# We have mapping $\left \langle x,x \right \rangle = x^{T}Ax$. Prove that $\left \langle x,x \right \rangle > 0$ iff...

Let's suppose we have the mapping $\left \langle x,y \right \rangle = x^{T}Ay$ ,

A is the symmetric matrix $\begin{pmatrix} a & b\\ b & c \end{pmatrix}$, $a,c >0$

And I need to proof that $\left \langle x,y \right \rangle$ is a scalar product iff $ac-b ^{2} > 0$.

I've already proved that this mapping is linear and $\left \langle x,y \right \rangle = \left \langle y,x \right \rangle$ for every $x$ and $y$.

Now I want to show that $\left \langle x,x \right \rangle > 0$ iff $ac-b ^{2} > 0$, for $x\neq 0$ .

Let's suppose that $x=\binom{x_1}{x_2}$

So I came to this

$$x_{1}^{2}a + 2x_{1}x_{2}b + x_{2}^{2}c > 0$$

$$x_{1}^{2}a + b^{2}\left \langle x,x \right \rangle + x_{2}^{2}c > 0 .$$

Now I see that $x_{1}^{2}+x_{2}^{2}= \left \| x \right \|^{2}$ but I stack here, because I don't know how to use this fact.

• Do you know that $x^TAx$ is an inner product if and only if $A$ is symmetric positive definite?
– Surb
Feb 25 '17 at 14:46

As mentioned in the comments and the other answer, this problem can easily be answered using eigenvalues of $A$.
However, if you don't know what eigenvalues are, you can simply multiply the inequality $x_{1}^{2}a + 2x_{1}x_{2}b + x_{2}^{2}c > 0$ by $a>0$, which gives you $$x_{1}^{2}a^2 + 2x_{1}x_{2}ab + x_{2}^{2}ac > 0.$$ Now rewrite the right side as $$x_{1}^{2}a^2 + 2x_{1}x_{2}ab +x_2^2b^2+ x_{2}^{2}(ac-b^2)=(ax_1+bx_2)^2+(ac-b^2)x_2^2.$$
Now if $ac-b^2\leq 0$, this can be nonpositive for $(x_1,x_2)=(b,-a)\neq (0,0)$. On the other hand, if $ac-b^2>0$, then it is nonnegative, and can only be $0$ when $x_2=x_1=0$.
You can make your life easier by writing $M$ with respect to a basis of eigenvectors. $M$ then becomes a diagonal matrix, with diagonal entries equal to its eigenvalues.
Now think about what condition the eigenvalues must satisfy in order that $\langle x, x \rangle > 0$ for all $x$.