# positive semidefinite, positive definite? [duplicate]

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

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

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

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

4. for all $(a_1,\dots,a_n)$, $0$ is an eigen value of $A$

-
Consider the case $n = 2$. Then $A = \begin{pmatrix} a_1^2 & a_1 a_2 \\ a_1 a_2 & a_2^2 \end{pmatrix}$ is always singular because $\det(A) = 0$. In fact, as defined, the matrix $A$ always has rank $\leq 1$ and so is singular for all $n > 1$. –  Michael Joyce Jun 13 '12 at 15:06
Note that $A = \begin{pmatrix}a_1 \\ \vdots \\ a_n\end{pmatrix} \begin{pmatrix} a_1 & \cdots & a_n\end{pmatrix}$. –  Dylan Moreland Jun 13 '12 at 15:08
@DylanMoreland Could you please tell me what does that indicate? –  Bunuelian Trick May 25 '13 at 4:37

## marked as duplicate by Lord_Farin, azimut, Tom Oldfield, Chris Godsil, Davide GiraudoMay 25 '13 at 11:22

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.

-
Small nit: $v^Tx$ is a scalar, so $(v^Tx)^T=(v^Tx)$ and $(v^Tx)^Tv^Tx=(v^Tx)^2$. And that's really what establishes the non-negativity. –  Michael C. Grant May 31 '13 at 7:20
@MichaelC.Grant Thank you! –  Bunuelian Trick Jun 18 '13 at 6:55

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.

-
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.