Find the precise conditions on $a,b,c$ in order that the matrix $\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ will be nonnegative definite.

Let $0\neq x=\begin{bmatrix}x_1 \\x_2 \end{bmatrix}\in \mathbb{R}^2$, then $\begin{bmatrix}x_1 \\x_2 \end{bmatrix}^{T}\begin{bmatrix}a & b\\ c & d\end{bmatrix}\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=x_1^2a+x_2^2d+x_1x_2(c+b)$, and so $x_1^2a+x_2^2d+x_1x_2(c+b)\geq 0$. What conditions must $a,b,c$ have? Could $a\geq 0$ and $b+c=0$ be taken? Thank you

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    $\begingroup$ Do you assume that the matrix is Hermitian or symmetric? The you could use Sylverster's criterion. $\endgroup$ – Dietrich Burde Apr 12 '18 at 18:50
  • $\begingroup$ @DietrichBurde No, I'm not assuming that. $\endgroup$ – user424241 Apr 13 '18 at 3:23

For matrices, being positive definite or positive semi-definite include the "symmetric/Hermitian" as part of their definition. If one want to talk about non-negative definite matrices, one is really dealing with matrices whose associated quadratic form is positive semi-definite.

Let us restrict ourselves to real matrices. Let $f = \frac12(b+c)$ and $Q(x,y)$ be the quadratic form $$Q(x,y) \stackrel{def}{=} ax^2 + dy^2 + (b+c)xy = ax^2 + dy^2 + 2fxy$$ In order for $Q(x,y)$ to be psd (positive semi-definite), i.e. $$Q(x,y) \ge 0\quad\text{ for all }\quad(x,y) \in \mathbb{R}^2$$

We need $a = Q(1,0) \ge 0$ and $d = Q(0,1) \ge 0$.

If $a = d = 0$, then $\pm 2f = Q(1,\pm1)$ tell us $Q(x,y)$ is psd if and only if $f = 0$.

Otherwise, at least one of $a , d \ne 0$. Let's say $a > 0$, we find

$$aQ(x,y) = a^2 x^2 + 2afxy + ady^2 = (ax + fy)^2 + (ad-f^2)y^2$$

If $ad - f^2 \ge 0$, above identity tell us $Q(x,y)$ is psd.

On the other direction, if $Q(x,y)$ is psd, we have $ad-f^2 = aQ\left(-\frac{f}{a},1\right) \ge 0$.

Combine these, we find under the assumption $a > 0$, $Q(x,y)$ is psd if and only if $ad - f^2 \ge 0$.
Same conclusion can be reached when we replace the assumption $a > 0$ by assumption $d > 0$.

Notice when $a = d = 0$, $Q(x,y)$ is psd $\iff f = 0 \iff ad - f^2 \ge 0$. This means in general

$Q(x,y)$ is positive semi-definite when and only when $$a \ge 0,\quad d \ge 0,\quad\text{ and }\quad ad - f^2 \ge 0 \iff 4ad - (b+c)^2 \ge 0$$

The is essentially the Sylvester's criterion pointed out by Dietrich Burde in comment.


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