Computation of $(\textbf{u}\cdot \nabla)\textbf{u}$ in the Navier stokes equation: inner or outer product? The Navier Stokes equation for an
incompressible flow of a homogenenous fluid reads
    \begin{align*}
 \nabla\cdot \textbf{u} &= 0\\
 \rho \frac{\partial }{\partial t}\textbf{u} + \rho(\textbf{u}\cdot \nabla)\textbf{u} &= -\nabla p + \mu \nabla ^2 \textbf{u}
 \end{align*}
Regarding, 
$$(\textbf{u}\cdot \nabla)\textbf{u}
$$
the two components of this term are: 
$$
\textbf{u} = \begin{bmatrix}
u \\ v
\end{bmatrix}  \quad \text{and}  \quad 
\nabla = \begin{bmatrix}
\frac{\partial }{\partial x} \\ \frac{\partial }{\partial y}
\end{bmatrix}
$$
and the sign $\cdot$ being the dot product. However in the example that I was given, the computation has been done as follows:
\begin{align*}
(\textbf{u}\cdot \nabla)\textbf{u} &= \Big(\begin{bmatrix}
u \\ v
\end{bmatrix} \begin{bmatrix}
\frac{\partial }{\partial x} & \frac{\partial }{\partial y}
\end{bmatrix}\Big)\begin{bmatrix}
u \\ v
\end{bmatrix}
\end{align*}
where $\begin{bmatrix}
u \\ v
\end{bmatrix} \begin{bmatrix}
\frac{\partial }{\partial x} & \frac{\partial }{\partial y}
\end{bmatrix}$ is an outer product (Notation: $\otimes$ and not a dot product $\cdot$) and hence a matrix. So for me, the notation $\textbf{u}\cdot \nabla$ is misleading. What am I getting wrong? 
Example
Compute the acceleration of a fluid at $(x,y) = (1,2)$ at time $t = 2$. with velocity $(u,v) = (1, x^2t)$
Fluid acceleration
$$
 \frac{D\textbf{u}}{Dt} = \frac{\partial \textbf{u}}{\partial t} +(\textbf{u} \cdot  \nabla) \textbf{u}
 $$
solution (given to me): 
    \begin{align*}
 \frac{D\textbf{u}}{Dt} &=\frac{\partial \textbf{u}}{\partial t} +(\textbf{u} \cdot  \nabla) \textbf{u}\\
 &=\frac{\partial }{\partial t}\begin{bmatrix}
 1\\x^2t
 \end{bmatrix} +
 \Big( 
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix} 
 \cdot 
 \begin{bmatrix}
 \frac{\partial}{\partial x}&\frac{\partial}{\partial y}
 \end{bmatrix}\Big)
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
 &=\begin{bmatrix}
 0\\x^2
 \end{bmatrix} +
 \begin{bmatrix}
 \frac{\partial 1}{\partial x} &  \frac{\partial 1}{\partial y}\\
 \frac{\partial x^2t}{\partial x} &  \frac{\partial x^2t}{\partial y}
 \end{bmatrix}\cdot
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
 &=\begin{bmatrix}
 0\\x^2
 \end{bmatrix} + 
 \begin{bmatrix}
 0 &  0\\
 2xt &  0
 \end{bmatrix}\cdot
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
 &=\begin{bmatrix}
 0\\x^2
 \end{bmatrix} + 
 \begin{bmatrix}
 0\\2xt
 \end{bmatrix}\\
 &=
 \begin{bmatrix}
 0\\x^2 + 2xt
 \end{bmatrix}\\
 \end{align*}
Now if I try to compute firs the dot product inside the parenthesis (instead of the outer product, as it was proposed in the answers), I get
\begin{align*}
&=\frac{\partial }{\partial t}\begin{bmatrix}
 1\\x^2t
 \end{bmatrix} +
 \Big( 
 \begin{bmatrix}
 1&x^2t
 \end{bmatrix} 
 \cdot 
 \begin{bmatrix}
 \frac{\partial}{\partial x}\\\frac{\partial}{\partial y}
 \end{bmatrix}\Big)
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
&=\frac{\partial }{\partial t}\begin{bmatrix}
 1\\x^2t
 \end{bmatrix} +
 \Big( \frac{\partial 1}{\partial x} + \frac{\partial x^2t}{\partial y}\Big)
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
&=\frac{\partial }{\partial t}\begin{bmatrix}
 1\\x^2t
 \end{bmatrix} +
 \Big( 0 + 0 \Big)
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
&=\begin{bmatrix}
 0 \\x^2
 \end{bmatrix} +
 \Big( 0 \Big)
 \begin{bmatrix}
 1\\x^2t
 \end{bmatrix}\\
&=\begin{bmatrix}
 0\\x^2
 \end{bmatrix}
\end{align*}
 A: Either the computation has been done incorrectly or you have misunderstood it. It should be
$$
\begin{align*}
(\textbf{u}\cdot \nabla)\textbf{u} 
&= \Big(\begin{bmatrix}
u & v
\end{bmatrix} 
\begin{bmatrix}
\frac{\partial }{\partial x} \\ 
\frac{\partial }{\partial y}
\end{bmatrix}
\Big)
\begin{bmatrix}
u \\ v
\end{bmatrix}
\\
&= \Big(
u \frac{\partial }{\partial x} + v \frac{\partial }{\partial y}
\Big)
\begin{bmatrix}
u \\ v
\end{bmatrix}
\\
&= 
\begin{bmatrix}
u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}
\\ 
u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}
\end{bmatrix}
\end{align*}
$$
A: $\mathbf{u} \cdot \nabla$ is the dot product of a vector $\mathbf{u}$ and a "vector" $\nabla$. Thus it should be interpreted as a "scalar". Therefore $(\mathbf{u} \cdot \nabla) \mathbf{u}$ is a scalar multiplication of the "scalar" $\mathbf{u} \cdot \nabla$ and the vector $\mathbf{u}$. In other words,
\begin{align}
((\mathbf{u} \cdot \nabla) \mathbf{u})_i
&= (\mathbf{u} \cdot \nabla) \mathbf{u}_i \\
&= \left( \sum_j \mathbf{u}_j \nabla_j \right) \mathbf{u}_i \\
&= \sum_j \mathbf{u}_j \nabla_j \mathbf{u}_i
\end{align}
Notice that $(\mathbf{u} \cdot \nabla) \mathbf{u} = \mathbf{u} \cdot (\nabla \mathbf{u})$, where $\nabla \mathbf{u}$ is an outer product between the "vector" $\nabla$ and the vector $\mathbf{u}$:
\begin{align}
((\mathbf{u} \cdot \nabla) \mathbf{u})_i
&= (\mathbf{u} \cdot (\nabla \mathbf{u}))_i \\
&= \left(\sum_j \mathbf{u}_j (\nabla \mathbf{u})_j\right)_i \\
&= \left(\sum_j \mathbf{u}_j (\nabla_j \mathbf{u})\right)_i \\
&= \sum_j (\mathbf{u}_j (\nabla_j \mathbf{u}))_i \\
&= \sum_j \mathbf{u}_j \nabla_j \mathbf{u}_i
\end{align}
A: Let
$$\eqalign{
x &= \pmatrix{x\\y},\quad u = \pmatrix{u\\v}=\pmatrix{1\\tx^2} \\
}$$
Then the matrix-valued gradient $\,G=(\nabla u)^T$ has components
$$\eqalign{
G_{ij}
 &= \frac{\partial u_i}{\partial x_j} 
\;=\; \pmatrix{
  \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y} \\
  \frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}
 }
  = \pmatrix{0&0\\2xt&0} \\
}$$
and can be used to write the material derivative as
$$\eqalign{
\frac{Du}{Dt}
  &= \frac{\partial u}{\partial t} + Gu \\
  &= \pmatrix{0\\x^2} + \pmatrix{0&0\\2xt&0}\pmatrix{1\\tx^2} \\
  &= \pmatrix{0\\x^2} + \pmatrix{0\\2xt} \\
  &= \pmatrix{0\\x^2+2xt} \\
}$$
which can be evaluated at $\;(t,x,y)=(2,1,2)$
$$\eqalign{
\frac{Du}{Dt} &= \pmatrix{0\\1^2+2(1)(2)} = \pmatrix{0\\5} \\
}$$
