How to prove the necessary conditions for checking the positive semidefiniteness of a $2$ by $2$ matrix?

Question

How to prove that the following conditions are necessary and sufficient for checking the positive semidefiniteness of a $2 \times 2$ matrix $A$?

$$ a_{11}\geq 0, \qquad a_{22}\geq 0, \qquad a_{11}a_{22} \geq a_{21}^2 $$

Answer 1

Consider a symmetric matrix $$ A=\left( \begin{array}{cc} a_{11}&a_{12}\\ a_{12}&a_{22} \end{array} \right). $$ $A$'s being positive semi-definite is defined as for all $$ \mathbf{x}=\left( \begin{array}{c} x_1\\ x_2 \end{array} \right), $$ it satisfies $$ \mathbf{x}^{\top}A\mathbf{x}\ge 0\iff a_{11}x_1^2+2a_{12}x_1x_2+a_{22}x_2^2\ge 0. $$

Necessity

Fix $x_2=0$, and the equality reduces to $$ a_{11}x_1^2\ge 0. $$ This leads to $$ a_{11}\ge 0. $$ Similarly, fixing $x_1=0$ would lead to $$ a_{22}\ge 0. $$

To show that $a_{11}a_{22}\ge a_{12}^2$, let us first show that, e.g., $a_{12}=0$ as long as $a_{22}=0$. In fact, if $a_{22}=0$, the inequality reduces to $$ a_{11}x_1^2+2a_{12}x_1x_2\ge 0. $$ If $a_{12}\ne 0$, the arbitrariness of $x_2$ would have this last inequality violated. Hence it is a must that $a_{12}=0$. Thanks to this result,let $x_1=a_{22}$ and $x_2=-a_{12}$, and the inequality becomes $$ a_{11}a_{22}^2+2a_{12}a_{22}\left(-a_{12}\right)+a_{22}a_{12}^2\ge 0, $$ or equivalently, $$ a_{22}\left(a_{11}a_{22}-a_{12}^2\right)\ge 0. $$ If $a_{22}\ne 0$, the above inequality gives $$ a_{11}a_{22}-a_{12}^2\ge 0; $$ if $a_{22}=0$, as is proven from above, $a_{12}=0$, and $$ a_{11}a_{22}-a_{12}^2=0\ge 0. $$ Thus $a_{11}a_{22}-a_{12}^2\ge 0$ unconditionally.

Sufficiency

We shall show that $$ a_{11}x_1^2+2a_{12}x_1x_2+a_{22}x_2^2\ge 0 $$ holds for all $x_1$ and $x_2$, as long as $$ a_{11}\ge 0,\quad a_{22}\ge 0,\quad a_{11}a_{22}-a_{12}^2\ge 0. $$

If $a_{11}=0$, the three constraints reduce to $$ a_{22}\ge 0,\quad a_{12}=0, $$ which gives $$ a_{22}x_2^2\ge 0. $$ Thus the desired inequality holds true when $a_{11}=0$.

When $a_{11}\ne 0\iff a_{11}>0$, we have \begin{align} a_{11}x_1^2+2a_{12}x_1x_2+a_{22}x_2^2&=a_{11}\left(x_1^2+2\frac{a_{12}}{a_{11}}x_1x_2+\frac{a_{22}}{a_{11}}x_2^2\right)\\ &=a_{11}\left[\left(x_1+\frac{a_{12}}{a_{11}}x_2\right)^2+\frac{a_{22}}{a_{11}}x_2^2-\frac{a_{12}^2}{a_{11}^2}x_2^2\right]\\ &=a_{11}\left[\left(x_1+\frac{a_{12}}{a_{11}}x_2\right)^2+\frac{a_{11}a_{22}-a_{12}^2}{a_{11}^2}x_2^2\right]. \end{align} Obviously, the constraint $a_{11}a_{22}-a_{12}^2\ge 0$ would make the right-hand-side non-negative, i.e., $$ a_{11}x_1^2+2a_{12}x_1x_2+a_{22}x_2^2\ge 0. $$

To sum up, $A$ is positive semi-definite if and only if $$ a_{11}\ge 0,\quad a_{22}\ge 0,\quad a_{11}a_{22}-a_{12}^2\ge 0. $$

Answer 2

Let $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}\in M_2(\mathbb{R})$ be a symmetric matrix. By definition, $A$ is $\geq 0$ iff for every vector $z=[x,y]^T$, $f(z)=z^TAz=ax^2+2bxy+cy^2\geq 0$.

$\textbf{Proposition.}$ $A\geq 0$ iff $\{a\geq 0,c\geq 0,ac-b^2\geq 0\}$.

$\textbf{Proof}$. i) If $a\not= 0$, then $f(z)$ is a polynomial of degree $2$ in $x$; the NSC (necessary and sufficient condition for $A\geq 0$) is $a> 0$ and, for every $y$, $(b^2-ac)y^2\leq 0$, that is $a> 0,ac-b^2\geq 0$, that is $a>0,c\geq 0,ac-b^2\ge 0$.

ii) If $c\not= 0$, then $f(z)$ is a polynomial of degree $2$ in $y$; in the same way as above, we obtain that the NSC is $a\geq 0,c>0,ac-b^2\geq 0$.

iii) If $a=c=0$, then the NSC is $b=0$ ($xy$ has not a constant signum), that is $a=c=0,ac-b^2\geq 0$.

We deduce the required equivalence. $\square$