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By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that \begin{equation} X=\left[ \begin{array}{cc} x & y \\ y & z \end{array} \right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge0, \quad xz\ge y^2,\tag{1} \end{equation} where $$ S_+^2 = \left\{X\in S^2 | X \succeq 0\right\}. $$

I have failed to prove the $(1)$.

$$ \begin{align} \forall \alpha,\beta,\quad \left[ \begin{array}{cc} \alpha & \beta \end{array} \right] \left[ \begin{array}{cc} x & y \\ y & z \end{array} \right] \left[ \begin{array}{cc} \alpha \\ \beta \end{array} \right] &\ge 0\\ \alpha^2x + 2\alpha\beta y + \beta^2z &\ge 0\\ \left(\alpha\sqrt{x} + \beta\sqrt{z}\right)^2 + 2\alpha\beta(y-\sqrt{xz}) &\ge 0 \\\therefore \alpha\beta(y-\sqrt{xz})\ge0\\ \text{if}\quad\alpha\beta\ge0,\quad \text{then} \quad y^2\ge xz\\ \text{if}\quad\alpha\beta\le0,\quad \text{then} \quad y^2\le xz \end{align} $$

How can I reach to $$ x\ge0,\quad z\ge0,\quad xz\ge y^2\qquad? $$

Danny_Kim
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2 Answers2

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This is essentially the implication from the Sylvester's criterion extended to the case of positive semi-definite matrices.

All of the leading principal minors must be nonnegative. So you immediately get $x\ge 0, x z\ge y^2, z\ge 0$.

Following your proof for $2\times2$ matrices, take consequently $\alpha = 0$ and $\beta = 0$ to get $x\ge 0, z\ge 0$. The last step is to take $\alpha = \beta = 1$

$$2y \ge -x-z\Rightarrow 4y^2 \ge (x+z)^2 \ge 4xz$$

Andrei Kulunchakov
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For the 2x2 case, you can form the characteristic polynomial, obtaining an expression for the eigenvalues using the quadratic formula. Setting the eigenvalues greater than or equal to zero yields the inequalities:

$2 \lambda_{+} = (x + z) + \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$

$2 \lambda_{-} = (x + z) - \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$

Observe that $(x+z)$ less than zero implies that $\lambda_{-}$ will always be negative. Since we require both eigenvalues to be positive, $x+z \ge 0$.

$x+z \ge 0 \implies \lambda_{+} \ge 0$

(the square roots will not yield complex numbers, because real symmetric matrices always have real eigenvalues)

$\lambda_{-} \ge 0 \implies (x+z) - \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$

$\implies (x+z) \ge \sqrt{(x+z)^2 + 4(y^2-xz)}$

$\implies (x+z)^2 \ge (x+z)^2 + 4(y^2-xz)$

$\implies 0 \ge y^2-xz$

$\implies xz \ge y^2$

We thus have $x+z \ge 0$ and $xz \ge y^2$ being the conditions which ensure that $X$ is positive semidefinite.

Now observe that $x$ and $z$ cannot have opposing signs if their product is greater than or equal to the (always positive) $y^2$. It follows that $x \ge 0$ and $z \ge 0$.

Putting it altogether, $X$ is positive semidefinite if and only if $x \ge 0$, $z \ge 0$ and $xz \ge y^2$.

GingerBreadMan
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