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.



$x \in \ker BB^\ast \cap \ker A^\ast A, \tag 1$


$BB^\ast x = 0 = A^\ast A x; \tag 2$


$Cx = BB^\ast x + A^\ast A x = 0, \tag 3$


$x \in \ker C, \tag 4$


$\ker A^\ast A \cap \ker B B^\ast \subseteq \ker C, \tag 5$

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

$B^\ast x, Ax = 0; \tag{16}$


$BB^\ast x = A^\ast Ax = 0, \tag{17}$

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),

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


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

  • $\begingroup$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \ne 0$. 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.