(Linear Algebra - newbie) How to show that some function is an internal product I need some help here. It's hard to me to tackle this kind of question and I'm not used to write math proofs. I need to find values of $t$ that make $$\langle(x_1, x_2), (y_1, y_2)\rangle = x_1y_1 + tx_2y_2$$ an internal product in $\Bbb R^2$.
I have showed that for $3$ of the $4$ properties, t does not matter at all. But for the property that $$\langle u,u\rangle \gt 0, u \neq 0$$ I have $$x_1^2 +tx_2^2 \gt 0$$ and them $$t \gt -((x_1/x_2)^2)$$ The answer seems to be $t \gt 0$ and I'm lost in this. Could anyone give me the right direction to complete the proof?
 A: You committed a silly mistake.$$x_1^2+tx_2^2>0\implies tx_2^2>-x_1^2\implies t\gt-{x_1^2\over x_2^2}$$
A: When t must satisfy the following for all $(x_1, x_2) \in R^2$
$$t\gt-{x_1^2\over x_2^2}$$
$t$ should be meet this condition, $t \gt 0$
A: To demonstrate that a function, say $f$, taking two vectors to a scalar is an internal product (a.k.a. and more commonly referred to as an inner product) is to demonstrate that it satisfies the axioms; that is, $f$ satisfies


*

*linearity (i.e. $f(\alpha x+ \beta y, x) = \alpha f(x,z) + \beta f(y,z)$ for vectors $x,y,z$ and scalars $\alpha, \beta$)

*symmetry (i.e. $f(x,y) = f(y,x)$)

*positive definite (i.e. $f(x,x)\geq 0$ for all vectors $x$ with equality if and only if $x = 0$)


you'll find that your condition requiring $t > 0$ comes from the third axiom about positive definiteness. More specific to your question, don't forget the additional part of positive definite regarding "with equality if and only if $x = 0$". 
A: The matrix associate to the bilinear form $$\langle(x_1, x_2), (y_1, y_2)\rangle = x_1y_1 + tx_2y_2$$
is 
$$
\begin{pmatrix}
1&0\\
0&t
\end{pmatrix}, 
$$
then you have an internal product if and only if is positive define if and only if the determinant is strictly positive, and all principal minor have strictly positive determinant. In your specific case this condition become $t>0$.
A: You have verified the symmetry and the bilinearity of this new product. When it comes to positive definiteness we require that
$$x_1^2 + t\>x_2^2>0\qquad\forall (x_1,x_2)\ne(0,0)\ .\tag{1}$$
That's how far you got, but now you are entangled in the logic. Note that the value of $t$ is given, and we have to check whether this given $t$ is fine, i.e., satisfies $(1)$. If $t\leq0$  then it is easy to produce  vectors $(x_1,x_2)\ne(0,0)$ such that $x_1^2+x_2^2\leq0$. It follows that $t\leq0$ is forbidden. On the other hand, if $t>0$ then $\tau:=\min\{t,1\}>0$, and
$$x_1^2+tx_2^2\geq\tau(x_1^2+x_2^2)>0$$
whenever $(x_1,x_2)\ne(0,0)$.
