 Solution

The assertion uses the positive parameters required for $M$ to be a preconditioner: assume $\alpha >0$ and $\beta >0$. For $x\in {\mathbb{R}}^n$ and $z\in {\mathbb{R}}^m$, complete the square:The matrix 2-norm bound $\Vert Ax{\Vert }_2\le \Vert A{\Vert }_2\Vert x{\Vert }_2$ shows thatfor $x\ne 0$ when $\alpha \beta >\Vert A{\Vert }_2^2$. If $x=0$ and $z\ne 0$, the square contributes $\beta \Vert z{\Vert }_2^2>0$. Thus $\boxed{M\text{ is positive definite}}$. Equivalently, the Schur complement of the lower-right block is $\alpha I-A^{T}A/\beta \succ 0$.

Taken literally without the positivity inherited from part b, the product condition alone is insufficient: $A=0$ and $\alpha =\beta =-1$ is a counterexample. Thus $\alpha ,\beta >0$ is a necessary implicit hypothesis.