On Sylvester’s criterion for $2 \times 2$ matrices

Question

In my electrical engineering text, I have an expression of the form: $$E_m(x_1, x_2) = \frac{1}{2}(A_{11}x_{1}^2+2Mx_1x_2+A_{22}x_{2}^2) > 0$$ where $$A = \begin{bmatrix} A_{11} & M \\ M & A_{22} \end{bmatrix}$$ and so the expression can be written in matrix form as $$E_m(x_1, x_2) = \frac{1}{2}\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} A_{11} & M \\ M & A_{22} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} >0 $$ The claim is that $E_m(x_1, x_2) >0$ if and only if $A_{11}>0$, $A_{22} >0$, and $M^2 < A_{11}A_{22}$ . But I'm having a hard time seeing why? $x_1$ and $x_2$ are arbitrary and can be positive or negative independently. Can someone show me how we know these sign constraints about the elements of $A$?

Answer 1

(In this answer, I assume that $x_1$ and $x_2$ are not zero simultaneously.)

Let us first exclude the case where both $A_{11}$ and $A_{22}$ are both zero because $M x_1 x_2 >0$ cannot hold for arbitrary values of $x_1$ and $x_2$.

Without loss of generality, since the expression is symmetric with respect to $x_1$ and $x_2$, let us assume that $A_{11} \neq 0$.

For $x_2=0$ one can easily see that$$E_m(x_1,0)=\frac{1}{2}(A_{11}x_1^2 + 2M x_10+ A_{22}0)=\frac{1}{2}A_{11}x_1^2 >0$$holds if and only if $A_{11} >0$. Similarly, for the case $A_{22} \neq 0$ one concludes that $A_{22} >0$.

So, let us assume that $x_2 \neq 0$. By factoring out $\frac{1}{x_2^2}$, we have$$E_m(x_1,x_2)= \frac{1}{2x_2^2}\left (A_{11} \left ( \frac{x_1}{x_2} \right )^2 + 2M \left ( \frac{x_1}{x_2} \right ) + A_{22} \right ).$$Since $\frac{1}{2x_2^2}>0$, the above expression is greater than zero if and only if$$A_{11} \left ( \frac{x_1}{x_2} \right )^2 + 2M \left ( \frac{x_1}{x_2} \right ) + A_{22} >0.$$Since $x_1$ and $x_2$ are arbitrary, the above relation holds if and only if$$A_{11}x^2+2Mx + A_{22} >0,$$where $x=\frac{x_1}{x_2}$ is any arbitrary real number.

Now, one can easily conclude (see "Addendum") the last inequality holds if and only if \begin{gather}(2M)^2 < 4A_{11}A_{22}, \quad A_{11} >0 \\ \Rightarrow \quad M^2 < A_{11}A_{22}, \quad A_{11}>0. \end{gather} Considering all possible cases, we conclude that$$E_m(x_1, x_2) = \frac{1}{2}(A_{11}x_1^2+ 2Mx_1x_2+A_{22}x_2^2) > 0$$if and only if$$A_{11}>0, \quad A_{22}>0, \quad M^2< A_{11}A_{22}.$$

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Addendum

Proposition. The inequality $ax^2+bx+c>0$, where $a \neq 0$, holds if and only if$$b^2<4ac, \qquad a>0.$$

Proof.

The following fact can be easily obtained: \begin{align} & b^2<4ac \\ \Leftrightarrow \quad & \frac{4ac-b^2}{4a^2} >0, \qquad (a\neq 0) \\ \Leftrightarrow \quad & \left ( x+ \frac{b}{2a} \right ) ^2 + \frac{c}{a} - \frac{b^2}{4a^2}>0 \\ \Leftrightarrow \quad & x^2 + \frac{b}{a}x + \frac{c}{a} > 0 \\ \Leftrightarrow \quad & ax^2+bx+c >0 \qquad (a>0). \end{align}