Show that given $A=\left( \begin{array}{cc}\alpha &\beta\\\beta&\delta\\\end{array}\right) \in M_2(\mathbb{R}^2)$, $\langle x,y \rangle = x^TAy, (x,y \in \mathbb{R}^2)$, defines an inner product on $\mathbb{R}^2$ if and only if $\alpha >0$ and det$(A)>0$.
I tried to prove the $\Leftarrow$ direction first:
($\Leftarrow$) Assume that $\alpha > 0, det(A) > 0 \Rightarrow \alpha \delta > \beta^2 \Rightarrow \delta > 0$.
Now I need to show that $x^TAy$ defines an inner product on $\mathbb{R}^2$, so I need to show that $\langle x,y \rangle$ is bilinear, symmetric and positive definite $\forall x,y \in \mathbb{R}^2$
Bilinearity: Let $x_1,x_2,y_1,y_2,x,y \in \mathbb{R}^2, \lambda, \mu \in \mathbb{R}$, then $$\langle \lambda_1 x_1 + \lambda_2 x_2, y \rangle = (\lambda_1 x_1 + \lambda_2 x_2)^TAy= \lambda_1 x_1^TAy + \lambda_2 x_2^TAy = \lambda_1 \langle x_1,y \rangle + \lambda_2 \langle x_2,y \rangle$$
$$ \langle x, \mu_1y_1 + \mu_2y_2 \rangle = x^TA( \mu_1y_1 + \mu_2y_2) = x^TA\mu_1y_1+xA\mu_2y_2= \mu_1 \langle x,y_1 \rangle + \mu_2\langle x, y_2 \rangle$$
Symmetry follows just as easily and is also not affected by $\alpha > 0, det(A)>0$
But I have encountered some problems when trying to prove positive definiteness.
If $ x = \left(
\begin{array}{c}
\xi_1 \\
\xi_2 \\
\end{array}
\right)$, then $$x^TAx = \alpha \xi_1^2 + 2\beta \xi_1 \xi_2 + \delta \xi_2^2$$
Now I'm not sure why this has to be greater than 0 if $\alpha > 0, det(A)>0$, if I plug in numbers it works every time, but I can't do the math to prove this. Can anybody help me please?
I also don't really know how to prove the other direction, so would appreciate some help on that as well.
Best Answer
This is a general theorem: a real symmetric square matrix defines an inner product iff it is positive definite, which means $\;\alpha\,,\,\det A >0\;$ , as given:
$\implies:\;\;$ taking $\;x=y=\binom10\;$ , we have that
$$x^tAx=(1\;0)\begin{pmatrix}\alpha&\beta\\\beta&\delta\end{pmatrix}\binom10=(1\;0)\binom\alpha\beta=\alpha$$
Since $\;x\neq 0\;$ the above must be positive and we have thus that $\;\alpha>0\;$
Taking $\;x=(0,1)\;$ and doing something similar as above, we get $\;\delta>0\;$ .
Finally, taking $\;x=\binom\delta{-\beta}\neq0\;$ ,we get
$$x^tAx=(\delta\;-\beta)\binom{\alpha\delta-\beta^2}{0}=\delta(\alpha\delta-\beta^2)$$
Again, the above must be positive, and since $\;\delta>0\;$ we get $\;\det A:=\alpha\delta-\beta^2>0\;$