Proving a function is an inner product on $\mathbb{R}^2$ if and only if $b^2-ac < 0$ and $a > 0$.

Question

Show that the function $$ \langle \textbf{x} , \textbf{y} \rangle = \begin{bmatrix} x_1 & x_2 \\ \end{bmatrix} \begin{bmatrix} a & b \\ b & c \\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix}$$ is an inner product on $\mathbb{R}^2$ if and only if $b^2-ac < 0$ and $a > 0$. The properties of the inner product are:

(1) $\langle \textbf{x} , \textbf{y} \rangle = \langle \textbf{y} , \textbf{x} \rangle$.

(2) $\langle \textbf{x} + \textbf{z} , \textbf{y} \rangle = \langle \textbf{x} , \textbf{y} \rangle + \langle \textbf{z} , \textbf{y} \rangle$.

(3) $\langle c\textbf{x} , \textbf{y} \rangle = c\langle \textbf{x} , \textbf{y} \rangle = \langle \textbf{x} , c\textbf{y} \rangle$.

(4) $\langle \textbf{x} , \textbf{x} \rangle > 0$ if $\textbf{x} \neq \textbf{0}$.

$\textbf{My solution:}$ Properties (1), (2) and (3) are clear, and are satisfied no matter the values of $a, b$ and $c$. So lets prove property (4) holds if and only if $b^2-ac < 0$ and $a > 0$.

Suppose $b^2-ac < 0$ and $a > 0$. Then $c>0$ and $\sqrt{ac} > b > -\sqrt{ac}$. Let $\textbf{x}=(x_1,x_2) \neq \textbf{0}$. If $x_1=0$, then $\langle \textbf{x} , \textbf{x} \rangle = ax_2^2 >0$; similarly if $x_2 =0$. So suppose $x_1$ and $x_2$ are non-zero. There are two cases, $x_1x_2>0$ or $x_1x_2<0$. If $x_1x_2>0$, then $$ \langle \textbf{x} , \textbf{x} \rangle = ax_1^2+cx_2^2 +2bx_1x_2 = (\sqrt{a}x_1 - \sqrt{c}x_2)^2+2x_1x_2(b+\sqrt{ac})>0$$ as $b>-\sqrt{ac}$. Now, if $x_1x_2<0$, then $$ \langle \textbf{x} , \textbf{x} \rangle = ax_1^2+cx_2^2 +2bx_1x_2 = (\sqrt{a}x_1 + \sqrt{c}x_2)^2+2x_1x_2(b-\sqrt{ac})>0$$ as $b < \sqrt{ac}$. Therefore, for all $\textbf{x} \neq 0$ we have $\langle \textbf{x} , \textbf{x} \rangle > 0$.

However, I am not sure how to approach the other direction. I tried to consider $ax_1^2+cx_2^2 +2bx_1x_2$ as a polynomial in $x_1$ and $x_2$ separately, and look at the discrimant, and I got $$\triangle = b^2- ac$$ in both cases. However, this says $b^2 \geq ac$ gives real roots, and $b^2< ac$ gives complex roots.

I appreciate any help or comment.

Answer 1

Try just writing out what $$ \langle \textbf{x} , \textbf{x} \rangle = \begin{bmatrix} x_1 & x_2 \\ \end{bmatrix} \begin{bmatrix} a & b \\ b & c \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} $$ being nonnegative means. To me, that says $$ a x_1^2 + 2b x_1 x_2 + c x_2 ^ 2 \ge 0.$$

for all $x_1, x_2$.

If you fix $x_2 = 1$, for instance, that becomes

$$ a x_1^2 + 2b x_1 + c \ge 0. $$

A quadratic like that is nonzero everywhere on the real line only if its discriminant is nonpositive, i.e., only if $(2b)^2 - 4ac \le 0$, i.e., $b^2 - ac \le 0$.

For it to be nonnegative everywhere, we also need $a > 0$.

Answer 2

The value of $x^T A x$ is positive for all $x\neq 0$, iff A is positive definite (by definition). To check, if a matrix is positive definite, you must check all eigenvalues to be positive or you check the determinant of the major-minors (in German: Hauptminoren, I cannot find a english translation) of the matrix to be positive. The determinant of $a$ is just $a$ and the determinant of your matrix block is $ac-b^2$, which is your assumption.

If either value is negative, the matrix is not positiv definite.