Is Lagrange multiplier useful in this Constrained optimization problem? I'm trying to numerically solve the following optimization problem.
Let $u \in R^{n \times n_c}$ be the variable we wish to find.
Let $\phi$ be the objective:
$\phi = Tr(u^\top L u) + 1/2 ||u - u^{obs} ||_2^2 $   st. $u e_{n_c} = 0_n$
where $u^{obs}$ is given, $L \in R^{n \times n}$ is a known graph Laplacian, and $e_{n_c}$ is the vector of ones, with length $n_c$. 
We can rewrite this using Lagrange multipliers as:
$\phi = Tr(u^\top L u) + 1/2 ||u - u^{obs} ||_2^2 + \lambda_n^\top u e_{n_c}$
where $\lambda_n$ is the lagrange multiplier, which is a vector of length $n$.
The theory now says to take the derivative of the objective with regards to $u$ and $\lambda$, which gives me:
$\frac{ \partial \phi }{\partial u} = 2 L u + (u - u^{obs}) + \lambda_n e_{n_c}^\top = 0_{n \times n_c}$
$\frac{\partial \phi} {\partial \lambda} = u e_{n_c} = 0_n$
Now the idea is to combine these two equations and eliminate $\lambda$ if possible, but I don't see any way I can combine this and get something I could solve algebraically or numerically.
Does anyone have any suggestions as to how I could proceed?
 A: The centering matrix is constructed from the identity matrix and the all-ones vector
$$\eqalign{
C = I - \frac{ee^T}{e^Te} \quad\implies\quad
 C^2=C=C^T,\quad Ce = 0
}$$
Construct the matrix $U$ from $C$ and an unconstrained matrix variable $X$.
$$\eqalign{
U &= XC \quad\implies\quad 
Ue &= 0,\quad
UC &= U \\
}$$
Write the objective function and calculate its gradient wrt $X$.
$$\eqalign{
\phi &= L:UU^T + \tfrac{1}{2}(U-U_{obs}):(U-U_{obs}) \\
d\phi
 &= L:(U\,dU^T+dU\,U^T) + (U-U_{obs}):dU \\
 &= \Big((L+L^T)U + U-U_{obs}\Big):dU \\
 &= \big(LU+L^TU + U-U_{obs}\big):dX\,C \\
 &= (LU + L^TU + U - U_{obs}C):dX \\
 &= ((L + L^T + I)U - U_{obs}C):dX \\
\frac{\partial \phi}{\partial X}
 &= (L + L^T + I)U - U_{obs}C \\
}$$
Set the gradient to zero and solve for the optimal matrix.
$$\eqalign{
 0 &= (L + L^T + I)U - U_{obs}C \\
 U &= (L + L^T + I)^{-1}U_{obs}C \\
}$$
NB: A colon is being used as a convenient product notation for the trace, i.e.
$$A:B = {\rm Tr}(A^TB)$$
Terms in such products can be rearranged via the cyclic property of 
the trace, e.g. 
$$\eqalign{A:BC = B^TA:C = AC^T:B}$$
Update
There was a question about where $C$ came from.
Consider the general solution of a linear equation
$$AX=B \quad\implies\quad A = BX^+ + Y(I-XX^+)$$
where $X^+$ denotes the pseudoinverse of $X$, the matrix $Y$ is arbitrary, and 
$(I-XX^+)$ is a projector into the nullspace, i.e. $\;(I-XX^+)X=0$.
For real vectors, the pseudoinverse can be written in terms of the transpose, i.e. $\;e^+=\frac{e^T}{e^Te}$. 
Now consider the constraint as a linear equation to be solved for $U$, and
note that $C$ is the nullspace projector for $e$.
A: I think I found an answer to my question, by using the Laplace multipler I can solve  the combined system in vectorized form, which for $n_c=2$ looks like:
$$
    \begin{pmatrix}
    \alpha L + I  & 0  & I \\
    0 & \alpha L + I & I \\
    I & I & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    \bar{u}_1 \\
    \bar{u}_2  \\
    \lambda \\
    \end{pmatrix} =     
\begin{pmatrix}
    \bar{u}^{obs}_1 \\
    \bar{u}^{obs}_2  \\
    0 \\
    \end{pmatrix}
$$
where
$ u = \begin{pmatrix}
    \vdots & \vdots  \\
    \bar{u}_1 & \bar{u}_2 \\
    \vdots & \vdots \\
    \end{pmatrix}
$
