Definiteness of quadratic form by completing the square

Question

I want to check my thinking here. We're asked to determine the definiteness of the matrix

$ \begin{pmatrix} 3 & -1 & 1\\\ -1 & 1 & 2\\ 1 & 2 & 9 \end{pmatrix} $

We can write down the quadratic form as:

$3x_1^2 - 2 x_1 x_2 + x_2^2 + 4 x_2 x_3 + 6x_3^2 + 2 x_1 x_3 $

So by completing the square we arrive at

$(x_1 - x_2)^2 + (x_2 + 2x_3)^2 + (x_1 + x_3)^2 + x_1^2 - x_2^2 + x_3^2 $

But now it seems quite difficult to determine the definiteness, how would this usually be done?

Answer 1

There is a quick way to check positive definiteness called Sylvester's criterion which states that a matrix is positive definite if and only if the upper-left hand matrices have positive determinant, i.e.

Applying it to your matrix:

$\begin{vmatrix} 3 \end{vmatrix} >0; \begin{vmatrix} 3 & -1 \\ -1 & 1\end{vmatrix} = 3 -1 = 2>0; \begin{vmatrix} 3 & -1 & 1 \\ -1 & 1 & 2 \\ 1 & 2 & 9 \end{vmatrix} = 15-11-3 = 1 > 0$

Therefore your matrix is positive definite.

Answer 2

fairly understandable algorithm for completing the square. Without finding the eigenvalues, given a symmetric matrix $H,$ find a nonsingular matrix $P$ such that $D=P^THP$ is diagonal. Then Sylvester's Law of Inertia says the diagonal elements of $D$ have the same $\pm$ signs as the eigenvalues.

I put the outcome first...

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$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 \\ - \frac{ 3 }{ 2 } & - \frac{ 7 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & - \frac{ 3 }{ 2 } \\ 0 & 1 & - \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 7 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) $$

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Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) $$

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

$$ E_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 3 & 0 & 1 \\ 0 & \frac{ 2 }{ 3 } & \frac{ 7 }{ 3 } \\ 1 & \frac{ 7 }{ 3 } & 9 \\ \end{array} \right) $$

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

$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & - \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & \frac{ 7 }{ 3 } \\ 0 & \frac{ 7 }{ 3 } & \frac{ 26 }{ 3 } \\ \end{array} \right) $$

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

$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & - \frac{ 3 }{ 2 } \\ 0 & 1 & - \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) $$

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

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$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 \\ - \frac{ 3 }{ 2 } & - \frac{ 7 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & - \frac{ 3 }{ 2 } \\ 0 & 1 & - \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 7 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & \frac{ 7 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & - 1 & 1 \\ - 1 & 1 & 2 \\ 1 & 2 & 9 \\ \end{array} \right) $$

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Answer 3

To the other answers I'll add for completeness a "completing the square" answer. \begin{align*} 3x_1^2 -2x_1x_2 +x_2^2 + 4x_2x_3 + 6x_3^2 + 2x_1x_3 &= 3(x_1 - x_2/3 + x_3/3)^2 +\frac23 x_2^2 + \frac{14}3x_2x_3 + \frac{26}3x_3^2 \\ &= 3(x_1 - x_2/3 + x_3/3)^2 + \frac23(x_2 + 7x_3/2)^2 +\frac12x_3^2, \end{align*} which is clearly positive definite. The key difference from your completion of the square is that I've fixed the $x_1x_2$ and $x_1x_3$ cross terms with the first square.