# Conditions for a real binary quadratic form to be positive definite

Since this question was heavily downvoted, I would like to change the presentation of the question as follows. I hope those of you who downvoted this question would be satisfied with the change.

In trying to solve this problem, I came up with the following result which is necessary to solve the problem.

My intent of posting the question is as follows.

1. To use the result myself to answer other questions in this site.

2. To provide a useful result which can be used by the users to answer other questions in this site.

3. To have as many diffferent proofs of the result as possible, each of which can be understood by most undergraduate students of mathematics.

I beg you allow me to put my proof in the spoiler box. I don't want to spoil your pleasure of solving a problem. I welcome you to provide as many different proofs of the result as possible. Please provide full proofs which can be understood by most undergraduate students of mathematics.

Let $f = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{R}$. $D = b^2 - 4ac$ is called the discriminant of $f$.
We say $f$ is positive definite if $f(x, y) \gt 0$ for every $(x, y) \ne (0, 0)$ in $\mathbb{R}^2$.

My Question Is the following proposition correct? If yes, how do you prove it?

Proposition Let $f = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{R}$, $D$ its discriminant.
Then $f$ is positive definite if and only if $D \lt 0$ and $a \gt 0$.

• Have you tried completing the square? – Gerry Myerson Nov 1 '13 at 22:18
• For non-zero $y$, check $f=y^2 \left[a\left(\frac xy\right)^2 + b\left(\frac xy\right) + c\right]$ – peterwhy Nov 1 '13 at 22:22
• What are your thoughts on the question? – Bruno Joyal Nov 1 '13 at 22:23
• "it's perfectly legitimate in this site to ask a question whose answer the poster knows" --- yes, provided the author is upfront with that information. Otherwise, someone could waste her time, and yours, telling you things you already know. Makoto, you have been here long enough to know better. If you ever wonder why people sometimes vote you down for no reason --- well, this is the reason. – Gerry Myerson Nov 2 '13 at 3:40
• @MakotoKato if you look at pages 258-261 of supermath.info/MultivariateCalculus2011Chapter5.pdf I derive a result which includes the one you desire here from the method of Lagrange Multipliers. I think the argument is a bit unusual, I do it because the students it's targeted for do not yet have linear algebra with which to properly tackle the problem. Perhaps you'll find it interesting. – James S. Cook Nov 4 '13 at 2:28

Here is an alternate proof, as requested by the OP.

The quadratic form corresponds to the matrix $$M = \left[\begin{array}{cc} a & b/2\\ b/2 & c\end{array}\right]$$ with $D = -4\det M.$

If $M$ is positive-definite, then it has two positive eigenvalues, and $\det M >0$. Moreover $(1,0)^TM(1,0) = a$ so $a>0$.

Now suppose $a>0$ and $D<0$. These two facts imply that also $c>0$. Since $M$ is symmetric, it has two real eigenvalues, and since $D<0$ they are either both positive or both negative. Since their sum is equal to the trace of $M$, which is positive, they must both be positive, and $M$ is positive-definite.

Spoiler

Suppose $f$ is positive definite. Since $f(1, 0) = a, a \gt 0$. Hence we have $f = ax^2 + bxy + cy^2 = a(x + \frac{b}{2a}y)^2 + \frac{4ac - b^2}{4a}y^2 = a(x + \frac{b}{2a}y)^2 + \frac{-D}{4a}y^2$
Since $f(-\frac{b}{2a}, 1) = \frac{-D}{4a}$, $D \lt 0$. The converse is clear.

• Why doesn't the spoiler box work? – Makoto Kato Nov 3 '13 at 2:32
• Spoilers have problems with multiple lines. You can either put everything on the same line or you can make separate spoiler box for each line. This was mentioned on meta a few times, see here and the posts linked from that question. – Martin Sleziak Nov 3 '13 at 8:01
• @MartinSleziak Thanks! – Makoto Kato Nov 3 '13 at 8:11