Quadratic function but with matrix not positive definite. [duplicate]

Let $f(x) = \frac{1}{2} x^TQx-b^Tx$, where $x \in \mathbb{R^n}$, and $Q$ is positive definite. I know that $x_0$ is a unique minimizer for $f$ if and only if $Qx_0 = b$.
However, what if $Q$ is just positive semi-definite? Would $f$ still be quadratic, is $f$ necessarily convex? if not then what other things we can conclude? Does its minimizer $x_0$ still needs to satisfy $Qx_0 = b$?

Solution 1:

Suppose that $Q$ is positive semi-definite. Then $f$ is convex because its Hessian matrix is positive semi-definite everywhere ($\nabla^{2}f(x) = Q$ in this case).

Now, since $f$ is convex, $x_{0}$ is a (global) minimiser of $f$ if and only if $\nabla f(x_{0}) = 0 \iff Qx_{0} = b$. If $Q$ is positive semi-definite but is not strictly positive definite, then it is a square matrix with less than full rank, so the linear system $Qx_{0} = b$ has either no solutions or infinitely many solutions (i.e. there are either no minimisers or infinitely many minimisers of $f$).