Show that matrix $ M=\alpha^2 I +AA^T$ is nonsingular

Question

Let $A$ be a real $m \times n$ matrix. Show that the matrix $M$ defined as $$M = \alpha^2 I_m + A A^T$$ is nonsingular, where $\alpha$ is a nonzero real number.

Answer 1

Suppose that $Mv = 0$. Then $$ \langle v, Mv \rangle = 0, $$ so $$ \langle v, \alpha^2v \rangle = - \langle v, AA^Tv\rangle, $$ so $$ \alpha^2 \langle v, v \rangle = - \langle A^Tv, A^Tv \rangle. $$ But the inner product of a vector and itself is non-negative, so both sides have to be zero. Since $\alpha$ isn't zero, $\langle v , v \rangle$ and hence $v$ itself, is zero. Thus $M$ has no kernel and is non-singular.