Quadratic forms, change of variables

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We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $\det P = \pm 1.$ If it is imperative to have diagonal elements restricted to $\pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.

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
$$ H = \left( \begin{array}{rr} 1 & 1 \\ 1 & 2 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rr} 1 & 1 \\ 1 & 2 \\ \end{array} \right) $$

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

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

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Luke Mathwalker
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Luke Mathwalker

Updated on August 01, 2022

Comments

  • Luke Mathwalker
    Luke Mathwalker over 1 year

    If one has a symmetric matrix $A$, one can diagonalize it with an orthonormal change of basis vectors, e.g. $S^TAS$ is diagonal. Now lets consider the following matrix $$A=\begin{bmatrix} 1&1\\ 1&2 \end{bmatrix}. $$ This matrix corresponds to the symmetric form $$x_1^2+2x_1x_2+2x_2^2=(x_1+x_2)^2+x_2^2.$$

    For me this looks like there has to be a way of determining some $S$ as above (without having to diagonalize etc., just by completing of the square) by taking some change of variables $x_1\leadsto x_1+x_2,\ x_2\leadsto x_2$. This would be done by the matrix $$S=\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}$$ but this doesn't work out for me...

    Any help will be gratefully appreciated.

    Edit: Let me reformulate my question. By Sylvester's law of inertia there exists for every symmetric matrix $A$ some basis such that $S^TAS$ is diagonal with only 1,-1 and 0 on the diagonal, where $S$ is a (not necessarily orthogonal) invertible matrix. I want to determine $S$ without having to calculate all the eigenvalues and diagonalize $A$, because the eigenvalues dont occur in the wanted form.

  • Luke Mathwalker
    Luke Mathwalker over 5 years
    Excellent answer! There actually was some link you referenced where you did exactly what I meant. Thanks, that's really a lot easier to compute than to diagonalize.