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I need some help regarding this example

$$ q(x)= x_1^2 + 7x_2^2 + 2x_3^2 +6x_1x_2 - 4x_1x_3 - 4x_2x_3 $$ $$ =x_1^2 + (6x_2 - 4x_3)x_1 + 7x_2^2 - 4x_2x_3 + 2x_3^2 $$ $$ = (x_1 + (3x_2 - 2x_3))^2 - (3x_2 - 2x_3)^2 + 7x_2^2 - 4x_2x_3 + 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2x_2^2 + 8x_2x_3 - 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2(x_2 - 2x_3)^2 - 8x_3^2 - 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2(x_2 - 2x_3)^2 - 10x_3^2$$

So I get the eigenvalues -2, 1 and -10 while my teacher gets -2, 1 and 6 and I'm wondering how does he get 6?

And also, I sort of understand the process but not completely so if anyone could explain how we complete squares on quadratic forms then i'd be grateful, I know it has to do with Sylvester's Law of Inertia but I didn't understand that well from the book.

Myzanthros
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    Assuming your diagonalisation is correct, 1,-2 and -10 are the entries of the diagonal matrix to which the matrix is diagonalised, which are not necessarily equal to the eigenvalues. By the way, what you mention as eigenvalues are not really the eigenvalues. – StubbornAtom Oct 29 '17 at 16:43
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    When you solve matrix $Q^T D Q = H$ as you did, with $D$ diagonal, the diagonal entries of $D$ are not the eigenvalues and change depending on choices made. – Will Jagy Oct 29 '17 at 17:34
  • Also, $$ (x_1 + 3x_2 - 2x_3)^2 - 2(x_2 - 2x_3)^2 - 10x_3^2 = x_1^2 + 7x_2^2 -14x_3^2 +6x_1x_2 - 4x_1x_3 - 4x_2x_3 $$ – Will Jagy Oct 29 '17 at 17:45
  • Thanks, yeah I know, just felt easier to write eigenvalues but I guess a more correct term or something would be representations of a non-orthogonal diagonal matrix? – Myzanthros Oct 30 '17 at 08:08

1 Answers1

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method discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr

Note that the matrix $H$ has characteristic polynomial $x^3 - 10x^2 + 6x + 12$ and (irrational) eigenvalues

$$-0.8119658377479428698001875737, \; \; \; \; 1.605232411407321548171091093, \; \; \;9.206733426340621321629096481 $$

$$ H = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$ $$ D_0 = H $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

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

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

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

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 4 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$ Q^T D Q = H $$ $$ H = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$

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$$\left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & - 2 & 4 \\ - 2 & 4 & 2 \\ \end{array} \right) $$

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$$\left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 4 \\ 0 & 4 & - 2 \\ \end{array} \right) $$

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$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 4 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$

Will Jagy
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