How to show that this matrix is positive semidefinite?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.