Formulating Linear Constraints on a Matrix

[JeSuisConf]

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

<br /> p_{11} = 2<br />
<br /> p_{22} = 5<br />
<br /> p_{33}+2p_{12}=-1<br />
<br /> 2p_{13} =2<br />
<br /> 2p_{23} = 0<br />

And these are for the symmetric matrix

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

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

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

If anyone's curious, I'm trying to solve for \mathbf{P} 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 :'(

 

[JeSuisConf]

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.

 

[HallsofIvy]

Well, I'm glad you answered it. It looks simple to me: I you let p_{12}= p then p_{33}= -1- 2p and all other values are essentially given:
\begin{pmatrix}2 &amp; p &amp; 1 \\ p &amp; 5 &amp; 0 \\1 &amp; 0 &amp; -1-2p\end{pmatrix}

 

[JeSuisConf]

Well, I'm glad you answered it. It looks simple to me: I you let p_{12}= p then p_{33}= -1- 2p and all other values are essentially given:
\begin{pmatrix}2 &amp; p &amp; 1 \\ p &amp; 5 &amp; 0 \\1 &amp; 0 &amp; -1-2p\end{pmatrix}

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.