Using the definition, show that the following matrix is positive semidefinite.
$$\begin{pmatrix} 2 & -2 & 0\\ -2 & 2 & 0\\ 0 & 0 & 15\end{pmatrix}$$
In other words, if the quadratic form is $\geq 0$, then the matrix is positive semidefinite.
The quadratic form of $A$ is
$$2x_1^2 + 2x_2^2 + 15x_3^2 – 4x_1x_2$$
After modifying it a little bit, I get
$$(\sqrt2 x_1 – \sqrt2 x_2)^2 + 15x_3^2$$
Both parts are positive and the only way the quadratic form is $0$ is when $x_1,x_2,x_3$ are $0$. So isn't this matrix positive definite?
Best Answer
Here is a way to show that it is not positive definite.
Let $x_1=x_2=1$ and $x_3=0$.
As for showing that it is positive semidefinite, you have shown that quadratic form is nonnegative.