How to prove $\sum_{i=1}^{n}{(x_i-\bar{x}_n)(x_i-\bar{x}_n)^2}$ can be expressed as $XCX^\top$ with $C = I_n -\frac{1}{n} 11^\top $?
I read this answer but it did not lead me anywhere because I couldn't grasp the concept of the Kronecker product and also its beyond the syllabus.
Expressions involving sums of indexed vectors can be very difficult to comprehend. My "goto" identity for unraveling them is the following $$\eqalign{ S &= \sum_{m=1}^M\sum_{n=1}^N A_{mn}\,y_m\,z_n^T \\ &= \sum_{m=1}^M\sum_{n=1}^N (Ye_m)A_{mn}(Ze_n)^T \\ &= Y\left(\sum_{m=1}^M\sum_{n=1}^N A_{mn}\;e_me_n^T\right)Z^T \\ &= YAZ^T }$$ where $y_m$ is the $m^{th}$ column of matrix $Y$, and $z_n$ is the $n^{th}$ column of matrix $Z$, and $A_{mn}$ are the elements of the matrix $A$.
In the current problem, $A$ is the identity matrix and $\,y=z=(x-\mu),\;$ where $\,\mu=\tfrac 1N(X{\tt1})$ is the row-mean vector of $X$.
For typing convenience, define the matrices $$\eqalign{ J &= \tfrac 1N {\tt1}{\tt1}^T \quad\implies\quad J^2=J=J^T \\ U &= \big[\mu\;\;\mu\;\;\ldots\;\mu \big] \;=\; \mu{\tt1}^T \;=\; XJ }$$ Then the sum can be expanded as $$\eqalign{ \sum_{n=1}^N (x_n-\mu)\,(x_n-\mu)^T &= \sum_{m=1}^M\sum_{n=1}^N \delta_{mn}\,(x_m-\mu)\,(x_n-\mu)^T \\ &= (X-U)I(X-U)^T \\ &= XX^T - UX^T - XU^T + UU^T \\ &= XX^T - (XJ)X^T - X(J^TX^T) + XJ^2X^T \\ &= XX^T - XJX^T \\ &= X(I-J)X^T \\ &= XCX^T \\ }$$