If $Q_1$ and $Q_2$ be two positive definite quadratic forms.Then the folowing must be a positive definite quadratic forms

Question

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

1. $Q_1 + Q_2$
2. $Q_1 - Q_2$
3. $Q_1 \cdot Q_2$
4. $Q_1 / Q_2$

I think the first option is correct.

Answer 1

The first one $Q_1+Q_2$ will be again a positive definite quadratic form.

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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 $\left(Q_1 \bot (-Q_2) \right)(0, X_2)=-Q_2(X_2) < 0$.

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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:

$$ \left(Q_1 \cdot Q_2\right)(\lambda X_1,\lambda X_2) = Q_1 (\lambda X_1) \cdot Q_2(\lambda X_2) \\ = \left(\lambda ^2 \cdot Q_1(X_1)\right) \cdot \left(\lambda ^2 \cdot Q_2(X_2)\right) = \lambda ^4 \cdot Q_1(X_1) Q_2(X_2) = \lambda ^\color{Blue}{4} \left(Q_1 \cdot Q_2\right)( X_1, X_2) . $$

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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:

$$ \left(\dfrac{Q_1} {Q_2}\right)(\lambda X_1,\lambda X_2) = \dfrac{Q_1 (\lambda X_1)} {Q_2(\lambda X_2)} \\ = \dfrac {\left(\lambda ^2 \cdot Q_1(X_1)\right)} {\left(\lambda ^2 \cdot Q_2(X_2)\right)} = \dfrac{Q_1(X_1)} {Q_2(X_2)} = \left(\dfrac{Q_1}{Q_2}\right)( X_1, X_2) . $$