In my quantum mechanics course, the lecturer do the following definition for outer product, then equate it a matrix. Then, she want us to show this equality. $$\sum \limits_{i}^{n}\sum \limits_{j}^{n} \text{out}(\vec e_{i},\vec e_{j}) \equiv E_{ij} $$
However, when I perform the calculations, I found a unit matrix as you can see from the following steps. For the sake of simplicity, I did not write the summation signs. $$\begin{equation*} \begin{split} &= \vec e_{i} \, \vec e_{j}\,^{T} \\ &= \vec e_{i} \, \vec e_{i}\,^{T} \delta_{ij} \\ &= \vec e_{i} \, \vec e_{i}\,^{T} \\ &= I \end{split} \end{equation*}$$
I is an nxn unit matrix. I want to also mention that $e_i,e_j$ are linearly independent unit vectors.
Where is my mistake?
Consider this example: $$v = (1,2,3),\ w= (4,5,6)^\top$$ Then the outer product would be $$ \begin{pmatrix} 1*4 & 1*5 & 1*6\\ 2*4 & 2*5 & 2*6 \\ 3*4 & 3*5 & 3*6 \end{pmatrix} $$
Considering $e_1 =(1,0,0), e_3 = (0,0,1)$ you will get $$E_{13} = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} $$ $E_{13}$ is the matrix which just zeros expect at the position $(1,3)$
In general you get $1$ for the outer product of two unity vectors when $i=j$ and $0$ if $i\neq j$