# How to show that this matrix is positive definite

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.

$$x_1^2 + 2x_1x_2 + 2x_2^2=(x_1+x_2)^2+x_2^2\geq 0$$ with equality iff $$x_2=0$$ and $$x_1+x_2=0$$, i.e. $$x_1=x_2=0$$.