Write in the canonical form the expression $$ab+bc+ca-a+b$$
First, I tried to eliminate the lonely terms, but seems like quite impossible: $$a(b'+1) + b(c'-1) + (c'-1)a - a + b = ab'+bc'+c'a-a$$ It is going to get me into an "infinite loop" if I keep trying to get rid of them.
Still if I am to write this as follows $$(a+b+c)^2-ab-bc-ca-a+b=(a+b+c)^2+\frac{1}{2}(a-b)^2+\frac{1}{2}(b-c)^2+\frac{1}{2}(c-a)^2-\\-a^2-b^2-c^2-a+b$$ I don't get anything useful.
Any help is appreciated.
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
your original can be written as half of the diagonal matrix below with $Q$ as coefficients, thus
$$ \left( \frac{a}{2} + \frac{b}{2} + c \right)^2 - \frac{1}{4} (-a+b -2)^2 - c^2 + 1 $$
The thing cannot be accomplished without a constant inside one of the squares as in $(-a+b-2)^2$
What we actually do is add in a new variable to make the thing homogeneous quadratic,
$$ab+bc+ca-ad+bd$$
My matrix $H$ is the Hessian matrix of second partial derivatives
$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 1 & - 1 & 1 & 0 \\ - 1 & 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 1 & - 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 1 & - 1 \\ 1 & \frac{ 1 }{ 2 } & - 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & - 2 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 1 & 0 & - 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 1 & 1 & - 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ \end{array} \right) $$