I was studying a literature where in an equation the term $$\nabla p_i$$ is given and further it is given that $$\nabla$$ is a column wise gradient and $$p_i = p_i(x,t)$$, what exactly meaning of column-wise gradient?.

Further it is seen in literature that $$p_i$$ is a scalar quantity and I found one definition of matrix calculus on Wikipedia (Scalar by Vector), where it can observed that derivative of a scalar $$p_i$$ with respect to independent vector is a row vector $$\left[\frac{\partial p_i}{\partial x} \ \ \ \frac{\partial p_i}{\partial t}\right]$$ And further, as in general $$\nabla p_i$$ is a Jacobian of scalar $$p_i$$ and we can find definition Jacobia Matrix and determinants in which it can seen that $$\nabla p_i = \frac{\partial p_i}{\partial x_j} \ \ \ \text{for} \ \ j = 1,2 \ \ \ \text{and} \ \ \ x_1 = x, \ \ x_2 = t$$ if $$i = 1,2,...,n$$ then $$\nabla p_i$$ will be an $$n \times 2$$ matrix. And $$\nabla p_i = \begin{bmatrix} \frac{\partial p_1}{\partial x} & \frac{\partial p_1}{\partial t} \\ \frac{\partial p_2}{\partial x} & \frac{\partial p_2}{\partial t} \\ \vdots & \vdots \\ \frac{\partial p_3}{\partial x} & \frac{\partial p_3}{\partial t} \end{bmatrix}$$

Can we call it column-wise gradient? If not, then what is column-wise gradient and how we can express it in general form?

My next question is here about Symmetric Gradient. While googling, I found one expression of symmetric gradient showing curl curl of... where symmetric gradient $$\epsilon$$ of a two component vector $$u(x,t) = (u_1(x,t),u_2(x,t))$$ is given as (with some modification form) $$\epsilon(u) = \begin{bmatrix} \frac{\partial u_1}{\partial x} & \frac{1}{2}\left(\frac{\partial u_1}{\partial t} + \frac{\partial u_2}{\partial x}\right) \\ \frac{1}{2}\left(\frac{\partial u_1}{\partial t} + \frac{\partial u_2}{\partial x}\right) & \frac{\partial u_2}{\partial t} \\ \end{bmatrix}$$

But, how we can write symmetric gradient in more generalized form e.g. for $$u(x,t) = (u_1(x,t),u_2(x,t),...,u_n(x,t))$$ for $$n > 2$$

Will symmetric gradient always be a square matrix?

If the dimensionality of domain matches the dimensionality of the vector, the gradient is a square matrix and can be symmetrized, resulting in symmetric gradient: $$\frac{\pmb\nabla\mathbf{p}+(\pmb\nabla\mathbf{p})^\intercal}2,\qquad \text{or in index form}\qquad \frac{\nabla_jp_i+\nabla_ip_j}2=\nabla_{(j}p_{i)}$$