# Problem about positive definiteness in the context of matrices

1. Apr 2, 2013

### s3a

1. The problem statement, all variables and given/known data
The problem:
Attached as TheProblem.jpg.

Solution:
2. Relevant equations
xT A x

3. The attempt at a solution
I computed the product and got 9x22 + 4 x1x1 + 4x12.

I'm thinking that I might need to show that that obtained polynomial must be below zero since we want to show that matrix A is not a positive definite matrix but, I don't know where to go from here.

Any input would be appreciated!

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2. Apr 4, 2013

### Mindscrape

That's right, and you know the squares certainly won't be negative so it comes down to the cross term (which I think you mislabeled) to cause a negative number.

3. Apr 10, 2013

### s3a

Yes, I did mislabel it. I meant: 9x22 + 4 x1x2 + 4x212.

However, I am still confused as to where I should go from here, algebraically.

4. Apr 10, 2013

### vela

Staff Emeritus
How did you get that polynomial from the 3x3 matrix?

5. Apr 10, 2013

### s3a

I computed the xTA x product.

6. Apr 10, 2013

### vela

Staff Emeritus
The polynomial you wrote down looks like it came from a 2x2 matrix. The matrix in the problem is 3x3. You should have x1, x2, and x3 appearing.

7. Apr 10, 2013

### s3a

Oops.

Is this the correct polynomial?:
5 x32 + 4 x2 x3 + 2 x1 x3 + 2 x1 x2 + 4 x12

P.S.
That seems to make my initial algebraic challenge from the beginning of this thread even greater.

8. Apr 10, 2013

### vela

Staff Emeritus
In this particular case, the lack of an $x_2^2$ term gives you the freedom to choose a large value for $x_2$ to get the overall sign you want. That's the easiest way to deal with this problem.

In the general case, one way would be to find linear combinations $y_i$ of the $x_j$'s so that you can write the polynomial in the form $\lambda_1 y_1^2+ \lambda_2 y_2^2 + \lambda_3 y_3^2$. Then from the signs of the coefficients, you can determine whether the matrix is positive definite or not.

For example, suppose you have the polynomial $4x_1^2 + 20x_1x_2 + 4x_2^2$. If you let $y_1 = x_1 + x_2$ and $y_2 = x_1 - x_2$, you'd have
$$4x_1^2 + 20x_1x_2 + 4x_2^2 = 7y_1^2 - 3y_2^2.$$ From the second form, it's clear the polynomial will be negative when $y_1=0$ and $y_2=1$.

The linear combinations correspond to the eigenvectors of the matrix, and the coefficients, to the eigenvalues. So finding the eigenvalues of the matrix is enough to tell you whether it's positive definite or not.

9. Apr 10, 2013

### s3a

Is that how this problem is asking to be answered or is that just a neat trick? I'm asking because I don't want to miss anything that's more standard, if it exists.

10. Apr 10, 2013

### vela

Staff Emeritus
Finding the eigenvalues? It's a standard way of determining whether a matrix is positive-definite. For this particular problem, however, I think you're just supposed to look at the polynomial and reason out how you can make it negative with the correct choice of $x_1$, $x_2$, and $x_3$.

11. Apr 11, 2013

### s3a

So, is there any significance to the particular answers given in the solution or do I just need to show that the polynomial is able to, at least, reach a negative value?

12. Apr 11, 2013

### vela

Staff Emeritus
It's just a counterexample to show the matrix isn't positive-definite.

13. Apr 11, 2013

### s3a

Okay so, that's essentially what I said, right?

14. Apr 12, 2013

### s3a

If I choose $x_1 = 1$, $x_2 = -5$ and $x_3 = 0$, do I accomplish what the problem is asking for?

15. Apr 12, 2013

### vela

Staff Emeritus
What is the problem asking you to show?

16. Apr 12, 2013

### s3a

The problem wants me to show that the definition of what makes a matrix positive definite fails. I believe what I did accomplishes this but, could you please confirm?

17. Apr 12, 2013

### vela

Staff Emeritus
What condition must a matrix satisfy if it is positive-definite?

18. Apr 12, 2013

### s3a

The $x^T$$A$ $x$ product (which is the polynomial we were working with just now) needs to be positive.

19. Apr 12, 2013

### vela

Staff Emeritus
That statement is not quite complete. You need to show $x^T A x > 0$ for all x. If you've shown this statement is false, you're done.

20. Apr 12, 2013

### s3a

Oh, thanks for correcting me/adding that. Now, that I see it, it does make sense but, to ask again, what I did is satisfactory, right?

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