Kernel of sum of linear self-adjoint operators Let $A$, $B$ be linear operators on vector spaces. (We may take $A$, $B$ to be matrices).
Let $AB=0$, and $B^*A^*=0$, where $A^*$ denotes the adjoint of $A$ (we may just take it to be conjugate transpose of the matrix.)
Then, we can show that $\text{Im}\ BB^*\subseteq\ker A^*A$ and $\text{Im}\ A^*A\subseteq\ker BB^*$.
If we define $C=BB^*+A^*A$, can we show that $\ker C=\ker BB^*\cap \ker A^*A$?
One direction is clear to me, $\ker BB^*\cap \ker A^*A\subseteq \ker C$, but I am not sure how to show the other direction.
Thanks for any help.
 A: If 
$x \in \ker BB^\ast \cap \ker A^\ast A, \tag 1$
then
$BB^\ast x = 0 = A^\ast A x; \tag 2$
then
$Cx = BB^\ast x + A^\ast A x = 0, \tag 3$
so
$x \in \ker C, \tag 4$
i.e.,
$\ker A^\ast A \cap \ker B B^\ast \subseteq \ker C, \tag 5$
as our OP yoyostein has affirmed.  On the other hand, if (4) binds, then so does (3), and
$BB^\ast x + A^\ast Ax = 0, \tag 6$
from which we have
$x^\ast BB^\ast x + x^\ast A^\ast Ax = 0, \tag 7$
and since
$x^\ast B = (B^\ast x)^\ast, \; x^\ast A^\ast  = (A x)^\ast, \tag 8$
(7) becomes
$(B^\ast  x)^\ast B^\ast x + (A x)^\ast Ax = 0; \tag 9$
we now note that for any vector
$y = (y_1, y_2, \ldots, y_n)^T \tag{10}$
we have
$\Bbb R \ni y^\ast y = (y_1^\ast, y_2^\ast, \ldots, y_n^\ast)(y_1, y_2, \ldots, y_n)^T = \displaystyle \sum_1^n y_i^\ast y_i \ge 0; \tag{11}$
the fact that $y^\ast y \in \Bbb R$ may also be seen thusly:
$(y\ast y)^\ast = y^\ast (y^\ast)^\ast = y^\ast y; \tag{12}$
also, (11) shows that
$y^\ast y = 0 \Longleftrightarrow y = 0; \tag{13}$
now from (11)-(13) we find
$\Bbb R \ni (B^\ast  x)^\ast B^\ast x,  (A x)^\ast Ax \ge  0; \tag{14}$
and thus the only way (9) may bind is if
$(B^\ast  x)^\ast B^\ast x,  (A x)^\ast Ax = 0, \tag{15}$
which by (13) forces
$B^\ast x, Ax = 0; \tag{16}$
then
$BB^\ast x = A^\ast Ax = 0, \tag{17}$
so that
$x \in \ker BB^\ast \cap \ker A^\ast A, \tag{18}$
and finally we see that
$\ker C \subseteq \ker BB^\ast \cap A^\ast A; \tag{19}$
thus, combining (5) and (19), 
$\ker C = \ker BB^\ast \cap A^\ast A. \tag{20}$
A: The statement is true even without assuming $AB = 0$.
For any matrix $A$ it is true that $\operatorname{Ker} A^*A = \operatorname{Ker} A$ and $\operatorname{Im} A^*A = \operatorname{Im} A^*$. For a proof see here. Thus, your claim $\operatorname{Ker} C \subseteq \operatorname{Ker} BB^* \cap \operatorname{Ker} A^*A$ is in fact equivalent to $\operatorname{Ker} C \subseteq \operatorname{Ker} B^* \cap \operatorname{Ker} A$.
Notice that we have:
$$\langle A^*Ax, x\rangle = \langle Ax, Ax\rangle = \|Ax\|^2\ge 0$$
$$\langle BB^*x, x\rangle = \langle B^*x, B^*x\rangle = \|B^*x\|^2\ge 0$$
Assume $x \in \operatorname{Ker} C$.
We have:
$$0= \langle Cx, x\rangle = \langle (BB^* + A^*A)x, x\rangle = \langle BB^*x, x\rangle + \langle A^*A, x\rangle = \|B^*x\|^2 +\|Ax\|^2 $$
Therefore it must be $ \|B^*x\| = \|Ax\| = 0$. Hence $B^*x = 0$ and $Ax = 0$, so $$x \in \operatorname{Ker} B^* \cap \operatorname{Ker} A = \operatorname{Ker} BB^* \cap \operatorname{Ker} A^*A$$ as discussed.
We can conclude $\operatorname{Ker} C \subseteq \operatorname{Ker} BB^* \cap \operatorname{Ker} A^*A$.
