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This question already has an answer here:

Let be $Y$ the matrix of observations with $n$ lines and $m$ columns. Let be $X$ the centered matrix, where $X_{i,j} = Y_{i,j} - \overline{Y_{.,j}}$ , $i = 1:n$, $j = 1:m$

Edit : $\overline{Y_{.,j}}$ is the mean of each column

$X^{t}$ is the transpose matrix

What is this matrix $\frac{1}{n-1} X^{t}X$ ?

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marked as duplicate by Tim, JohnK, kjetil b halvorsen, gung - Reinstate Monica, Michael R. Chernick Apr 26 '17 at 12:46

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    $\begingroup$ The squared deviations from the median? $\endgroup$ – JohnK Apr 26 '17 at 11:30
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    $\begingroup$ Looks like a sample variance covariance matrix where the centering has been done using the median $\endgroup$ – IcannotFixThis Apr 26 '17 at 11:38
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It is the estimated variance-covariance matrix of the Y's. It is a square, symmetric matrix with the variances on the diagonal (for i = 1, 2, ..., n). The covariances are in the off-diagonal cells where the covariance of Yi and Yj is in the (i, j) cell.

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