0
$\begingroup$

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 $X^T X$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.