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$$Q(x,y)= x^{2}+4xy+3y^{2}$$

Is $Q(x,y)$ positive definite, negative definite or neither?

This is the last part of the question. I was asked to find an othogonal change of basis which diagonalizes this quadratic form, that was fine.

I found that this has a positive and a negative eigenvalue. $$\lambda_{1} = 2+\sqrt{5} $$ $$\lambda_{2} = 2-\sqrt{5}$$ I know that if all the eigenvalues are strictly positive then the quadratic form is positive definite and if all the eigenvalues are strictly negative then the quadratic form is negative definite.

I said that $Q$ is neither. Because of the eigenvalues being of opposite signs.

(Side Question) When I was looking online I also saw on MathWorld that you can use this formula $d\equiv 4ac-b^{2}>0$ implies positive definite. Is this correct?

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Write the quadratic form matrixwise:

$$q(x,y)=(x\;y)\begin{pmatrix}1&2\\2&3\end{pmatrix}\binom xy$$

and the above matrix is neither positive definite nor negative definite.

Another way: complete the square:

$$x^2+4xy+3y^2=(x+2y)^2-y^2$$

and we can see at once that the above is not the sum of only positive/negative squares.

One further way: calculate the above matrix's eigenvalues:

$$\begin{vmatrix}t-1&-2\\-2&t-3\end{vmatrix}=t^2-4t-1$$

and the above quadratic's roots are

$$t_{1,2}=\frac{4\pm2\sqrt5}2=2\pm\sqrt5$$

and thus we have one positive eigenvalue and one negative, so the quadratic form is neither positive nor negative definite.

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  • $\begingroup$ Yeah I was just about to try $Q=pAp^{T}$ where $p = (x,y)$ Thanks very much for taking the time to help. $\endgroup$ – Patrick Moloney Nov 14 '16 at 11:02
  • $\begingroup$ @PatrickMoloney As you wish, yet knowing how to check these things in different way can be helpful as sometimes a method can be very difficult of lengthy, and other one may be more friendly. $\endgroup$ – DonAntonio Nov 14 '16 at 11:04
  • $\begingroup$ I prefer the eigenvalue way. It's a lot easier and less time consuming since I've already found the eigenvalues in part (b) of the question. Again, thanks. $\endgroup$ – Patrick Moloney Nov 14 '16 at 11:07
  • $\begingroup$ @PatrickMoloney Then it is, yet in this particular case, without having to check the eigenvalues, I thing completing the square is much easier and fast. $\endgroup$ – DonAntonio Nov 14 '16 at 11:07

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