Referring to this post: https://stats.stackexchange.com/questions/164223/proof-of-loocv-formula

The line which says $\sum_{t=1}^TX_t'X_t=X'X$ is the result I'm trying to interpret.

Or in perhaps more standard notation:

Question. Prove that $ X^{T} X = \sum_{i=1}^{n} \mathbf{x}_i \mathbf{x}_i^T $ where $ \mathbf{x}_i $ is the $ i^{th} $ row of a matrix $ X $ (as a column vector).

If someone could please explain why this is true, it would be very helpful. I'm not quite sure how I would come up with this "decomposition" of $X^T X$.

  • $\begingroup$ I believe you can prove the entries of both sides are equal by using the matrix multiplication formula. $\endgroup$ – WeakestTopology Nov 19 '19 at 11:55
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    $\begingroup$ If $x_i$ is a row vector, $x_i x_i^+$ is a scalar and the identity does not make much sense. Are you sure that $x_i $ is a row vector or that the transpose operations are well reported ? $\endgroup$ – Thomas Nov 19 '19 at 12:02
  • $\begingroup$ @Thomas Yes you're correct, thanks for the comment, clarified in edit $\endgroup$ – user523384 Nov 19 '19 at 12:04

With implicit summation over indices,$$(X^TX)_{jk}=X^T_{ji}X_{ik}=(x_i)_j(x_i)_k=(x_ix_i^T)_{jk}.$$

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