The problem is as follows:

For what values of c is the quadratic form

$$Q(x,y) = 3x^2-(5+c)xy+2cy^2$$

positive definite, positive semidefinite, or indefinite?

Ok. My approach was to find the associated matrix and look at its eigenvalues. For the matrix, I got the following:

$$A = \begin{pmatrix}3&-\dfrac{5+c}{2}\\-\dfrac{5+c}{2}&2c\end{pmatrix}$$

and the following characteristic polynomial:

$$x^2-x(3+2c)-c^2+\dfrac{7c}{2}-\dfrac{c^2}{4}$$ (where the $x$'s represent the eigenvalues)

And this is where I believe my approach might've not been correct. However, I don't really know another way to go about this. Is there some sort of approach I'm overlooking? This is my first experience with quadratic forms, so I have very few techniques at my disposal.

Thanks!

• Your approach is correct, you stopped half-way. Compute the zeros of the characteristic polynomial, obtain the eigenvalues, then decide about definiteness. – daw Feb 7 '15 at 18:35

There's another approach: the Gauss decomposition consists to write the quadratic form on sum of squares. We have

$$Q(x,y)=3x^2-(5+c)xy+2cy^2=3\left(x-\frac{5+c}{6}y\right)^2+\left(2c-\frac{(5+c)^2}{12}\right)y^2$$ so the discussion is on the sign of $\alpha(c)=2c-\frac{(5+c)^2}{12}$

• if $\alpha(c)>0$ then $Q$ is positive definite
• if $\alpha(c)<0$ then $Q$ is indefinite
• if $\alpha(c)=0$ then $Q$ is positive semidefinite.
• You're welcome:) – user63181 Feb 8 '15 at 21:45

At least one eigenvalue will be positive ( why? ) For a 2D quadratic form you can look at the determinant. It can be positive, negative or 0. The determinant is the product of the eigenvalues.

$6c -(-5-c)^2/4 = \det(A)$, now complete the square and finish it.

• Ah, yes. That's true! Thanks, I got it! – thisisourconcerndude Feb 8 '15 at 21:45