Convexity of $f(x)=\frac{1}{2}x^TAx$?

Question

For a function $f(x)=\frac{1}{2}x^TAx$, where $A \in \mathbb{R^n}$ is a symmetric matrix, and its largest eigenvalue $\lambda_{max}(A) >0$ and its smallest eigenvalue $\lambda_{min}(A) <0$,

is $f(x)$ a convex function?

What I tried is:

I found the Hessian $$\nabla^2f(x) = A$$But is this positive semi-definite? How do I use the fact that its largest eigenvalue $\lambda_{max}(A) >0$ and its smallest eigenvalue $\lambda_{min}(A) <0$ from here?

Answer 1

a positive semidefinite matrix has all eigenvalues non-negative. However, $A$ has a negative eigenvalue $\lambda_{min}$.

Answer 2

In this case, when $A$ has both positive and negative eigenvalues, the function $f(x)$ is neither convex nor concave. One way to look at it is also as a combination of concave and convex functions.

Now in case you want to extract the convex and concave part of it, you can perform decomposition of the matrix $A$ as $A = A^p + A^n$ where $A^p$ is PSD and $A^n$ is NSD. In such a case function will be $$f(x)= f_p(x)+f_n(x) = x^TA^px+x^TA^nx$$ .

For decomposition refer: How to extract the positive semidefinite part of a matrix