Would require some help on this

$\sum{r_i^2 * e_i^2}$

would it be equivalent to

$(re)^T(re)$. It does seem to be equivalent but i cannot seem to prove it.

Assuming r and e are both vectors.


  • $\begingroup$ If $r$ and $e$ are both vectors, what does $re$ mean? $\endgroup$ – NickD Feb 9 at 3:52
  • $\begingroup$ @NickD r dot product e $\endgroup$ – aceminer Feb 9 at 3:56
  • $\begingroup$ Then what is $(re)^T$? $\endgroup$ – NickD Feb 9 at 4:52
  • $\begingroup$ Hmmm. The transpose of the dot product of r and e $\endgroup$ – aceminer Feb 9 at 4:56
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    $\begingroup$ Let $R={\rm Diag}(r),$ i.e. a diagonal matrix with the vector $r$ as the diagonal elements. Then your function can be written as $$f = (Re)^T(Re) = e^TR^TRe$$ $\endgroup$ – greg Feb 9 at 6:33

Just write it out componentwise for two dimensional vectors.

If $re$ is the dot product as you say in the comments it is not. $\sum r_i^2*e_i^2=r_1^2e_1^2+r_2^2e_2^2$ while $re$ is a scalar $(re)=r_1e_1+r_2e_2$ and $(re)^T(re)$ is the square of this which includes a term $2r_1e_1r_2e_2$

If $(re)$ is a vector that is the componentwise product of $r$ and $e$ it is true. Write it out for yourself.

  • $\begingroup$ So how do I express this summation in wuadaratic form? $\endgroup$ – aceminer Feb 9 at 4:27
  • $\begingroup$ I don't know the word wuadaratic. When I put it into a famous search engine it changes it into quadratic. $\endgroup$ – Ross Millikan Feb 9 at 4:28
  • $\begingroup$ Yes it was a typo i realized my mistake $\endgroup$ – aceminer Feb 9 at 4:36
  • $\begingroup$ I don't understand what "express a summation in quadratic form" means.Your summation is a product of squares. $\endgroup$ – Ross Millikan Feb 9 at 4:38
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    $\begingroup$ Your first expression is a fine way to express it. I haven't seen a need for this, so I doubt there is an easier way. Componentwise multiplication of vectors is rarely seen because it doesn't transform nicely when you change the basis. Why do you need this? $\endgroup$ – Ross Millikan Feb 9 at 5:06

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