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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.

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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),

$\ker C = \ker BB^\ast \cap A^\ast A. \tag{20}$

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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 $\operatorname{Ker} C \subseteq \operatorname{Ker} BB^* \cap \operatorname{Ker} A^*A$.

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  • $\begingroup$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \ne 0$. Also, $A\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ etc. are vectors, so where does $1 - 1$ come from? $\endgroup$ – Robert Lewis Dec 25 '17 at 21:38
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    $\begingroup$ @RobertLewis Thanks for the correction. I wrote a solution somewhat similar to yours, but using inner products. $\endgroup$ – mechanodroid Dec 26 '17 at 10:56
  • $\begingroup$ Well done, amigo! $\endgroup$ – Robert Lewis Dec 26 '17 at 16:06

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