Completing the square (diagonal matrix/bilinear form)

Question

Let $A=\begin{pmatrix} 2 & 2 & 5 \\ 2 & 1 & 3 \\ 5 & 3 & \frac{17}{2} \end{pmatrix} \in M_3(\mathbb{R})$.

I want to find an invertible matrix $C$ such that $C^TAC=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, since there are three eigenvalues which are postive, negative and zero.

I tried to use completing the square with the bilinear form $s(v,v)=\langle v,Av \rangle$:

$s(v,v)= \langle v,Av \rangle=(v_1,v_2,v_3)\begin{pmatrix} 2 & 2 & 5 \\ 2 & 1 & 3 \\ 5 & 3 & \frac{17}{2}\end{pmatrix}\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}=2v_1^2+4v_1v_2+10v_1v_3+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$

Now I used completing the square with respect to $v_1$:

$2(v_1^2+2v_1v_2+5v_1v_3)+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$

$=2(v_1^2+v_1(2v_2+5v_3))+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$

$=2(v_1^2+v_1(2v_2+5v_3)+\frac{1}{2}(2v_2+5v_3)^2-\frac{1}{2}(2v_2+5v_3)^2)+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$

$=2(v_1+(2v_2+5v_3))^2-\frac{1}{2}(2v_2+5v_3)^2+v_2^2+6v_2v_3+\frac{17}{2}v_3^2$

$=2(v_1+2v_2+5v_3)^2-\frac{1}{2}(2v_2+5v_3)^2+(v_2+3v_3)^2-\frac{1}{2}v_3^2$

Now there is one summand too much so I can't set a matrix $C$ which is the inverse of the matrix with the entries in the braces.

I think I made a mistake with the completing the square but I don't see where.

Answer 1

$$2x^2+4xy+10xz+y^2+6yz+\frac{17}2z^2=2\left(x+y+\frac52z\right)^2-y^2-4yz-4z^2=$$

$$=2\left(x+y+\frac52z\right)^2-\left(y+2z\right)^2$$

and there you go: one positive, one negative, one zero...

Answer 2

I doubled your matrix to get all integers:

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 4 & 10 \\ 4 & 2 & 6 \\ 10 & 6 & 17 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

Divide back by 2, you have diagonal matrix with diagonal $2, -1,0.$ So, as a final step, multiply on the far left and far right by $$ \left( \begin{array}{rrr} \frac{1}{\sqrt 2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \; , \; $$ which is its own transpose.

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 4 & 4 & 10 \\ 4 & 2 & 6 \\ 10 & 6 & 17 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 4 & 4 & 10 \\ 4 & 2 & 6 \\ 10 & 6 & 17 \\ \end{array} \right) $$

==============================================

$$ E_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 4 & 0 & 10 \\ 0 & - 2 & - 4 \\ 10 & - 4 & 17 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 5 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 5 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 2 & - 4 \\ 0 & - 4 & - 8 \\ \end{array} \right) $$

==============================================

$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 4 & 10 \\ 4 & 2 & 6 \\ 10 & 6 & 17 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ \frac{ 5 }{ 2 } & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 4 & 10 \\ 4 & 2 & 6 \\ 10 & 6 & 17 \\ \end{array} \right) $$