If $Q_1$ and $Q_2$ be two positive definite quadratic forms.Then the folowing must be a positive definite quadratic forms If $Q_1$ and $Q_2$ be positive definite quadratic forms.
Then which of the folowings is a positive definite quadratic form?


*

*$Q_1 + Q_2$ 

*$Q_1 - Q_2$ 

*$Q_1 \cdot Q_2$ 

*$Q_1 / Q_2$ 


I think the first option is correct.
 A: The first one $Q_1+Q_2$ will be again a positive definite quadratic form. 

The second one; 


*

*If we think of $Q_1-Q_2$ as $Q_1 \bot (-Q_2)$;
then in this sense it is clearly an indefinite quadratic form.
Note that $\left(Q_1 \bot (-Q_2) \right)(X_1, 0)=Q_1(X_1) > 0$;
also 
note that $\left(Q_1 \bot (-Q_2) \right)(0, X_2)=-Q_2(X_2) < 0$. 



The third one; $Q_1 \cdot Q_2$ is not even a quadratic form 
(except the cases $Q_1 \equiv 0$ and $Q_2 \equiv 0$);
it is a $\color{Blue}{\text{quartic}}$ form.
Because for every arbitrary $\lambda$; we have: 
$$ 
\left(Q_1 \cdot Q_2\right)(\lambda X_1,\lambda X_2) 
= 
Q_1 (\lambda X_1) \cdot Q_2(\lambda X_2) 
\\ 
= 
\left(\lambda ^2 \cdot Q_1(X_1)\right) 
\cdot 
\left(\lambda ^2 \cdot Q_2(X_2)\right) 
= 
\lambda ^4 \cdot Q_1(X_1) Q_2(X_2) 
= 
\lambda ^\color{Blue}{4} \left(Q_1 \cdot Q_2\right)( X_1, X_2) 
. 
$$

The last one $Q_1 / Q_2$ is not even a quadratic form 
(except the case $Q_1 \equiv 0$);
it will be a form of degree zero over the set 
$\{(X_1, X_2) : Q_2(X_2)\neq0\}$.
[In our special case 
$\{(X_1, X_2) : Q_2(X_2)\neq0\}= \{(X_1, X_2) : X_2\neq0 \}$ 
.]  
Because for every arbitrary $\lambda$; we have: 
$$ 
\left(\dfrac{Q_1} {Q_2}\right)(\lambda X_1,\lambda X_2) 
= 
\dfrac{Q_1 (\lambda X_1)} {Q_2(\lambda X_2)} 
\\ 
= 
\dfrac 
{\left(\lambda ^2 \cdot Q_1(X_1)\right)} 
{\left(\lambda ^2 \cdot Q_2(X_2)\right)} 
= 
\dfrac{Q_1(X_1)} {Q_2(X_2)} 
= 
\left(\dfrac{Q_1}{Q_2}\right)( X_1, X_2) 
. 
$$
