# Sylvester's criterion

1. I have a 3 x 3 matrix [a+8,1,0;1,a+2,6;0,6,1] with two unknown variables. I have to use Sylvester's criterion to find set values of parameter a where the matrix is positive definite.

What I thought to do is to find the determinant of the matrix and a would be the boundaries between the two values?

1. Letting a = -2, the matrix is [6,1,0;1,0,6;0,6,1] which is indefinite. Without finding the eigenvectors, I have to find vector x such that x^t Ax > 0 and vector y y^t Ay<0

What I thought is that you make a 1 x 3 matrix for both x and y then mutliply with A. From that you calculate for det < 0 and det >0

Would this be the correct way of thinking for this question

https://i.stack.imgur.com/MFKtN.png

• "2 unknown variables" ? Where is the second one ? – Jean Marie May 30 at 0:48
• How do you propose to compute the determinant of a non-square matrix in part 2? – amd May 30 at 1:20
• Not sure that was what I thought you would do. Mind sharing how you would do it? – Mr A May 30 at 1:39

If your matrix is $$\begin{pmatrix}a+8 & 1 & 0 \\ 1 & a+2 & 6 \\ 0 & 6 &1 \end{pmatrix},$$Sylvester's criterion says that the above will be positive-definite if and only if $$\det(a+8) > 0, \qquad \det\begin{pmatrix} a+8 & 1 \\ 1 & a+2\end{pmatrix}>0 \qquad\mbox{and}\qquad \det\begin{pmatrix}a+8 & 1 & 0 \\ 1 & a+2 & 6 \\ 0 & 6 &1 \end{pmatrix}>0.$$Now it boils down to an algebra/pre-calc problem: find the values of $$a$$ satisfying all three inequalities at the same time.
For the first problem, we note that Sylvester's Criterion states that a matrix is positive definite if and only if each of its principal minors are positive. So it suffices to compute formulas for each of the principal minors and restrict $$a$$ to take values only within the range that would yield the desired positive determinants.