I came across this problem when going over some material related to shear stress vector.
As far as I know the symbol $\nabla$ has a couple of different meanings.
Let $\vec{i},\vec{j},\vec{k}$ be the unit normal vectors in the $x,y$ and $z$ directions.
Gradient:
When applied to a scalar field $u(x,y,z)$ it gives a vector: that is
$$\nabla u=\frac{\partial u}{\partial x}\vec{i}+\frac{\partial u}{\partial y}\vec{j}+\frac{\partial u}{\partial z}\vec{k}$$
Divergence:
When applied to a vector $\vec{u}=u\vec{i}+v\vec{j}+w\vec{k}$:
$$\nabla\cdot\vec{u}= \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}$$
Curl:
Again applied for the $\vec{u}=u\vec{i}+v\vec{j}+w\vec{k}$
$$\nabla\times\vec{u}=
\begin{vmatrix}
\vec{i}&\vec{j} &\vec{k}\\
\frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\
u&v &w \\
\end{vmatrix}$$
However, for the statement (obtained from: here):
$$\vec{u}=u_x\vec{i}_x+u_y\vec{i}_y+u_z\vec{i}_z$$ with $\vec{n}=\vec{i}_y$
I cannot understand why:
$$(\vec{\nabla}\vec{u})^T\cdot \vec{n}=\frac{\partial v_y}{\partial y}\vec{i_y}$$
Can someone please explain this to me.
Appreciate your help
$$\left(\nabla \vec{u}\right)^T = \partial_i u_j \vec{e}_i \otimes \vec{e}_j= \begin{pmatrix} \partial u_x /\partial x \quad \partial u_y /\partial x \quad \partial u_z /\partial x \\ \partial u_x /\partial y \quad \partial u_y /\partial y \quad \partial u_z /\partial y \\ \partial u_x /\partial z \quad \partial u_y /\partial z \quad \partial u_z /\partial z \\ \end{pmatrix} $$ (in the middle the $e_i$ are the unit vectors, summation convention used)