# How to show that this matrix is positive semidefinite?

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?

• As to your last sentence: if $x_1=x_2$ and $x_3=0$ then the form vanishes, for instance when $x=(3,3,0)$. That's why it's only positive semi definite. – kimchi lover Feb 8 at 17:52

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.

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.

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.

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.