# Dot product of an identity matrix

I am doing some work for my fluid dynamics class. I was trying to do a simple proof, but hit a road block. I have an expression where a 3-vector say $$\vec{u}$$ dotted with a 3x3 identity matrix $$\bf{I}$$, and I am not sure what the result would be. I guess that it should just be $$\vec{u}$$ however I was taught (perhaps incorrectly) that the dot product results in a scalar thus meaning this operation maybe invalid altogether, I could be wrong but if I am right I would like to perhaps see a simple proof of why this works. Thanks for all your time!

question:

$$\vec{u} \cdot \bf{I} \stackrel{?}{=} \vec{u}$$

Context:

I am doing a short proof in fluid dynamics. Show that the Eulerian operator $$\frac{D}{Dt} \equiv \frac{\partial}{\partial{t}} + \vec{u} \cdot \nabla$$ acting on a position vector $$\vec{r}$$ (where $$\vec{r} = (x,y,z)$$) gives the fluid velocity $$\vec{u}$$. That is:

$$\frac{D \vec{r}}{Dt} = \frac{\partial \vec{r}}{\partial{t}} + \vec{u} \cdot \nabla \vec{r} \stackrel{?}{=} \vec{u}$$ , for $$\vec{r} = (x,y,z)$$

So first step I simply apply the Eulerian operator to the position vector:

$$\rightarrow \frac{D \vec{r}}{Dt} = \frac{\partial \vec{r}}{\partial{t}} + \vec{u} \cdot \nabla \vec{r}$$

since $$\vec{r}$$ is a position vector its partial derivative with respect to time vanishes, and we are left with:

$$\rightarrow\frac{D \vec{r}}{Dt} = \ \vec{u} \cdot \nabla \vec{r}$$

doing matrix multiplication with $$\nabla$$ and $$\vec{r}$$ I get the 3x3 identity matrix $$\bf{I}$$:

$$\rightarrow\frac{D \vec{r}}{Dt} = \vec{u} \cdot \bf{I}$$

This is the point I get stuck on, I have no familiarity with the subject of fluid dynamics prior so I am unsure of how to proceed. Since the proof requests me to show that the Eulerian operator of a position vector gives the fluid velocity, I can take a pretty good guess that $$\vec{u} \cdot \bf{I} = \vec{u}$$ but someone please show and confirm. Thanks!

• Is the dot here just matrix multiplication? – Lord Shark the Unknown Sep 29 '18 at 4:26
• There is not enough context here to determine what the notation even means. – Hans Lundmark Sep 29 '18 at 8:54
• I have added context @HansLundmark – QuantumPanda Sep 29 '18 at 15:50
• How did you end up with a matrix for $\nabla\vec r$? – amd Sep 29 '18 at 20:50
• Matrix multiplcation, 3x1 vector times a 1x3 vector gives a matrix. Also known as the outer product or diadic product as my professor has told me – QuantumPanda Sep 29 '18 at 22:45

I believe everything will be clear if you think of $$\vec{u} \cdot \nabla$$ as single operator $$\vec{u} \cdot \nabla = u_1 \frac{\partial}{\partial x_1} + u_2 \frac{\partial}{\partial x_2} + u_3 \frac{\partial}{\partial x_3}$$ and then apply the whole thing to $$\vec{r} = (x_1,x_2,x_3)$$ instead of doing the gradient first and then the dot product. Like this: $$\left( u_1 \frac{\partial}{\partial x_1} + u_2 \frac{\partial}{\partial x_2} + u_3 \frac{\partial}{\partial x_3} \right) (x_1,x_2,x_3) = u_1 (1,0,0) + u_2 (0,1,0) + u_3 (0,0,1) .$$