# Proving that matrix is positive definite

I have a set of real variables $\{a_1,a_2,a_3,...,a_n\}$.
Let $u=\sqrt{a_1^2+a_2^2+...+a_n^2}$. I am trying to prove that the following matrix is positive definite: $A\in \mathbb{R}^{n\times n}$,
$$A_{ij}=-a_i a_j$$ $$A_{ii}=u+u^2-a_i^2$$ where $i,j=1,2,...,n$, and $i\neq j$.
I can't find a method to prove it. I tried proving directly using $x^TAx$, but I didn't get anywhere. Calculating eigenvalues seem to be getting too complicated too fast. Same with determinants.
What I understood so far, is that matrix is symmetric, and that all values in its main diagonal are non-negative. All values on the main diagonal are positive if there is at least one $a_i$ that is nonzero, for some $i$.

## 1 Answer

The matrix isn't necessarily positive definite, since it's the zero matrix if all $a_i$ are zero. However, it's positive semidefinite, and is positive definite if at least one $a_i$ is non-zero.

Let $a$ denote the vector $(a_1,\dotsc,a_n)$. The projection matrix $aa^\top$ has one eigenvalue $a^\top a=u^2$ and $n-1$ eigenvalues $0$. Your matrix is $(u+u^2)I-aa^\top$, so it has one eigenvalue $u$ and $n-1$ eigenvalues $u^2+u$, all of which are non-negative, and all of which are positive if and only if all $a_i$ are non-zero.

• "..,so it has one eigenvalue $u$ and $n-1$ eigenvalues $u^2+u$". How did u calculate this? – Sunny88 Feb 25 '13 at 15:41
• @Sunny88: Adding $\lambda I$ to $-aa^\top$ shifts all its eigenvalues by $\lambda$. The one eigenvalue $-u^2$ is shifted to $(u+u^2)-u^2=u$, and the $n-1$ eigenvalues $0$ are shifted to $(u+u^2)+0=u+u^2$. – joriki Feb 25 '13 at 15:52
• I think that in your answer "if and only if all $a_i$ are non-zero" should be changed to "if and only if not all $a_i$ are zero". – Sunny88 Dec 4 '13 at 23:13