1
$\begingroup$

in my optimization course, we are given the following function:

$f = E^{T} C E - \lambda(E^T E - 1)$,

where $E, C$ are matrices, and $\lambda$ is a real number. In class, the lecturer wrote: $\partial L / \partial E = 0$, and then gets:

$CE + E^T C - 2\lambda E = 0$,

which is then simplified to:

$CE = \lambda E$.

Could someone explain how these last 2 lines are obtained? The partial derivative is with respect to $E$, but, how does one do it with respect to $E^T$?

Thanks! Thomas

$\endgroup$
  • $\begingroup$ have to tried to write it component wise? If not, you should. $\endgroup$ – Surb Jun 18 '17 at 23:15
  • $\begingroup$ No, how does one do that? $\endgroup$ – Thomas Moore Jun 18 '17 at 23:25
  • 1
    $\begingroup$ well $f=g-\lambda h$ with $g=E^TCE$ and $h=(E^TE-1)$. So $f_{i,j}=g_{i,j}-\lambda h_{i,j}$ and $g_{i,j} = (E^TCE)_{i,j}=\sum_{k} (E^TC)_{i,k}E_{k,j}=\sum_{k,l}E_{l,i}C_{l,k}E_{k,j}$ etc... and then you just use what you know about computing derivatives of polynomials. $\endgroup$ – Surb Jun 18 '17 at 23:33
  • 1
    $\begingroup$ This is maybe not the shortest way, but definitely the safest and once you have done several such computations you will become quick and efficient at them. $\endgroup$ – Surb Jun 18 '17 at 23:35
0
$\begingroup$

I don't know why you have to do the partial derivative in $E^T$ but in case you are interested in it, $\partial/\partial E^T = (\partial/\partial E)^T$.

$\endgroup$
  • $\begingroup$ There is no partial derivative in $E^T$, it is the expression to differentiate which contains an $E^T$ $\endgroup$ – Surb Jun 19 '17 at 10:52
  • $\begingroup$ It is asked in the question (the second to last paragraph) how to do a partial derivative in $E^T$, thus my answer. However I do not see other questions contained in this post. $\endgroup$ – Yining Wang Jun 19 '17 at 13:44
0
$\begingroup$

I believe there are some things in the problem that you misinterpreted, i.e. $E$ is actually a vector $e$, and $C$ is a symmetric matrix.

In that case the Lagrangian, its differential & gradient are $$\eqalign{ L &= e^TCe - \lambda(e^Te-1) \cr dL &= de^TCe + e^TC\,de - 2\lambda e^T\,de \cr &= e^TC^T\,de + e^TC\,de - 2\lambda e^T\,de \cr &= 2e^TC\,de - 2\lambda e^T\,de \cr &= 2(Ce-\lambda e)^T\,de \cr \frac{\partial L}{\partial e} &= 2(Ce-\lambda e) \cr }$$ Setting the gradient to zero yields $$\eqalign{ Ce &= \lambda e \cr }$$

$\endgroup$

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.