positive semidefinite, positive definite? 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 $A$
Please help. I can't even make a guess.
 A: If $v=\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix}$ then $A=(a_{ij})=vv^T$
(1) is false for such a matrix always has rank no more than 1
(2) is false for positive definite matrices better be invertible!
(3) is true, for $x^TAx=x^Tvv^Tx=(v^Tx)^Tv^Tx$ which is a non-negative real number for all vectors $x$ as it is just self inner product.
(4) is true if $n>1$ for then $A$ is singular.
A: If $n=1$ the question is trivial, so assume $n>1$.
The determinant of $A$ is given by the Laplace rule
$$\det (A)=\sum_{\sigma\in S_n}sgn(\sigma)\prod a_{i,\sigma(i)}$$
In your definition $a_{i,\sigma(i)}=a_ia_{\sigma(i)}$ and since $\sigma$ is bijective we have
$$\prod a_{i,\sigma(i)}=\prod a_i^2.$$
For $n\geq 2$ there are equally many positive and negative permutations and therfore the sum vanishes. So the matrix is always singular, has thus the eigenvalue $0$ and is not positive definite.
Edit: For the positive semidefiniteness observe that a matrix is positive semidefinite if and only if it is the Gram matrix of some set of vectors (not necessarily linear independent). Choose $b_i=a_ie_1$ (possibly trivial). We are done.
A: Hint for 2: For any such matrix, the $2\times 2$ matrix in the upper left corner has determinant zero. Look up Sylvester's criterion.
Hint for 4: A real square matrix is singular iff it has zero as an eigenvalue. Add this to the fact other commentors have pointed out the matrix has rank one and...
