You are Here: Home >< Maths

# Positive definite matrix watch

1. 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

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

and

So A is pos def iff

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

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?
2. 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 . For this to be positive, .

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

I also got a different cubic equation than you did from evaluating the 3x3 determinant.
3. (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 . For this to be positive, .

So I don't know where that 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.
4. 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?
5. (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 , which I can't get it to do.
6. 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 and show that for the values of a you obtained.
7. (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 and show that for the values of a you obtained.
So we'd be taking any generic v, finding out Av and then finding and then checking it for values of a between -3 and 4?
8. Yep. Although that isn't really necessary as Sylvester's Criterion proves the matrix is positive definite.
9. 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.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 27, 2011
The home of Results and Clearing

### 3,037

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Sheffield Hallam University
Tue, 21 Aug '18
2. Bournemouth University
Wed, 22 Aug '18
3. University of Buckingham
Thu, 23 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams