Sample covariance mean-corrected vector proof

Question

Prove that $$(n-1)S = X^TX -{1\over{n}}(X^T\vec1)(\vec1^TX) = X^TX-n\vec{\bar x}\vec{\bar x}^T$$

My attempt so far goes like this

$$S = {1\over{n-1}}X_m^TX_m$$

Edit: Where $X_m$ is the mean corrected matrix. So

$$(n-1)S = (X-\vec1\vec{\bar x^T})^T(X-\vec1\vec{\bar x^T})$$ $$=(X^T-\vec{\bar x}\vec{1^T})(X-\vec1\vec{\bar x^T})$$ $$=X^TX -X^T\vec1\vec{\bar x^T}-\bar x\vec1^TX + \vec {\bar x}\vec1^T\vec1\vec {\bar x^T}$$

I think the last term can be written as $n\vec {\bar x}\vec {\bar x^T}$ but beyond that I'm not sure where to go from there (if I'm even right so far).

Answer 1

After the OP's clarifications, $X$ is not a vector but a matrix, and $S$ is a variance covariance matrix.

We have that $\vec{1^T} \vec1= n$, a scalar.

Also, $X^T\vec1$ produces a $p \times 1$ column vector with $\sum_{i=1}^n x_{ji}$ in each position, for the $j=1,...,p$ variables. Multiply and divide by $n$ to get

$$X^T\vec1\vec{\bar x^T} = n\cdot \left(\frac 1n X^T\vec1\right)\vec{\bar x^T} =n\vec{\bar x}\vec{\bar x^T}$$

I guess the OP can take it from here.