Derivative of an expression with matrix transpose

Question

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

Answer 1

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

Answer 2

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 $$\eqalign{ Ce &= \lambda e \cr }$$