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How do you determine whether the quadratic form $Q(x,y) = 2x^2 - 4xy + 5y^2$ is positive definite, negative definite, or indefinite?

Could someone show step by step with explanations? Thank you

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Diagonalize. In this case, it comes down to completing the square.

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  • $\begingroup$ Could you show me the steps? $\endgroup$ – Kenneth Hend May 3 '13 at 21:00
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    $\begingroup$ For example, $2x^2-4xy=2((x-y)^2 -y^2)$, so our function is $2(x-y)^2+3y^2$, clearly positive definite. You can use more formal diagonalization also, useful in more variables. $\endgroup$ – André Nicolas May 3 '13 at 21:05
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We can use the method of gauss to reduce the quadratic form: $$Q(x,y) = 2x^2 - 4xy + 5y^2=2(x-y)^2+3y^2$$ hence the signature is $(2,0)$ and the quadratic form is positive definite.

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If I remember correctly, it is equivalent to check the corresponding property of its Hessian matrix ($2\times 2$ symmetric matrix in this case), e.g. by looking at its eigenvalues.

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By computing the bilinear form $B((x_1,y_1),(x_2,y_2))=\frac{1}{2}[Q((x_1,y_1)+(x_2,y_2))-Q((x_1,y_1))-Q((x_2,y_2))]$

and testing it with the points $(1,0)$ and $(0,1)$, you can discover that this is the matrix for the form is

$$ \begin{bmatrix}2&-2\\-2&5\end{bmatrix} $$

This matrix has eigenvalues 1 and 6, so it is positive definite.

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  • $\begingroup$ No doubt there are ways to simplify the connection... I would like to see them! $\endgroup$ – rschwieb May 3 '13 at 20:57
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Compute its discriminant (in this case $\Delta = (-4)^2 - 4 \cdot 2 \cdot 5 = -24$). If it's positive then the form is indefinite, if it's zero then the form is semi-definite, if it is negative the form is definite. Positive or negative (semi-)definite can be seen from either coefficient of $x^2$ or $y^2$.

This is a special case of Sylvester's criterion.

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