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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.

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1 Answer 1

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

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