How to show sample correlation is sample covariance for standardized values?

Question

Given a matrix $X$ and the resulting sample correlation matrix $R$, consider the standardized observations:

$$\frac{(x_{jk} - \bar x)} {\sqrt{S_{kk}}} \quad k=1,2,...,p \quad j=1,2,...,n$$

Show that these standardized quantities have sample covariance matrix $R$.

Sample covariance is defined as:

$$\frac{1}{n} \sum\limits_{j=1}^n{(x_{ji} - \bar x_i)(x_{ji} - \bar x_k)}$$

Sample correlation coefficient:

$$r_{ik}=\frac{S_{ik}} {\sqrt{S_{ii}}\sqrt{S_{kk}}} = \frac{\sum\limits_{j=1}^n(x_{ji}-\bar x_i)(x_{jk}-\bar x_k)}{\sqrt{\sum\limits_{j=1}^n(x_{ji}-\bar x_i)^2\sum\limits_{j=1}^n(x_{jk}-\bar x_k)^2}}$$

It is also known that: $$R (correlation) = D^{-1/2}SD^{-1/2}$$ S = variance-covariance matrix

D = sample standard deviation matrix

How do I show that for standardized quantities sample correlation is just the sample covariance?

Attempted answer: For standardized variable, variance is defined as: $$\frac{1}{n}\sum_j\frac{(x_{jk} - \bar x_k)^2} {S_{kk}}=\frac{S_{kk}}{S_{kk}}=1, \quad k=1,2,...,p,$$

thus D = diagonal matrix with variance satisfies: $$ D = I $$ and since $$R (correlation) = D^{-1/2}SD^{-1/2}$$ we can conclude that $$ R=D $$

Answer 1

The variance of a standardized variable is $$\frac{1}{n}\sum_j\frac{(x_{jk} - \bar x_k)^2} {S_{kk}}=\frac{S_{kk}}{S_{kk}}=1, \quad k=1,2,...,p,$$ and so is, therefore, its standard deviation. Hence, $D$, which is a diagonal matrix with the variances (not the standard deviations, that is given by $D^{1/2}$) on the main diagonal, satisfies $D=I$, and therefore, $D^{1/2}=D^{-1/2}=I$, the identity matrix.

Notice we do not need to subtract the sample average, as a demeaned variable has mean zero by construction: $\sum_{j=1}^n(x_j-\bar x)=\sum_{j=1}^nx_j-n\bar x=0$.