Finding where the proof that if $A$ and $B$ are positive-definite matrices, then $AB$ is positive-definite is wrong.

Question

I know the claim is wrong, yet I found an interesting 'proof' that shows the claim is right, and I haven't been able to show why it's wrong.

It goes as follows:

I'll use the claim that $\langle Ax, y \rangle$ is inner product iff A is positive-definite.

Since, A and B are positive-definite, then $\langle Ax, y \rangle$, $\langle Bx, y \rangle$ are inner products.

From here, we can deduce that $\langle ABx, y \rangle = \langle Bx, y \rangle_{A} = \langle x, y \rangle_{AB}$, and from the claim above we get that $AB$ is positive-definite.

What is wrong in the proof?

Answer 1

The mistake is in

From here, $\color{red}{\text{we can deduce that}}$ $\langle ABx, y \rangle = (Bx, y) = \{x, y\}$, where $\{x, y\}$ $\color{red}{\text{is an inner product}}$, and from the claim above we get that $AB$ is positive-definite.

Avoiding notation, you are saying that $(Ax)^Ty$ and $(Bx)^Ty$ being inner products (which is correct) implies that $(ABx)^Ty$ is also an inner product. Why is that? It would certainly imply that $AB$ is positive definite, but you have to show that $(ABx)^Ty$ is also an inner product (which is false in general).

For example, $A=B=\begin{pmatrix} 2 & 3 \\ 0 & 2\end{pmatrix}$ define the inner product $(Ax)^Ty=(Bx)^Ty$ since $A$ is positive definite as $$x^T A x=2x_1^2+3x_1x_2+2x_2^2=\frac{1}{2}(x_1^2+x_2^2)+\frac{3}{2}(x_1+x_2)^2>0\text{ whenever }x\neq0$$ but $AB=A^2=4\begin{pmatrix}1 & 3\\ 0 & 1\end{pmatrix}$ doesn't define an inner product since, taking $x=\binom{1}{-1}$, you get $(A^2x)^Tx=-4<0$.