Let $a_{i,j} =a_ia_j$ , $1 ≤ i, j ≤ b$, where $(a_1,a_2,\ldots,a_n)$, are real numbers. Multiple choice question.

Question

Let $a_{i,j} =a_ia_j$ , $1 ≤ i, j ≤ b$, where $(a_1,a_2,\ldots,a_n)$, are real numbers. Let $A =(a_{i,j})$ be the $n ×n$ matrix.

Then

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

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

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

4. For all $(a_1,a_2,\ldots,a_n)$, zero is an eigenvalue of $A$.

For second order matrix I checked that (1) and (2) are not correct. But I am not sure about the others that they are true/false. Thanks to help me.

Answer 1

Hint: The matrix $A$ looks like $\left( a_1 \left( \begin{array}{c} a_1 \\ a_2 \\ \ldots \\ a_n \\ \end{array} \right), a_2 \left( \begin{array}{c} a_1 \\ a_2 \\ \ldots \\ a_n \\ \end{array} \right), \ldots, a_n \left( \begin{array}{c} a_1 \\ a_2 \\ \ldots \\ a_n \\ \end{array} \right) \right)$.

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Assume $n > 1$. Denote the vector $(a_1, ..., a_n)^T$ by $\vec{w}$. Then $A\vec{v} = \left< \vec{w}, \vec{v} \right> \vec{w}$. This is just the orthogonal projection on $\mathrm{span}\{\vec{w}\}$. The image of $A$ is one-dimensional (if $\vec{w} \neq 0$) or zero dimensional (if $\vec{w} = 0$). Thus, it is not possible to make the matrix non-singular and it has zero as an eigenvalue. We have $$\left< A \vec{v}, \vec{v} \right> = \left< \left< \vec{w}, \vec{v} \right> \vec{w}, \vec{v} \right> = \left< \vec{v}, \vec{w} \right>^2 \geq 0 $$ so the matrix is positive semi definite but not necessarily positive definite.