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"Determine if the following quadratic form is positive definite, negative definite or undefinite
$Q:\mathbb R^3\to \mathbb R, \,Q(u)=x_1^2+4x_1x_2-2x_2^2+2x_1x_3-2x_3^2$"

$Q=\begin{bmatrix} 1&2&1 \\\ 2&-2&0 \\\ 1&0&-2 \end{bmatrix}$

  1. I tried to compute the diagonal matrix but the eigenvalues are not integers, thus it's a bit hard to calculate by hand. UPDATE: Seemingly, I've done something wrong previously.
  2. I tried to group them to form squares, however there is nothing that guarantees is either positive or negative. Plugging in numbers results in both positive and negative results.
  3. What else to try?
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    $@$Andrew: this is really a topic for textbooks rather than papers: you can consult Chapter 1 of any number of texts on quadratic forms (I recommend Lam's book and Cassels' book in particular). I do have a treatment of this material in $\S$ 5 of my own lectures notes: http://math.uga.edu/~pete/quadraticforms.pdf. However I discuss the algorithm rather sketchily and give no examples, so I think other sources would serve you better.2012-01-19

4 Answers 4

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Notice that if you add two times the identity matrix to $Q$ then the bottom-right $2\times 2$-submatrix will be $0$.

This shows that $-2$ is an eigenvalue--a corresponding eingenvector is $[0,1,-2]$.

Now use polynomial division to divide the characteristic polynomial by $(\lambda+2)$.

The zeros of the resulting quadratic polynomial are the two remaining eigenvalues.

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    Well, by taking an arbitrary vector, say $[x,y,z]$, setting the result equal to zero and seeing what $x,y$ and $z$ needed to fulfill.2012-01-19
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The first vector of the canonical basis being positive and the second negative, the form is indefinite.

EDIT A. More careful phrasing:

The restriction of $Q$ to the first coordinate axis being positive definite, and its restriction to the second coordinate axis being negative definite, $Q$ is indefinite.

EDIT B. If the matrix of a quadratic form on $\mathbb R^n$ has a positive ( > 0) diagonal entry and a negative ($ < 0$) diagonal entry, then it is indefinite.

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    Yes, I agree with what you said.2012-01-20
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The principal determinant method is easy to apply, the eigenvalue method is more tedious

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    a$n$d there is another way to solve the problem , Theorem / Let A be a symmetric matrix of order m. Then A is positive definite iff its naturally ordered principal minors are all positive.2012-02-19