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?
$\endgroup$ 14 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$It's very easy to show whether your matrix is positive semidefinite without even going into quadratic form. For all positive semidefinite $m \times m$ matrices $A$,$$\lambda_{i} \geq 0 \space \space \space (\forall \space i=1,...,m)$$So all eigenvalues of a positive semidefinite matrix need to be nonnegative. Especially if you're dealing with small matrices or using software like Octave, this test is very quick to do.
Hope this helps.
$\endgroup$ $\begingroup$You stated : The QF of A is $2x_1^2 + 2x_2^2 + 15x_3^2 - 4x_1x_2$, so$$QF(x_1,x_2,x_3)= 2x_1^2 + 2x_2^2 + 15x_3^2 - 4x_1x_2 \\ = 2(x_1^2 - 2x_1x_2 + x_2^2 ) + 15x_3^2 \\= 2(x_1-x_2)^2 + 15x_3^2 \ge 0$$
so the QF is always non-negative, hence the Matrix is positive semidef.
$\endgroup$ $\begingroup$After congruent operations( elementary row operation followed by same Column operation) we get quadratic form as $$x_1^2+x_2^2\geq 0$$which is positive semi definite quadratic form.
$\endgroup$
