Show that $\langle x,y\rangle_A = \langle Ax,Ay\rangle$ is an inner product on $\mathbb R^n$

Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then $\langle x,y\rangle_A$ is an inner product iff

1) $\ker A = \{0\}$.

2) $\operatorname{rank} A= n$.

3) all eigenvalues of $A$ are positive.

4) all eigen value of $A$ are non negative.

Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.

I am unable to check the 3) and 4) option. Please help to check the other option. Thank you

If $\lambda=0$ is an eigenvalue of $A$ then there is a nonzero $v$ for which $Av=\lambda v=0,$ and you have already noted that in this case $\langle Av,Av \rangle=0.$ So that only leaves possibility 3. There would still be needed a proof that if all eigenvalues positive then one gets an inner product this way.

Added: You seem to have a logic error in treating option (1). That says, that iff kernel of $A$ is zero, then one gets an inner product. But what you did was to assume kernel nonzero, then get that one doesn't get an inner product. This is one direction of the iff statement (1), rather than a refutation of choice (1). So it may be that choice (1) is the correct one after all.

As for choice (3), it does not work since if $A=-I$ all eigenvalues of $A$ are $-1,$ yet one does get an inner product by the construction.

So my suggestion is to try to show the other direction for choice (1), i.e. that if kernel of $A$ is zero then one does get an inner product by the construction.

@user120386 Since we know that $Ker A =\{ 0 \}$ and $Rank$ $A = n $, for $option$ $3$ consider the forward direction that a matrix $A$ such that $\langle x,y\rangle_A$ is an inner product defined by $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , lets consider a matrix $A$ of order $2$ as $A =\begin{bmatrix}1&-2\\-2&1\end{bmatrix}$ , then $A$ still defines inner product as $Ker A = \{ 0 \}$ but the eigenvalues of $A$ are $-1$ and $3$ , violating the 3rd option that all eigenvalues are positive and since the if condition is violated there is no iff statement left . thus 3rd option is incorrect .

 Please comment if the above answer seem incorrect.