Let $a_{ij} = a_ia_j$ for $1 \le i,j \le n$ and $a_1,a_2,\dots,a_n$ are real numbers. Let $A = (a_{ij})$ be the $n \times n$ matrix. Then

1) It is possible to choose $a_1,a_2,\dots a_n$ so as to make the matrix $A$ non-singular.

2) The matrix $A$ is positive definite if $(a_1,a_2,\dots,a_n)$ is a non-zero vector.

3) The matrix $A$ is positive definite for all $(a_1,a_2,\dots,a_n)$.

4) For all $a_1,a_2,\dots,a_n$ we have zero is an eigen value of $A$.

Click 1) It is possible to choose $a_1,a_2,\dots a_n$ so as to make the matrix $A$ non-singular.

2) The matrix $A$ is positive definite if $(a_1,a_2,\dots,a_n)$ is a non-zero vector.

3) The matrix $A$ is positive definite for all $(a_1,a_2,\dots,a_n)$.

4) For all $a_1,a_2,\dots,a_n$ we have zero is an eigen value of $A$.

**(We assume $n>1$. otherwise it is stright forward.)****Solution**:**option 1**: (**False**)Let $v = (a_1,a_2,\dots,a_n)$, then the first row of the matrix $A$ is given by the vector $(a_1a_1,a_1a_2,a_1a_3,\dots,a_1a_n) = a_1 v$ ($v$ scalar multiplied by the scalar $a_1$). Similarly the $i$th row of $A$ is given by $a_i v$. This shows that, all the rows of $A$ are some scalar multiples $v$. Therefore the dimension of the row space of $A$ = row rank of $A$ = rank of $A \le 1$. Therefore $A$ is singular for all choices of $v$. (unless n=1 the matrix $A$ is an $1 \times 1$ matrix).**option 2**: (**False**) If $A$ is positive definite then all its eigen values are positive. Hence $\text{det }A \ne 0$. We have shown that $A$ has determinant zero for all $(a_1,a_2,\dots,a_n)$.**option 3**: (**True**) If $A$ is symmetric then $A$ is positive semi-definite if and only if all its principal minors are non-negative. We will prove that, all the principal minors of $A$ are non-negative. Let $ 1 \le k \le n$ and consider the principal $k$ minor $A(k)$ given by $A(k)_{ij} = a_{ij}$ for $1\le i,j \le k$. Let $v(k) = (a_1,a_2,\dots,a_k)$, then the first row of the matrix $A(k)$ is given by the vector $(a_1a_1,a_1a_2,a_1a_3,\dots,a_1a_k) = a_1 v(k)$ ($v(k)$ scalar multiplied by the scalar $a_1$). Similarly the $i$th row of $A(k)$ is given by $a_i v(k)$. This shows that, all the rows of $A(k)$ are some scalar multiples $v(k)$. This shows that dimension of the row space of $A(k)$ = row rank of $A(k)$ = rank of $A(k) \le 1$. Therefore $A(k)$ is singular. (unless k=1, this case $A(1)$ can be invertible). This shows that $A$ is positive semi definite.**option 4**:(**True**) We have shown that $A$ is not invertible and hence zero is an eigen value of $A$.**here**for more problems.

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