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    A = \begin{pmatrix} 2 & 1 & a \\a & 2 & 1 \\1 & a & 2 \end{pmatrix}

    I need to find the value(s) of a for which A is positive definite, then prove that A is indeed positive definite for those value(s).

    I know that I'm using sylvester's criterion which gives me

    4 - a > 0 \

a < 4

(for some reason the less than sign isn't bothering to show up)

    and

    8+1+a^3 - 2a - 2a - 2a > 0

\implies a^3 -6a + 9 > 0

\implies (a+3)(a^3-3a+3)>0

\implies a > -3

    So A is pos def iff -3 < a < 4

    However, proving that this is the case would require finding the quadratic form of A which is

    Q_A(x,y,z) = 2x^2 + 2y^2 + 2z^2 + (a+1)xy + (a+1)xz + (a+1)yz

    Whilst completing the square and putting all of this into quadratic form, I think I'm messing up somewhere or I'm using the wrong technique. Is there an easier way of doing this?
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    After a brief glance on Wikipedia, I gather that Sylvester's Criterion says a matrix is positive definite if the determinants of its upper left 1x1, 2x2 and 3x3 matrices are positive.

    The upper left 2x2 determinant is 2-a. For this to be positive, a < 2.

    So I don't know where that a < 4 is coming from.

    I also got a different cubic equation than you did from evaluating the 3x3 determinant.
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    (Original post by Dragon)
    After a brief glance on Wikipedia, I gather that Sylvester's Criterion says a matrix is positive definite if the determinants of its upper left 1x1, 2x2 and 3x3 matrices are positive.

    The upper left 2x2 determinant is 2-a. For this to be positive, a < 2.

    So I don't know where that a < 4 is coming from.

    I also got a different cubic equation than you did from evaluating the 3x3 determinant.
    Sorry about that, I mistyped the matrix - the top left 1 was supposed to be a 2.
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    OK so that looks all correct to me. Surely evaluating the determinants like you did is sufficient to show A is positive definite? Why is that not a valid proof?
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    (Original post by Dragon)
    OK so that looks all correct to me. Surely evaluating the determinants like you did is sufficient to show A is positive definite? Why is that not a valid proof?
    Well, I've found the values of a for which A is positive definite, but now all that remains is to prove that A is indeed positive definite for these values, that's what I'm having trouble with.

    I presumed that I'd have to use the fact that a positive matrix is pos def iff its quadratic form is strictly greater than 0, and I would expect that would simply yield that it was, as long as -3<a<4, which I can't get it to do.
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    Well according to the Wikipedia article, Sylvester's Criterion is an if and only if and thus a sufficient condition to show A is positive definite.

    I guess another, probably easier, way to show it is to take any 0 \ne v = (x, y, z) \in \mathbb{R}^3 and show that v^T A v > 0 for the values of a you obtained.
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    (Original post by Dragon)
    Well according to the Wikipedia article, Sylvester's Criterion is an if and only if and thus a sufficient condition to show A is positive definite.

    I guess another, probably easier, way to show it is to take any 0 \ne v = (x, y, z) \in \mathbb{R}^3 and show that v^T A v > 0 for the values of a you obtained.
    So we'd be taking any generic v, finding out Av and then finding v^T A v and then checking it for values of a between -3 and 4?
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    Yep. Although that isn't really necessary as Sylvester's Criterion proves the matrix is positive definite.
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    It's not actually *that* hard to show your expression for Q is positive definite for those values.

    Note that you know (x-y)^2 + (y-z)^2 + (z-x)^2 >=0 for all x,y,z.
 
 
 
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