Invertibility of real operator I have this problem:
If $\;A\;$ is an invertible real operator, then prove also $\;AA^t+A^tA\;$ is invertible.
Now, I know $\;A\;\;\text{invertible}\;\iff\;\;A^t\;\;\text{invertible}\;$ , for example using that the row rank equals the column rank or even using determinants: 
$$\det A=\det A^t\implies \det A\neq 0\iff \det A^t\neq 0$$
I can also see how being real is in order to avoid problems with positive characteristic (for example, if $\;A=A^t\;$ and the characteristic is $\;2\;$ ...).
But I can't manage to solve the problem beyond this point, though I tried
$$AA^t+A^tA=A\left(A^t+A^{-1}A^tA\right)=\ldots ?$$
 A: I'm assuming we "operate" in finite dimension, that is, $\text{size}(A) = N < \infty$; this assumption seems warranted to me by the OP's introduction of the notions of determinant and row/column rank, which are most often invoked in the context of finite dimension.  Under these circumstances, we note that $A$ invertible implies, for nonzero vectors $x$, that $Ax \ne 0$, so $\Vert Ax \Vert \ne 0$ as well.  Thus
$\langle x, A^tA x \rangle = \langle Ax, Ax \rangle = \Vert Ax \Vert^2 > 0, \tag{1}$
and, since $A^t$ is invertible, what is effectively the same argument shows that
$\langle x, AA^t x \rangle > 0 \tag{2}$
as well.  Thus we have that
$\langle x, (AA^t + A^tA)x \rangle = \langle x, AA^t x \rangle + \langle x, A^tAx \rangle > 0 \tag{3}$
for all vectors $x \ne 0$.  Next observe that $AA^t + A^tA$ is itself symmetric:
$(AA^t + A^tA)^t = (AA^t)^t + (A^tA)^t = AA^t + A^tA, \tag{4}$
and thus it may be diagonalized; there is a complete set of unit eigenvectors $e_j$, $\Vert e_j \Vert = 1$, $1 \le j \le N$ spanning the underlying (real) vector space  $V$ on which $A$ is defined, and to each $e_j$ there corresponds a real eigenvalue $\lambda_j$:
$(AA^t + A^tA)e_j = \lambda_j e_j. \tag{5}$
Now (3) shows $\lambda_j > 0$ for all $j$, since we can choose $x$ to be any of the $e_j$, yielding
$\lambda_j = \langle e_j, \lambda_j e_j \rangle = \langle e_j, (AA^t + A^tA)e_j \rangle > 0;  \tag{6}$
since all its eigenvectors are positive, we have
$\det(AA^t + A^tA) = \prod \lambda_j > 0, \tag{7}$
which shows that $AA^t + A^tA$ is itself invertible.  QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Solution synthesised from the answer by Robert Lewis and its comment by Timbuc.
Firstly $AA^t+A^tA$ is clearly symmetric (each term is individually). Also it is positive definite; for $x\neq0$:
$$
\langle x,(AA^t+A^tA)x\rangle
=\langle x,(AA^t)x\rangle+\langle x,(A^tA)x\rangle
=\langle A^tx,A^tx\rangle+\langle Ax,Ax\rangle>0,
$$
using that $Ax\neq0$ since $A$ is invertible (so again this is because each term is individually positive definite). Positive definite symmetric matrices are invertible. 
(More directly, the inequality obtained shows that $(AA^t+A^tA)x\neq0$ for all $x\neq0$, so symmetry and positive definiteness need not be evoked at all. But morally the proof just imitates the obvious proof that a sum of positive definite symmetric bilinear forms is again positive definite.)
