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I am studying a chapter on positive definites and there is this question in which I have to find whether the quadratic forms are positive definite or not. I have to confirm my answer for this quadratic form: $x^{2}+8xy+y^{2}$. I made it's matrix representation which is $$A = \begin{bmatrix} 1&4\\4&1 \end{bmatrix}$$ The eigenvalues of this matrix are 5 and -3. Since one of the eigenvalue is negative, it must not be a positive definite?

My second doubt is regarding this form: $x^{2}+6xy$. The matrix representation is $$B=\begin{bmatrix} 1&3\\3&0 \end{bmatrix}$$ The eigenvalues of this matrix are $$\lambda = \frac{1 \pm \sqrt{37}}{2}$$ So this quadratic form is also not positive definite, right?

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  • $\begingroup$ Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula. $\endgroup$ – Yadati Kiran Nov 23 '18 at 14:43
  • $\begingroup$ Please could you clarify a little bit. It would be of great help. $\endgroup$ – Himanshu Sharma Nov 23 '18 at 14:45
  • $\begingroup$ Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $k\times k$ matrix is positive for $1\leq k\leq n$. $\endgroup$ – Yadati Kiran Nov 23 '18 at 14:48
  • $\begingroup$ Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $\det (\lambda I - B) $ works just fine. $\endgroup$ – Eric Nov 23 '18 at 14:49
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    $\begingroup$ Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea. $\endgroup$ – Eric Nov 23 '18 at 15:02

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