Can someone please help clarify how to calculate the outer product of two basis vectors, assuming $\mathbb{R}3$? From the definition, I understand that $\textbf{a} \otimes \textbf{b}$ $= a_ib_j = c_{ij}$. Another way to write that is $\textbf{a} \otimes \textbf{b}$ $= a_ib_j\textbf{e}{_i}$ $ \otimes \textbf{e}{_j}$.
Where: $e_1 = \left[\matrix{1\cr0\cr0}\right] , e_2 = \left[\matrix{0\cr1\cr0}\right], e_3 = \left[\matrix{0\cr0\cr1}\right].$
When I evaluate the outer product of $\textbf{e}{_i} \otimes \textbf{e}{_j}$ I get the following matrix:
$$ \left[\matrix{e_1e_1 & e_1e_2 & e_1e_3\cr e_2e_1 & e_2e_2 & e_2e_3 \cr e_3e_1 & e_3e_2 & e_3e_3}\right]$$
My intuition tells me this should equal the identity matrix, but when I actually calculate the values, I get stuck doing the following:
$e_1e_1 = \left[\matrix{1\cr0\cr0}\right]\left[\matrix{1\cr0\cr0}\right]$
There isn't a dot product between the two so I am not sure how to evaluate. I could also be writing down the matrix incorrectly since the examples I am looking at are using the components of vectors, and not entire vectors.
Thank you!
I did some digging and found what I was missing. What you get is a 2nd order identity tensor if you have mutually orthogonal basis vectors. $e_1e_1$ is just the outer product again, but they are vectors, not components of vectors (yet). Here is a breakdown of what I found which may help people reading this in the future:
Notation: Outer product, Dyadic, Tensor product
All of these terms are equivalent.
Dyadic product of (2) vectors $\textbf{a}$ and $\textbf{b}$ denoted by $\textbf{ab}$
Outer product of (2) column vectors $\textbf{a} \otimes \textbf{b}$ or $\textbf{ab}^T$
Tensor product of (2) vectors $\textbf{a}$ and $\textbf{b}$ : $\textbf{a} \otimes \textbf{b}$
I will do a short example which will help show what a 2nd order identity tensor is.
$\textbf{a} = a_1\textbf{i} + a_2\textbf{j}+ a_3\textbf{k}$
$\textbf{b} = b_1\textbf{i} + b_2\textbf{j}+ b_3\textbf{k}$
$\textbf{i},\textbf{j},\textbf{k}$ can also be denoted by $\textbf{e}{_{(1)}},\textbf{e}{_{(2)}},\textbf{e}{_{(3)}}$
$\textbf{ab} = a_1b_1\textbf{ii} + a_1b_2\textbf{ij} + ... + a_3b_3\textbf{kk}$
or
$\textbf{ab} = a_1b_1\textbf{e}{_{(1)}}\textbf{e}{_{(1)}} + a_1b_2\textbf{e}{_{(1)}}\textbf{e}{_{(2)}} + ... + a_3b_3\textbf{e}{_{(3)}}\textbf{e}{_{(3)}}$
$\textbf{ii} = \textbf{e}{_{(1)}}\textbf{e}{_{(1)}} = \left[\matrix{1 & 0 & 0\cr 0 & 0 & 0 \cr 0 & 0 & 0}\right]$
$\textbf{jj} = \textbf{e}{_{(2)}}\textbf{e}{_{(2)}} = \left[\matrix{0 & 0 & 0\cr 0 & 1 & 0 \cr 0 & 0 & 0}\right]$
$\textbf{kk} = \textbf{e}{_{(3)}}\textbf{e}{_{(3)}} = \left[\matrix{0 & 0 & 0\cr 0 & 0 & 0 \cr 0 & 0 & 1}\right]$
$\textbf{ij} = \textbf{e}{_{(1)}}\textbf{e}{_{(2)}} = \left[\matrix{0 & 1 & 0\cr 0 & 0 & 0 \cr 0 & 0 & 0}\right]$
and so on.
If: $\textbf{A} = 2\textbf{ij} + 3\textbf{ji} - 8\pi\textbf{jk} + \sqrt{2}\textbf{kk}$
$ = 2\textbf{e}{_{(1)}}\textbf{e}{_{(2)}} + 3\textbf{e}{_{(2)}}\textbf{e}{_{(1)}} - 8\pi\textbf{e}{_{(2)}}\textbf{e}{_{(3)}} + \sqrt{2}\textbf{e}{_{(3)}}\textbf{e}{_{(3)}}$
Which equals: $\textbf{A} = \left[\matrix{0 & 2 & 0\cr 3 & 0 & -8\pi \cr 0 & 0 & \sqrt{2}}\right]$
This will become clearer when we show the following:
Lets show the original problem of: $\textbf{e}{_i} \otimes \textbf{e}{_j}$
$\textbf{e}{_i} \otimes \textbf{e}{_j}$ = $\textbf{e}{_i}\textbf{e}{_j}$
$\textbf{e}{_1} = \left[\matrix{1\cr0\cr0}\right] , \textbf{e}{_2} = \left[\matrix{0\cr1\cr0}\right], \textbf{e}{_3} = \left[\matrix{0\cr0\cr1}\right]$.
$\textbf{e}{_i}\textbf{e}{_j}$ = $\left[\matrix{\textbf{e}{_1}\textbf{e}{_1} & \textbf{e}{_1}\textbf{e}{_2} & \textbf{e}{_1}\textbf{e}{_3}\cr \textbf{e}{_2}\textbf{e}{_1} & \textbf{e}{_2}\textbf{e}{_2} & \textbf{e}{_2}\textbf{e}{_3} \cr \textbf{e}{_3}\textbf{e}{_1} & \textbf{e}{_3}\textbf{e}{_2} & \textbf{e}{_3}\textbf{e}{_3}}\right]$
These entries of the above matrix are the actual vectors $\textbf{e}{_1}, \textbf{e}{_2},$ and $\textbf{e}{_3}$.
Now we need to calculate each cell of the above matrix. Lets start with $\textbf{e}{_1}\textbf{e}{_1}$
$\textbf{e}{_1}\textbf{e}{_1} = \left[\matrix{e_1e_1 & e_1e_2 & e_1e_3\cr e_2e_1 & e_2e_2 & e_2e_3 \cr e_3e_1 & e_3e_2 & e_3e_3}\right]$
These are components of vector $\textbf{e}{_1}$.
This is equal to : $\left[\matrix{1*1 & 1*0 & 1*0\cr 0*1 & 0*0 & 0*0 \cr 0*1 & 0*0 & 0*0}\right] = \left[\matrix{1 & 0 & 0\cr 0 & 0 & 0 \cr 0 & 0 & 0}\right]$
Lets continue with the second cell: $\textbf{e}{_1}\textbf{e}{_2}$
I am going to rename $\textbf{e}{_2}$ for clarity because it can get a little confusing. $\textbf{e}{_2} = \textbf{b}{_2} = \left[\matrix{0\cr 1\cr 0}\right]$
$\textbf{e}{_1}\textbf{e}{_2} = \textbf{e}{_1}\textbf{b}{_2} = \left[\matrix{e_1b_1 & e_1b_2 & e_1b_3\cr e_2b_1 & e_2b_2 & e_2b_3 \cr e_3b_1 & e_3b_2 & e_3b_3}\right]$
These are components of vectors $\textbf{e}{_1}$ and $\textbf{b}{_1} $.
This is equal to : $\left[\matrix{1*0 & 1*1 & 1*0\cr 0*0 & 0*1 & 0*0 \cr 0*0 & 0*1 & 0*0}\right] = \left[\matrix{0 & 1 & 0\cr 0 & 0 & 0 \cr 0 & 0 & 0}\right]$
And we continue for all the other entries of the original matrix. What we get is the following 27x27 matrix:
$\textbf{e}{_i} \otimes \textbf{e}{_j}$ = $\textbf{e}{_i}\textbf{e}{_j} = \left[\matrix{[3\times3] & [3\times3] & [3\times3] \cr [3\times3] & [3\times3] & [3\times3] \cr [3\times3] & [3\times3] & [3\times3]}\right]$
For mutually orthogonal basis vectors ($\textbf{e}{_i} \cdot \textbf{e}{_j} = 0)$, this is a second order identity tensor of 3-dimensional space. You can see how it works by the example given above for $\textbf{A}$.