align*} \int \nabla \mathbf{u} : \nabla \mathbf{v} \; dx \end{align*}

Here I think the two vectors $\mathbf{u}$ and $\mathbf{v}$ should be column vectors, i.e. $\mathbf{u} = [u_1,u_2,...,u_n]^T$ and $\mathbf{v} = [v_1,v_2,...,v_n]^T$. Is that right?

So, here $\nabla \mathbf{u}$ will be a Jacobian matrix, and $\nabla \mathbf{v}$ as well. Right?

Then, does the operation $\nabla \mathbf{u} : \nabla \mathbf{v}$ mean the element-wise multiplication such that \begin{align*} \nabla \mathbf{u} : \nabla \mathbf{v} &= (\nabla \otimes \mathbf{u}) : (\nabla \otimes \mathbf{v}) \\ &= \left( \begin{bmatrix} \nabla_1 \\ \nabla_2 \\ \vdots \\ \nabla_n \end{bmatrix} \begin{bmatrix} u_1 & u_2 & \cdots & u_n \end{bmatrix} \right) : \left( \begin{bmatrix} \nabla_1 \\ \nabla_2 \\ \vdots \\ \nabla_n \end{bmatrix} \begin{bmatrix} v_1 & v_2 & \cdots & v_n \end{bmatrix} \right) \\ &= \begin{bmatrix} \nabla_1 u_1 \nabla_1 v_1 & \nabla_1 u_2 \nabla_1 v_2 & \cdots & \nabla_1 u_n \nabla_1 v_n \\ \nabla_2 u_1 \nabla_2 v_1 & \nabla_2 u_2 \nabla_2 v_2 & \cdots & \nabla_2 u_n \nabla_2 v_n \\ \vdots & \vdots & \vdots & \vdots \\ \nabla_n u_1 \nabla_n v_1 & \nabla_n u_2 \nabla_n v_2 & \cdots & \nabla_n u_n \nabla_n v_n \end{bmatrix}. \end{align*} Am I right?

I didn't ever see "$:$" before, but I think it is so-called the Frobenius inner product, though its Wiki page doesn't mention this notation.

• You are correct. In some branches of Physics, e.g. Continuum Mechanics, this operation is called the double-dot product or double-contraction product. In the context of matrices, it is sometimes referred to as the inner product or trace product. – lynn Dec 2 '17 at 19:41
• Your expansion uses the elementwise/Hadamard product and produces a matrix result. The trace/Frobenius product produces a scalar result; it is calculated by summing all the terms in the Hadamard result. A colon usually indicates the trace/Frobenius product. The Hadamard product symbol is either $(\circ)$ or $(\odot)$. – greg Nov 23 '18 at 3:15