Say we wish to show that the following matrix, is positive definite;
$$
A = \begin{bmatrix}
1 & 1 \\
1 & 2
\end{bmatrix}
$$
In this case, I know A is positive definite for two reasons; 1) the eigenvalues of A are both positive and, 2) both principal minors of A are positive.
I now wish to use the formal definition of positive deftness to show that A is positive definite. In other words, I want to show the following
$$
x'Ax > 0, \ x \neq0
$$
We then have;
$$
x_1^2 + 2x_1x_2 + 2x_2^2 \stackrel{?}{>}0
$$
The above inequality is clearly true if $(x_1>0 \text{ and } x_2>0)$ or $(x_1<0 \text{ and } x_2<0)$.
How do I show that this inequality holds if $x_1 \text{ and } x_2 $ have different signs? The $x_1x_2$ term will be negative in this case whereas the squared terms are obviously positive.
Best Answer
$x_1^2 + 2x_1x_2 + 2x_2^2=(x_1+x_2)^2+x_2^2\geq 0$ with equality iff $x_2=0$ and $x_1+x_2=0$, i.e. $x_1=x_2=0$.