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Let $U$ be a finite-dimensional unitary space and let $A,B\in\mathcal L(U)$ s. t. $B^*A=0$. Prove that: $$\ker(A+B)=\ker A\cap\ker B.$$

note: $B^*$ denotes the Hermitian adjoined operator of $B$.

My thoughts:

Since $\operatorname{im}B^*=(\ker B)^\perp$ as proven here, then we can also state $\operatorname{im} B=(\ker B^*)^\perp$.

$B^*A=0\implies\operatorname{im}A\leqslant\ker B^*=(\operatorname{im} B)^\perp\iff\ker A\geqslant\operatorname{im} B$. In case of $\operatorname{im}B\subset\ker A,\dim(\ker A\cap\ker B)\geqslant 1$.

If we take an arbitrary $x\in\operatorname{im}B$, then $(A+B)x=Ax+Bx=0+Bx\ne0$, so $\begin{aligned}(A+B)x=0&\iff Ax=0\ \land Bx=0\\&\iff x\in\ker A\cap\ker B,\end{aligned}$

but I don't think this sketch is concise enough to work as proof.

May I ask for advice on how to solve this task?

Thank you in advance!

PinkyWay
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1 Answers1

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Suppose $(A+B)x=0$ then $B^*Bx=B^*(A+B)x=0$, so it follows that $Bx=0$ and hence $Ax=0$.

Chrystomath
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  • @Chrystomath why does $B^*Bx=0$ imply $Bx=0$? Is the field $\mathbb{R}$ or $\mathbb{C}$? –  Aug 18 '20 at 14:18
  • @crystal_math ${B^}Bx=0\implies|Bx|^2=\langle x,{B^}Bx\rangle=0$ – Chrystomath Aug 18 '20 at 16:17
  • @Chrystomath Yeah, that's what I thought too but does Unitary space means that it's an inner product space? –  Aug 18 '20 at 16:23