What does it mean to take the gradient of a vector field? $\nabla \vec{v}(x,y,z)$? I only understand what it means to take the grad of a scalar field.
[Math] What does it mean to take the gradient of a vector field
multivariable-calculusVector Fields
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From my limited understanding you want the dyadic tensor product, not the outer product.
In more complex examples this is not always the case, but for ${\bf a},{\bf b}\in$ Euclidean space, the dyadic product is related to the outer product by $$ {\bf a}\otimes{\bf b}\equiv{\bf a}{\bf b}^T $$
Which agrees with your example.
When placed next to a scalar-valued function or vector field alike $\nabla$ is considered an operator. Subsequently, for the vector field case you want the dyadic product of $\nabla$ and the vector field it is acting on. Check out the section marked, three dimensional space on the wikipedia page for dyadics https://en.wikipedia.org/wiki/Dyadics
A tensor is just a multilinear, scalar-valued function. If I write $V \multimap W$, for the collection of linear functions from a vector space $V$ to a vector space $W$, then, over the reals, a rank-$n$ tensor is just $V^{\otimes n}\multimap\mathbb R$ where $V^{\otimes n}$ means the $n$-fold tensor product, e.g. $V^{\otimes 2} = V\otimes V$. A tensor field is just a smoothly indexed family of tensors.
For simplicity, I'll just talk about the manifold $\mathbb{R}^n$, but anywhere I explicitly write out $\mathbb{R}^n$ (as opposed to $V$), you could just as well use a submanifold of $\mathbb{R}^n$, e.g. a 1-dimensional curve in $\mathbb{R}^n$. A rank-$k$ tensor field on $\mathbb{R}^n$ is a (suitably smooth) function $\tau : \mathbb{R}^n \to (V^{\otimes k} \multimap \mathbb R)$ where $V$ is itself $\mathbb{R}^n$. Now say we write the directional derivative of $\tau$ in some direction $v \in V$ at $x \in \mathbb{R}^n$ as $D(\tau)(x; v)$. The result itself would be a rank-$k$ tensor, i.e. $D(\tau)(x; v) : V^{\otimes k} \multimap \mathbb R$. So what is the type of $D(\tau)$ itself. Well we know the type of $\tau$ and we know the type of $V$ and we know $D(\tau)(x; v)$ is linear in $v$ and non-linear in $x$. So we have $$D(\tau) : \mathbb{R}^n \to (V \multimap (V^{\otimes k}\multimap\mathbb{R}))$$ but it is easy to show that $(V \multimap (V^{\otimes k}\multimap\mathbb{R})) \cong (V\otimes V^{\otimes k} \multimap \mathbb R) = (V^{\otimes k+1}\multimap \mathbb R)$. Which is to say $D(\tau)$ as a function of $x$ alone — essentially currying $D(\tau)$ — is a rank-$(k+1)$ tensor field.
So to answer your question, you find the gradient of a tensor field by viewing the directional derivative as a linear function of the direction. When you have a basis, as you do for $\mathbb{R}^n$, this linear function can be represented by a vector of partial derivatives (which are directional derivatives along the basis directions). A bit more concretely, $$D(\tau)(x)_i = \frac{\partial\tau}{\partial x_i}(x)$$
And yes, even high school, single variable calculus was doing this for the $n = 1$ case. It's just that $\mathbb{R}^{\otimes k} \cong \mathbb{R}$ for any $k$, and $(\mathbb{R}\multimap\mathbb{R}) \cong \mathbb{R}$.
Best Answer
The gradient of a vector is a tensor which tells us how the vector field changes in any direction. We can represent the gradient of a vector by a matrix of its components with respect to a basis. The $(\nabla V)_{\text{ij}}$ component tells us the change of the $V_j$ component in the $\pmb{e}_i$ direction (maybe I have that backwards). You can check out the Wikipedia article for the details of calculating the components.
To get a physical picture of its meaning we can decompose it into 1) the trace (the divergence) 2) an anti-symmetric tensor (the curl) 3) a traceless symmetric tensor (the shear)
If the vector field represents the flow of material, then we can examine a small cube of material about a point. The divergence describes how the cube changes volume. The curl describes the shape and volume preserving rotation of the fluid. The shear describes the volume-preserving deformation.