Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$. Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$. Consider the $n \times n$ matrix $A=(a_{ij})$.
Then


*

*It is possible to choose $a_1.\dots,a_n$ such that $A$ is non-singular

*matrix $A$ is positive definite if $(a_1,\dots,a_n)$ is nonzero vector

*matrix $A$ is positive semi definite for all $(a_1,\dots,a_n)$

*for all $(a_1,\dots,a_n)$, $0$ is an eigen value of $A$
I have calculated upto $3\times 3$ that determimant is $0$ but I have no idea how to conclude rigoriously. please help?
 A: It depends.
With $a=(a_1,\ldots,a_n)$, we can more conveniently write $v\mapsto \langle a,v\rangle a$ instead of $v\mapsto Av$, which simplifies computations.


*

*If $n=1$, then $a_1=1$ is a valid choice and makes $A$ regular (the identity). For $n>1$, there exists nonzero $v$ with $\langle a,v\rangle = 0$, hence $A$ is singular. (To be explicit, either some $a_i$ is $=0$ and then the $i$th standard base vector $e_i$ is in the kernel; or $a_1,a_2$ are nonzero and $a_1e_2-a_2e_1$ is in the kernel. That is: Such a choice is possible if and only if $n=1$.

*$\langle Av,v\rangle = \langle \langle a,v\rangle a,v\rangle = \langle a,v\rangle^2\ge 0$, so we see that $A$ is positive semi-definite. But if $n>1$ or $a=0$, it is defnitely not positive definite, as follows from 1. Hence the claim is true only for $n=1$

*As shown in 2, this is correct

*As shown in 1, this is correct only if $n>1$.
So the only claim that is unrestrictedly correct, is 3.
A: Here are a few comments: 
First, 4. implies that $A$ is always singular, answering 1. Follow Gerry Myerson's advice to prove 4 (basically, you are trying to show that the first two columns are linearly dependent). Now 2. cannot be true since 0 is an eigenvalue. 
Also, it may be helpful to think of $A$ as the product $a a^T$ where $a$ is the column vector of the $a_i's$. Edit: you can use this to show 3.
