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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?

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    $\begingroup$ $x_1^2a=0$ when $x_1=0$. $\endgroup$
    – user1551
    Oct 26, 2022 at 22:42
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Oct 26, 2022 at 22:51

1 Answer 1

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

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