TITLE: Difficulty in linear algebra question.
QUESTION [7 upvotes]: Let $a_{ij}=a_ia_j$ $1\leq i,j\leq n$ where $a_1,... a_n$ are real numbers. Let $A=(a_{ij})$ be the $n \times  n $ matrix . Then

It is spoosible to choose $a_1,... a_n$ so as to mke A non singular.
The matrix is positive definite if $(a_1,...,a_n)$ is a nonzero vector.
The matrix A is positive definite for all $(a_1,...,a_n)$.
For all $(a_1,...,a_n)$ zero is an eigen value of A.

My attempt: 
Option 1 is false because if we choose $a_1=\frac{1}{\sqrt{2}} , 
  a_2=\sqrt{2}$ then $2\times 2$ matrix $A$ is singular.
Option 2 is also false because if we take $a_1=1  \ a_2=2 $ then $2\times2$ matrix $A$ is not positive definite.
Option 3 is also false. But I am not getting option 4. Could you please help me? Thanks in advance.

REPLY [4 votes]: As mentioned in the comments, I assume $n \ge 2$.
A key observation is that if $a$ denotes the vector $(a_1,\ldots,a_n)$, then
$$A = aa^\top.$$
1) To prove 1. is true, you need to provide an example. However, to prove 1. is false, you need to show that all choices of $a_1,\ldots,a_n$ result in $A$ being singular; you cannot just provide one example. Hint: compute the rank of $A$. Alternatively, this follows from 4) if you prove 4) first.
2) and 3) To check positive definiteness you need to check $v^\top A v > 0$ for all vectors $v$. If you write $A$ as $aa^\top$, this condition becomes $(a^\top v)^2 > 0$. No matter what $a$ is, this condition will never hold for all $v$, since there is always some $v$ orthogonal to $a$. Your counterexample suffices to answer 2) and 3) as well. Alternatively, if you prove 4) first, then 2) and 3) follow easily since you will have a zero eigenvalue.
4) Hint: if $v$ is orthogonal to $a$, what is $Av$?