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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.

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  • $\begingroup$ Do you know that $x^TAx$ is an inner product if and only if $A$ is symmetric positive definite? $\endgroup$
    – Surb
    Feb 25 '17 at 14:46
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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 $x_2=x_1=0$.

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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 $x$.

Also, do you know that the trace and determinant of a matrix is independent of the choice of basis? What is the trace and determinant of this matrix, as computed in the original basis? And now, working in the eigenvector basis, can you see how the trace and determinant are related to the eigenvalues?

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