= 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:
$$
\begin{aligned}
\binom{x}{z}^{\!T}
M\binom{x}{z}
&=\alpha\|x\|_2^2+2\langle Ax,z\rangle+\beta\|z\|_2^2\\
&=\beta\left\|z+\frac{Ax}{\beta}\right\|_2^2
+x^T\left(\alpha I-\frac{A^TA}{\beta}\right)x.
\end{aligned}
$$
The <matrix 2-norm> bound $\|Ax\|_2\leq\|A\|_2\|x\|_2$ shows that
$$
x^T\left(\alpha I-\frac{A^TA}{\beta}\right)x
\geq\left(\alpha-\frac{\|A\|_2^2}{\beta}\right)\|x\|_2^2>0
$$
for $x\ne0$ when $\alpha\beta>\|A\|_2^2$. If $x=0$ and $z\ne0$, the square contributes $\beta\|z\|_2^2>0$. Thus $\boxed{M\text{ is positive definite}}$. Equivalently, the <Schur complement> of the lower-right block is $\alpha I-A^TA/\beta\succ0$.
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.
Back to article page
