# Positive Definiteness

• Dec 9th 2010, 04:41 PM
AKTilted
Positive Definiteness
Hi,

I'm trying to show that the following matrix:

$\displaystyle A = \begin{bmatrix} 1&2&3\\2&5&1\\3&1&36 \end{bmatrix}$ is positive definite. Using the typical definitions I get to this inequality:

x^2 + 2x*(2y+3z) + 5y^2 + 2*y*z + 36z^2 > 0. I just don't know to show that this is true.

Also, I'm trying to show that this matrix:

$\displaystyle B = \begin{bmatrix} 1&2&3\\2&5&1\\3&1&34 \end{bmatrix}$ is NOT positive definite. I can't find a vector x that shows x^T*B*x > 0.

Thanks a lot for your help.
• Dec 9th 2010, 04:45 PM
dwsmith
Never mind
• Dec 9th 2010, 07:16 PM
tonio
Quote:

Originally Posted by AKTilted
Hi,

I'm trying to show that the following matrix:

$\displaystyle A = \begin{bmatrix} 1&2&3\\2&5&1\\3&1&36 \end{bmatrix}$ is positive definite. Using the typical definitions I get to this inequality:

x^2 + 2x*(2y+3z) + 5y^2 + 2*y*z + 36z^2 > 0. I just don't know to show that this is true.

Also, I'm trying to show that this matrix:

$\displaystyle B = \begin{bmatrix} 1&2&3\\2&5&1\\3&1&34 \end{bmatrix}$ is NOT positive definite. I can't find a vector x that shows x^T*B*x > 0.

Thanks a lot for your help.

Dou you know Sylvester's Criterion? A is pos. def. because all its principal minors are positive, whereas B doesn't fulfill this

condition (in fact, $\displaystyle \det B =0$ ).

So in order to find an element $\displaystyle x\in\mathbb{R}^3\,\,s.t.\,\,x^tBx\ngtr 0$ , just choose a non-trivial vector in the

kernel of B...

Tonio