# Why is the following matrix not positive-definite (for $a > 0$)?

Consider the matrix $$A$$ defined as follows $$\begin{pmatrix}a & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$, where $$a > 0$$.

According to the definition, $$A$$ is positive definite if for any $$x \in \mathbb{R}^{3} \backslash\{0\}$$, $$x^{T}Ax > 0$$. Calculating $$x^{T}Ax = x_{1}^{2}a$$ which is clearly bigger than zero. Hence the matrix must be positive-definite.

From the other hand, it has one positive and two zero eigen values which suggests that the matrix is not positive-definite but only positive semi-definite.

Wolfram alpha also says that this matrix is not positive-definite but positive semi-definite. What am I missing?

• $x_1^2a=0$ when $x_1=0$. Oct 26, 2022 at 22:42
• By the way, if only real vectors $x$ are used, the usual definition of (real) positive definite matrix requires the matrix to be symmetric. E.g. when $A=\pmatrix{1&-1\\ 1&1}$ we have $x^TAx>0$ for every nonzero real vector $x$, but this $A$ is not positive definite because it is not symmetric. Oct 26, 2022 at 22:51

## 1 Answer

If $$x=(0,1,1)^T\neq\vec0$$ then $$x^TAx=0$$.