Formulating Linear Constraints on a Matrix

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Hi everyone. I have this problem which I am trying to formulate. Basically, I have the following linear constraints:

[tex] p_{11} = 2[/tex]
[tex] p_{22} = 5[/tex]
[tex] p_{33}+2p_{12}=-1[/tex]
[tex] 2p_{13} =2[/tex]
[tex] 2p_{23} = 0[/tex]

And these are for the symmetric matrix

[tex] \mathbf{P} =<br /> \left( \begin{array}{ccc}<br /> p_{11} & p_{12} & p_{13} \\<br /> p_{12} & p_{22} & p_{23} \\<br /> p_{13} & p_{23} & p_{33} \end{array} \right)[/tex]

I would like to formulate a way to represent the linear constraints and [tex]\mathbf{P}[/tex] as a matrix at the same time.

I can do this using [tex]\mathbf{P}[/tex] or a vector of the entries of [tex]\mathbf{P}[/tex]. The linear constraints are easy if I use a vector ([tex]\mathbf{Ap}=\mathbf{b}[/tex], but then I don't know how to represent [tex]\mathbf{P}[/tex] as a matrix from the vector! And if I leave [tex]\mathbf{P}[/tex] as a matrix, all the constraints are easy to formulate except [tex]p_{33}+2p_{12}=-1[/tex]. Can anyone help me figure this out?

If anyone's curious, I'm trying to solve for [tex]\mathbf{P}[/tex] over the cone of PSD matrices using SDP. But I am entirely new to SDP and I'm scratching my head formulating this problem. I feel stupid right now :'(
 
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I answered my own question... and the answer has to do with algebraic dependence in the structure of the problem, and how SDP problems are formulated. I just thought I was missing something obvious ... hence my convoluted search.
 
Well, I'm glad you answered it. It looks simple to me: I you let [itex]p_{12}= p[/itex] then [itex]p_{33}= -1- 2p[/itex] and all other values are essentially given:
[tex]\begin{pmatrix}2 & p & 1 \\ p & 5 & 0 \\1 & 0 & -1-2p\end{pmatrix}[/tex]
 
HallsofIvy said:
Well, I'm glad you answered it. It looks simple to me: I you let [itex]p_{12}= p[/itex] then [itex]p_{33}= -1- 2p[/itex] and all other values are essentially given:
[tex]\begin{pmatrix}2 & p & 1 \\ p & 5 & 0 \\1 & 0 & -1-2p\end{pmatrix}[/tex]

That's exactly what came out in the end. I'm not sure why I didn't see it at first... I always seem to manage to do things in the most roundabout way.