TITLE: Let $A_{n\times n}$ be a real matrix. Is it true that $I+A^TA$ is invertible?
QUESTION [5 upvotes]: Let $A_{n\times n}$ be a real matrix. Is it true that $I+A^TA$ is always invertible?

REPLY [13 votes]: Let $M = I+A^tA$. Take $v$ in the kernel of $M$. Now,
$$
0 = [I+A^tA]v,
$$ 
and so we get $A^tAv = -v$. Right multiplying by $v^t$ once again, we have
$v^tA^tAv = -v^tv$ and so
$$
0 \leq \|Av\|^2 = (Av)^t(Av) = v^tA^tAv = -v^tv = -\|v\|^2 \leq 0
$$
which proves that $v = 0$. Thus, $M$ is injective and since it is square, it is invertible.