Did I spot a mistake in Riley, Hobson, Bence?

Question

Let $f({\bf x})\equiv f(x_1,x_2,\ldots,x_n)$ is a function of $n$ real variables ${\bf x}=(x_1,x_2,\ldots,x_n)$. The Taylor expansion of $f({\bf x})$ is about a local ${\bf x}^0=(x_1^0,x_2^0,\ldots,x_n^0)$, reads, $$f({\bf x})=f({\bf x}^0)+\frac{1}{2!}\sum_i\sum_j\left(\frac{\partial^2 f}{\partial x_i\partial x_j}\right)_{{\bf x}={\bf x}^0}\Delta x_i\Delta x_j+\ldots$$ where $\Delta x_i=(x_i-x_i^0)$.

Defining $$M_{ij}\equiv\left(\frac{\partial^2 f}{\partial x_i\partial x_j}\right)_{{\bf x}^0}\geq 0,$$

and after a little bit of manipulation, we find that $$\Delta f\equiv f({\bf x})-f({\bf x}^0)\approx \frac{1}{2}\sum_j\sum_j M_{ij}\Delta x_i \Delta x_j=\frac{1}{2}(\Delta {\rm x})^TM(\Delta{\bf x})=\frac{1}{2}\sum_{r=1}^{n}\lambda_r a_r^2$$ where $\lambda_r$ are the eigenvalues of the real symmetric matrix $M$, and therefore, real. $a_r^2$ are real positive constants.

Now, the condition of minimum requires $\Delta f>0$. This is certainly satisfied if all eigenvalues of positive. But this is also satisfied if some of the eigenvalues are zero and others are positive. But this book by Riley, Hobson, and Bence says that $\lambda_r$ must be positive. Is there a mistake in the book or do I get it wrong? See the fourth line from the top, page 167.

https://www.astrosen.unam.mx/~aceves/Metodos/ebooks/riley_hobson_bence.pdf