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At the moment I am trying to rewrite this term:

$2x^2-2xy+5y^2-4x+2y+2$

as a sum of squares. So I am trying to find an experession of $2x^2-2xy+5y^2-4x+2y+2=a^2+b^2+c^2$ (for example)

It looks easy, but everything I have tried failed so far. So I wonder if there is such expression. I know that this term is nonnegativ for every pair $(x,y)$.

One try might look like this:

$2x^2-2xy+5y^2-4x+2y+2=x^2-2xy+y^2+x^2+4y^2-4x+2y+2=(x-y)^2+(x-2)^2+4y^2+2y-2$

Here $4y^2+2y-2=4(y-\frac12)(y+1)$

Do you see a nice sequence of calculations?

Thanks in advance.

Cornman
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  • Does this answer help? On converting a conic section to the standard form. Setting $x_1=x-1$ yields $5 y^2 - 2 y x_1 + 2 x_1^2$, then rotate. – Alexey Burdin Jun 22 '20 at 19:28
  • @AlexeyBurdin Thanks for this suggestion. I have to study the answer, so I can not tell right away if it helps. It looks interesting for sure. But it seems to involve some multivariable calculus, which I initialy tried to avoid with this question. I am happy that such a simple question leads to great techniques and answers. – Cornman Jun 22 '20 at 19:32

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The second matrix identity below says $$ \frac{1}{2} (2x-y-2)^2 + \frac{9}{2} y^2 $$

The method is discussed at

reference for linear algebra books that teach reverse Hermite method for symmetric matrices

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & - 1 & - 2 \\ - 1 & 5 & 1 \\ - 2 & 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 9 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 9 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & - 1 & - 2 \\ - 1 & 5 & 1 \\ - 2 & 1 & 2 \\ \end{array} \right) $$

Will Jagy
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    Could you elaborate your approach more? Where do these matrix came from? How is this connected to the problem? I have never seen something like this, so I am interested in learning more about this technique. Thanks. Edit: I see your edit should do it. Thank you! – Cornman Jun 22 '20 at 19:24
  • After a while I have come back to this question and your answer, and dabbled around in the given links. There is a lot and indeed this is exciting stuff. I would like to learn more of this and are searching for a good reference, or expostion of this topic. Can you recommand something? Mostly I am interested in the application presented here, so for quadratic forms. – Cornman Jun 18 '21 at 18:33